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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0c10 | Consider the congruent segments $AB$, $BC$ and $AD$, with $D \in (BC)$. Show that the perpendicular bisector of the segment $DC$, the angle bisector of the angle $ADC$ and the line $AC$ are concurrent.
Mircea Fianu | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0gvs | In the plane, $2005$ points were marked, no three of which are collinear. Straight lines were drawn through all the pairs of marked points. Prove that all the marked points can be colored into two colors in such a way that for any two points of the same color the number of the drawn lines separating them is even. (We s... | [
"Крім того, покладемо\n$$\n\\delta(a,P,Q,R) = \\delta(a,P,Q) + \\delta(a,Q,R) + \\delta(a,P,R).\n$$\nРозглянемо довільні три відмічені точки $P$, $Q$, $R$. Очевидно, що\n$$\nn_{PQ} + n_{QR} + n_{PR} = \\sum_{a} \\delta(a,P,Q,R),\n$$\nде сума береться по всіх проведених прямих.\nЛегко бачити, що якщо пряма $a$ прохо... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0jho | Problem:
Compute
$$
\sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty} \frac{a_{1}+a_{2}+\cdots+a_{7}}{3^{a_{1}+a_{2}+\cdots+a_{7}}}
$$ | [
"Solution:\nAnswer: $\\frac{15309}{256}$\n\nNote that, since this is symmetric in $a_{1}$ through $a_{7}$,\n$$\n\\begin{aligned}\n\\sum_{a_{1}=0}^{\\infty} \\sum_{a_{2}=0}^{\\infty} \\cdots \\sum_{a_{7}=0}^{\\infty} \\frac{a_{1}+a_{2}+\\cdots+a_{7}}{3^{a_{1}+a_{2}+\\cdots+a_{7}}} & = 7 \\sum_{a_{1}=0}^{\\infty} \\s... | United States | HMMT | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | final answer only | 15309/256 | |
01zt | Let $k$ and $n$ be positive integers. An international company has connected $k$ cities of Armenia with $k$ cities of Belarus by direct two-way airlines. From each of these Belarusian cities there is a direct flight to exactly $n$ Armenian ones. It turned out that for any two Armenian cities there are exactly two Belar... | [
"Let's translate the problem into the language of graphs.\n\nGiven a bipartite graph $G$ with parts $A$ and $B$ having the same number of vertices: $|V(A)| = |V(B)| = k$. The degree of each vertex in $B$ is $n$. For any two vertices $u, v$ of the part $A$, there are exactly two vertices in $B$ that are adjacent to ... | Belarus | SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0jkq | Prove that there exists an infinite set of points
$$
\dots, P_{-3}, P_{-2}, P_{-1}, P_0, P_1, P_2, P_3, \dots
$$
in the plane with the following property: For any three distinct integers $a, b$ and $c$, points $P_a, P_b$ and $P_c$ are collinear if and only if $a + b + c = 2014$. | [
"**Solution 1** (by Razvan Gelca). We claim that defining $P_n$ to be the point with coordinates $(n, n^3 - 2014n^2)$ will satisfy the conditions of the problem. Recall that points $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ are collinear if and only if\n$$\n\\begin{vmatrix} x_1 & y_1 & 1 \\\\ x_2 & y_2 & 1 \\\\ x_... | United States | USAMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Linear Algebra > Determinants",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0g50 | Problem:
Determine all integer values that the expression
$$
\frac{p q + p^{p} + q^{q}}{p + q}
$$
can take, where $p$ and $q$ are both prime numbers. | [
"Solution:\nAnswer: The only possible integer value is $3$.\n\nIf both $p$ and $q$ are odd, then the numerator is odd while the denominator is even. Since an even number never divides an odd number, this does not lead to an integer value. Hence we can assume that one of our primes is even and therefore equal to $2$... | Switzerland | Second round 2023 | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 3 | |
0icd | Problem:
$$
b_{i}= \begin{cases}1 & \text{ if } i \text{ is a multiple of } 3 \\ 0 & \text{ otherwise }\end{cases}
$$
Let $\{a_{i}\}$ be a sequence of elements of $\{0,1\}$ such that
$$
b_{n} \equiv a_{n-1}+a_{n}+a_{n+1} \quad(\bmod 2)
$$
for $0 \leq n \leq 59$ ($a_{0}=a_{60}$ and $a_{-1}=a_{59}$). Find all possible v... | [
"Solution:\n\nTry the four possible combinations of values for $a_{0}$ and $a_{1}$. Since we can write $a_{n} \\equiv b_{n-1}-a_{n-2}-a_{n-1}$, these two numbers completely determine the solution $\\{a_{i}\\}$ beginning with them (if there is one).\n\nFor $a_{0}=a_{1}=0$, we can check that the sequence beginning $0... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 0, 3, 5, 6 | |
0huo | Problem:
Determine, with proof, whether there is a function $f(x, y)$ of two positive integers, taking positive integer values, such that
- For each fixed $x$, $f(x, y)$ is a polynomial function of $y$;
- For each fixed $y$, $f(x, y)$ is a polynomial function of $x$;
- However, $f(x, y)$ does not equal any polynomial f... | [
"Solution:\nThe answer is yes. Consider the following expression:\n$$\nf(x, y) = 1 + (x-1)(y-1) + (x-1)(y-1)(x-2)(y-2) + (x-1)(y-1)(x-2)(y-2)(x-3)(y-3) + \\cdots.\n$$\nHere, although the sum appears to be infinite, if we fix a value $y = y_{0}$, all but the first $y_{0}$ terms contain the factor $(y - y_{0})$ and t... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | Yes. For example, f(x, y) = 1 + (x−1)(y−1) + (x−1)(y−1)(x−2)(y−2) + (x−1)(y−1)(x−2)(y−2)(x−3)(y−3) + ⋯, which truncates for fixed x or y, is separately polynomial but not a polynomial in two variables. | |
07rx | Show that there are 21 consecutive composite four-digit numbers. | [
"The obvious solution, if we did not care about the bound, is something like $N = 22!$. Then, $N + k$ is divisible by $k$ for $k = 2, \\dots, 22$. However, $22!$ is far too large.\n\nInstead, we take $N$ to have several small prime factors in order that $N + k$ has small factors for most values of $k = 2, \\dots, 2... | Ireland | Irish | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof only | null | |
0bh2 | Let $a$ be a positive real number, $(a_n)_{n \ge 1}$ be a sequence of real numbers and $(x_n)_{n \ge 1}$ be the sequence defined by
$$
x_{n+1} = \left(1 - \frac{a}{n}\right) x_n + \frac{a_n}{n},
$$
where $x_1$ is an arbitrary real number. Prove that
$$
\lim_{n \to \infty} x_n = 0 \quad \text{if and only if} \quad \lim_... | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
08sa | Two diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at a point $P$ in the quadrilateral. If we have $AC = 2$, $BD = 3$ and $\angle APB = 60^\circ$, what is the smallest possible value of $AB + BC + CD + DA$? | [
"Take points $E$ and $F$ so that $ABEC$ and $ACFD$ are parallelograms. Then $AB = CE$, $DA = FC$, and by the triangle inequality we get $BC + CF \\ge BF$, $DC + DE \\ge DE$. Therefore $AB + BC + CD + DA \\ge BF + DE$.\n\nAnd if we consider a case $AC$ and $BD$ cross at their midpoints, we get $BC + CF = BF$, $DC + ... | Japan | Japanese Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Method... | null | proof and answer | sqrt(7) + sqrt(19) | |
0eha | Problem:
Kozarec valjaste oblike s polmerom $4~\mathrm{cm}$ in višino $9~\mathrm{cm}$ je do $\frac{2}{9}$ višine napolnjen z vodo. Mark se je odločil, da bo vso vodo prelil v kozarec stožčaste oblike s polmerom $5~\mathrm{cm}$ in višino $6~\mathrm{cm}$ (glej sliko). Pri prelivanju je $5\%$ vode polil. Koliko decilitro... | [
"Solution:\n\nIzračun količine vode v valjastem kozarcu:\n\n$$\nV = \\frac{2}{9} \\pi \\cdot r^2 \\cdot v = \\frac{2}{9} \\pi \\cdot 4^2 \\cdot 9 = \\frac{2}{9} \\pi \\cdot 16 \\cdot 9 = 2 \\pi \\cdot 16 = 32\\pi \\approx 100,53~\\mathrm{cm}^3\n$$\n\nIzračun količine vode v stožčastem kozarcu po polivanju:\n\n$$\n0... | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | final answer only | 0.955 dl; approximately 5.1 cm | |
0dtn | Suppose $p$ is a prime number and $x, y, z$ are integers satisfying $0 < x < y < z < p$. If $x^3, y^3, z^3$ have equal remainders when divided by $p$, prove that $x^2 + y^2 + z^2$ is divisible by $x + y + z$. | [
"Note that $p > 3$ and $p \\mid x^3 - y^3 = (x-y)(x^2 + xy + y^2)$. Since $x, y < p$, $p \\nmid x-y$. Therefore $p \\mid x^2 + xy + y^2$. Similarly $p \\mid x^2 + xz + z^2$, $p \\mid y^2 + yz + z^2$.\n\nThus $p \\mid (x^2 + xy + y^2) - (y^2 + yz + z^2) = (x-z)(x+y+z)$ and so $p \\mid x+y+z$.\n\nSince $x, y, z < p$ ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0e5o | In a triangle $ABC$, denote $D$ the orthogonal projection of the point $C$ onto the line $AB$, denote $E$ the orthogonal projection of the point $D$ onto the line $AC$, and denote $P$ the midpoint of the line segment $CD$. Let $K_1$ be the circumscribed circle of the triangle $ABC$, and let $K_2$ be the circle of radiu... | [
"Denote $F$ the orthogonal projection of the point $D$ onto the line $BC$.\n\nObviously, the points $E$ and $F$ lie on the sides $AC$ and $BC$, respectively. According to Euclid's theorem, $|EC| \\cdot |EA| = |CD|^2 - |EC|^2$ and $|FC| \\cdot |FB| = |CD|^2 - |FC|^2$,\n\nhence the powers of ... | Slovenia | Selection Examinations for the IMO 2012 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0l36 | The national debt of the United States is on track to reach $5 \times 10^{13}$ dollars by 2033. How many digits does this number of dollars have when written as a numeral in base 5? (The approximation of $\log_{10} 5$ as 0.7 is sufficient for this problem.)
(A) 18 (B) 20 (C) 22 (D) 24 (E) 26 | [
"The number of digits required to write the positive integer $n$ in base $b$ is $1 + \\log_b n$, rounded down to an integer. Therefore the required value is the floor of\n$$\n1 + \\log_5 (5 \\cdot 10^{13}) = 1 + \\log_5 5 + 13 \\log_5 10 = 1 + 1 + 13 \\cdot \\frac{1}{\\log_{10} 5} \\approx 2 + \\frac{13}{0.7} = 20.... | United States | 2024 AMC 12 B | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | MCQ | B | |
0i8p | Find all ordered triples of primes $(p, q, r)$ such that
$$
p \mid q^r + 1, \quad q \mid r^p + 1, \quad r \mid p^q + 1.
$$ | [
"We check that this is a solution:\n$$\n2 \\mid 126 = 5^3 + 1, \\quad 5 \\mid 10 = 3^2 + 1, \\quad 3 \\mid 33 = 2^5 + 1.\n$$\nNow let $p, q, r$ be three primes satisfying the given divisibility relations. Since $q$ does not divide $q^r + 1$, $p \\neq q$, and similarly $q \\neq r, r \\neq p$, so $p, q$ and $r$ are a... | United States | USA IMO 2003 | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | [(2, 5, 3), (5, 3, 2), (3, 2, 5)] | |
079v | $n$ is a positive integer. Let $A, B$ be two sets of $n$ points in the plane such that no three points of them are collinear. Denote by $T(A)$ the number of non-self-intersecting broken lines containing $n-1$ segments such that its vertices are in $A$. Define $T(B)$ similarly. If the elements of $B$ are the vertices of... | [
"We call such a broken line a *good path*.\n\n**Lemma.** Let $C$ be a set of $n \\ge 2$ points in the plane, no three of which are collinear and let $x_0$ be a vertex of the convex hull of $C$. The number of good paths with vertices of $C$ starting at $x_0$ is at least $2^{n-2}$. Equality holds only when $C$ is con... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | English | proof only | null | |
0icp | Problem:
How many ways can you mark 8 squares of an $8 \times 8$ chessboard so that no two marked squares are in the same row or column, and none of the four corner squares is marked? (Rotations and reflections are considered different.) | [
"Solution:\n\nIn the top row, you can mark any of the 6 squares that is not a corner. In the bottom row, you can then mark any of the 5 squares that is not a corner and not in the same column as the square just marked. Then, in the second row, you have 6 choices for a square not in the same column as either of the ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 21600 | |
0hkx | Problem:
A house has several rooms. There are also several doors, each of which connects either one room to another or a room to the outside. Suppose that every room has an even number of doors leaving it. Prove that the number of outside entrance doors is even as well. | [
"Solution:\nEvery door has two \"sides,\" one toward one room and one toward either another room or the outside. Clearly the total number of sides, being twice the number of doors, is even. However, for each room, the number of sides pointing to it is even. Since even subtracted from even gives even, the number of ... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0697 | Let $AB\Gamma$ be an equilateral triangle of side $k$ cm. We divide $AB\Gamma$ with parallel lines into $k^2$ small equilateral triangles of side $1$ cm. In this way, we create a grid (see figure for $k=7$). Inside every small triangle we put exactly one positive integer from $1$ to $k^2$, so that there are not two tri... | [
"The small triangles are divided in four categories.\n\n1st category: They have one vertex $A$ or $B$ or $\\Gamma$.\n\nThese are not members of any hexagon and so their numbers do not take part in the final sum of values of all hexagons.\n\n2nd category: Contains small triangles which belong only to one hexagon. On... | Greece | SELECTION EXAMINATION | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 3(k^4 - 14k^2 + 33k - 24)/2 | |
0ahb | Let the quadrangle $ABCD$ be inscribed in a circle of radius $1$. Prove that the difference between its perimeter and the sum of the lengths of its diagonals is positive and less than $4$. | [
"From the triangle inequality we have:\n$$\n2L = \\overline{AB} + \\overline{BC} + \\overline{CD} + \\overline{DA} + \\overline{AB} + \\overline{CD} + \\overline{DA} > \\overline{AC} + \\overline{BD} + \\overline{AC} + \\overline{BD}\n$$\nfrom which we get one of the inequalities. Let us denote the point of interse... | North Macedonia | XVI Junior Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
0c3d | Let $n$ be a positive integer and $A$ be a set of complex numbers with $2n + 1$ elements. Prove that there exists two sets $B, C$ so that $B \cup C = A$, $B \cap C = \emptyset$, $B$ has $n$ elements and $|\sum_{z \in B} z| \le |\sum_{z \in C} z|$. | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof only | null | |
0b4n | Problem:
Let $x$ and $y$ be integers satisfying $x^{2}+30x+25=y^{4}$. What is the largest possible value of $x+y$? | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 43 | |
0506 | How many positive integers are there that are divisible by $2010$ and that have exactly $2010$ divisors (1 and the integer itself included)? | [
"Let $N$ be a positive integer that is divisible by $2010$ and that has exactly $2010$ positive divisors. Since $2010 = 2 \\cdot 3 \\cdot 5 \\cdot 67$, also $N$ should be divisible by these four primes. Thus, $N = 2^a \\cdot 3^b \\cdot 5^c \\cdot 67^d \\cdot s$, where $a, b, c, d > 0$ and $s$ is not divisible by an... | Estonia | Selected Problems from Open Contests | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 24 | |
0jrp | Problem:
In a quadrilateral, the two segments connecting the midpoints of its opposite sides are equal in length. Prove that the diagonals of the quadrilateral are perpendicular. (In other words, let $M$, $N$, $P$, and $Q$ be the midpoints of sides $AB$, $BC$, $CD$, and $DA$ in quadrilateral $ABCD$. It is known that s... | [
"Solution:\n\nWe will use a well-known theorem from geometry. A midsegment in a triangle is called a segment that joins the midpoints of two of its sides.\n\nTheorem. The midsegment in a triangle connecting two sides in a triangle is parallel to the third side and half of its length. In other words, if $K$ and $L$ ... | United States | BAMO | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals"
] | null | proof only | null | |
0ds5 | Let $n$ be a positive integer and $a_1, a_2, \dots, a_{2n}$ be $2n$ distinct integers. Given that the equation
$$
|x - a_1| |x - a_2| \dots |x - a_{2n}| = (n!)^2
$$
has an integer solution $x = m$, find $m$ in terms of $a_1, \dots, a_{2n}$. | [
"We have\n$$\n|m - a_1| |m - a_2| \\cdots |m - a_{2n}| = (n!)^2.\n$$\nFirst we show that we cannot have distinct $i$, $j$, $k$ so that $|m - a_i| = |m - a_j| = |m - a_k|$. If so, then two of $(m - a_i)$, $(m - a_j)$, $(m - a_k)$ must be of the same sign, say $(m - a_i)$, $(m - a_j)$. Then $a_i = a_j$, a contradicti... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | (a_1 + a_2 + \cdots + a_{2n})/(2n) | |
0g9q | Find all triples $(p, x, y)$ consisting of a prime number $p$ and two positive integers $x$ and $y$ such that $x^{p-1} + y$ and $x + y^{p-1}$ are both powers of $p$.
試求所有質數 $p$ 與正整數對 $(x, y)$, 使得 $x^{p-1} + y$ 與 $x + y^{p-1}$ 皆為 $p$ 的幂次。 | [
"所有解為 $(p, x, y) \\in \\{(3, 2, 5), (3, 5, 2)\\} \\cup \\{(2, n, 2^k - n) \\mid 0 < n < 2^k\\}$.\n\n(1) 當 $p=2$ 時, 顯然所有和為 $2$ 的幂次的 $(x,y)$ 皆滿足題意, 因此我們只需考慮 $p > 2$ 即可。\n\n(2) 假設 $x^{p-1} + y = p^a$ 及 $x + y^{p-1} = p^b$. 不失一般性, 我們假設 $x \\le y$, 從而 $a \\le b$. 我們因此有\n$$\np^b = y^{p-1} + x = (p^a - x^{p-1})^{p-1} + x\... | Taiwan | 2015 Math Olympiad Second Stage Training Camp | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (p, x, y) ∈ {(3, 2, 5), (3, 5, 2)} ∪ {(2, n, 2^k − n) | 0 < n < 2^k} | |
0knw | Problem:
In a trapezoid, the midsegment has length $17$ and the distance between the midpoints of the diagonals is $7$. Find the lengths of the bases. | [
"Solution:\n\nLet $a$ and $b$ be the bases, with $a > b$. The length of the midsegment is the average of the bases, so\n$$\n\\frac{a + b}{2} = 17,\n$$\nand the distance between the midpoints of the diagonals is half their difference, so\n$$\n\\frac{a - b}{2} = 7.\n$$\nAdding the two equations gives $a = 24$, and su... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 24 and 10 | |
0ica | Problem:
Compute:
$$
\left\lfloor\frac{2005^{3}}{2003 \cdot 2004}-\frac{2003^{3}}{2004 \cdot 2005}\right\rfloor .
$$ | [
"Solution: 8\nLet $x=2004$. Then the expression inside the floor brackets is\n$$\n\\frac{(x+1)^{3}}{(x-1) x}-\\frac{(x-1)^{3}}{x(x+1)}=\\frac{(x+1)^{4}-(x-1)^{4}}{(x-1) x(x+1)}=\\frac{8 x^{3}+8 x}{x^{3}-x}=8+\\frac{16 x}{x^{3}-x} .\n$$\nSince $x$ is certainly large enough that $0<16 x /(x^{3}-x)<1$, the answer is 8... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 8 | |
08sf | Calculate the following number:
$$
877 \times 879 - 121 \times 123.
$$ | [
"Using $(a-b)(a+b) = a^2-b^2$, we obtain\n$$\n\\begin{aligned}\n877 \\times 879 - 121 \\times 123 &= (878 - 1)(878 + 1) - (122 - 1)(122 + 1) \\\\\n&= (878^2 - 1) - (122^2 - 1) = 878^2 - 122^2 \\\\\n&= (878 - 122)(878 + 122) = 756 \\times 1000 = 756000.\n\\end{aligned}\n$$\n\nAlternatively,\n$$\n\\begin{aligned}\n87... | Japan | Japan Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | final answer only | 756000 | |
04mu | Let $ABCD$ be a square with side length $1$. Let $X$ be a point on the side $AB$, and let $Y$ be a point on the side $AD$ such that $\angle CXY = 90^\circ$. Find the locus of the point $X$ for which the area of the triangle $CDY$ is the smallest possible. | [] | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | The midpoint of side AB | |
00k8 | Prove that there exist no positive real numbers $x$, $y$, $z$ such that
$$
(12x^2 + yz) \cdot (12y^2 + xz) \cdot (12z^2 + xy) = 2014x^2y^2z^2 .
$$ | [
"The AM-GM inequality gives us:\n$$\n12x^2 + yz = x^2 + x^2 + \\dots + x^2 + yz \\ge 13 \\sqrt[13]{x^{24}yz}\n$$\nApplying this idea to the other two expressions then yields\n$$\n\\begin{aligned}\n(12x^2 + yz) \\cdot (12y^2 + xz) \\cdot (12z^2 + xy) &\\ge 13^3 \\sqrt[13]{x^{24}yz \\cdot y^{24}xz \\cdot z^{24}xy} \\... | Austria | Austria 2014 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
03sm | Suppose an infinite sequence $\{a_n\}$ satisfies $a_0 = x$, $a_1 = y$, $a_{n+1} = \frac{a_n a_{n-1} + 1}{a_n + a_{n-1}}$, $n = 1, 2, \dots$.
(1) Find all real numbers $x$ and $y$ that satisfy the statement: there exists a positive integer $n_0$, such that, for $n \ge n_0$, $a_n$ is a constant.
(2) Find an explicit ex... | [
"(1) We have\n$$\na_n - a_{n+1} = a_n - \\frac{a_n a_{n-1} + 1}{a_n + a_{n-1}} = \\frac{a_n^2 - 1}{a_n + a_{n-1}}, \\quad n = 1, 2, \\dots \\quad \\textcircled{1}\n$$\nIf there exists a positive integer $n$ such that $a_{n+1} = a_n$, we get\n$$\na_n^2 = 1 \\quad \\text{and} \\quad a_n + a_{n-1} \\neq 0.\n$$\nIf $n=... | China | China Mathematical Competition (Extra Test) | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | (1) Exactly those initial pairs with either absolute value of y equal to one and x not equal to minus y, or absolute value of x equal to one and y not equal to minus x. In these cases the sequence is constant from the second index onward with value either one or minus one. (2) For all nondegenerate cases, a_n = [(x + 1... | |
0cgz | For any positive integer $n$, define $a_n = \{\frac{n}{s(n)}\}$, where $s(k)$ represents the sum of the digits of the natural number $k$, and $\{x\}$ is the fractional part of the real number $x$.
a) Prove that there exist infinitely many positive integers $n$ such that $a_n = \frac{1}{2}$.
b) Determine the smallest ... | [
"a.\nIf $s(n) = 2$ and $n$ is odd, then $a_n = \\frac{1}{2}$. The only solutions with these properties are of the form $n = 10^k + 1$, with $k \\in \\mathbb{N}^*$.\n\nb.\nLet $n$ be a positive integer such that $a_n = \\left\\{\\frac{n}{s(n)}\\right\\} = \\frac{1}{6}$.\nSince $\\frac{n}{s(n)} - \\lfloor \\frac{n}{s... | Romania | 74th NMO Selection Tests for JBMO | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 1899999 | |
0c3r | Problem:
Fie $\mathcal{F}$ mulțimea funcțiilor continue $f:[0,1] \rightarrow \mathbb{R}$, care îndeplinesc condiția $\max_{0 \leq x \leq 1}|f(x)|=1$, și fie $I: \mathcal{F} \rightarrow \mathbb{R}$,
$$
I(f)=\int_{0}^{1} f(x) \, \mathrm{d}x - f(0) + f(1)
$$
a. Arătați că $I(f)<3$, oricare ar fi $f \in \mathcal{F}$.
b.... | [
"Solution:\n\na. Fie $f$ o funcție din $\\mathcal{F}$. Din condiția $\\max_{0 \\leq x \\leq 1}|f(x)|=1$, rezultă că\n$$\nI(f) \\leq \\int_{0}^{1} 1 \\, \\mathrm{d}x + 1 + 1 = 3.\n$$\nInegalitatea este strictă, în caz contrar, $f(x)=1$, oricare ar fi $x \\in [0,1]$, și $f(0)=-1$, contradicție.\n\nb. Pentru $n \\geq ... | Romania | Olimpiada Naţională de Matematică Etapa Judeţeană şi a Municipiului Bucureşti | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Functions"
] | null | proof and answer | 3 | |
02jn | Problem:
Para encher de água um tanque em forma de um bloco retangular de $300~\mathrm{cm}$ de comprimento, $50~\mathrm{cm}$ de largura e $36~\mathrm{cm}$ de altura, um homem
utiliza um balde cilíndrico, de $30~\mathrm{cm}$ de diâmetro em sua base e $48~\mathrm{cm}$ de altura, para pegar águ... | [
"Solution:\n\nNesta solução todas as medidas de volume são dadas em $\\mathrm{cm}^3$.\n\nO volume $V$ do balde é dado pela fórmula habitual do volume de um cilindro, ou seja, $V = $ área da base $\\times$ altura. A base do balde é um círculo de diâmetro $30~\\mathrm{cm}$; seu raio é então $r = 15~\\mathrm{cm}$ e su... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Solid Geometry > Volume",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | final answer only | 17 | |
042t | Suppose set $X = \{1, 2, \dots, 20\}$. $A$ is a subset of $X$. The number of the elements of $A$ is at least $2$ and all the elements of $A$ can be arranged as consecutive positive integers. Then the number of such set $A$ is ______. | [
"Each set $A$ satisfying the above conditions can be uniquely determined by its minimum element $a$ and maximum element $b$, where $a, b \\in X$ and $a < b$. The total number of such ways of taking $(a, b)$ is $C_{20}^2 = 190$, so the number of such sets $A$ is $190$."
] | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 190 | |
0ae2 | Дадена е дропката $\frac{57}{71}$. Кој број треба да се одземе од броителот и истиот да се додаде на именителот па вредноста на дропката после скратувањето да е $\frac{1}{3}$? | [
"Треба да се реши следнава равенка: $\\frac{57-x}{71+x} = \\frac{1}{3}$. Значи $3(57-x) = 71+x$, т.е. $171-3x = 71+x$, $171-71 = x+3x$, $100 = 4x$, $x = 25$."
] | North Macedonia | Регионален натпревар по математика за основно образование | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Macedonian, English | final answer only | 25 | |
0345 | Problem:
Solve in integers the equation
$$
x^{3}+10 x-1=y^{3}+6 y^{2}
$$ | [
"Solution:\nIt is clear that $x$ and $y$ have different parity. Then $k = x - y$ is an odd number and\n$$\n(3k - 6) y^{2} + (3k^{2} + 10) y + k^{3} + 10k - 1 = 0\n$$\nThe discriminant of this equation is equal to\n$$\nD = -3k^{4} + 24k^{3} - 60k^{2} + 252k + 76\n$$\nand must be a perfect square. Since $D = -k^{2}(k... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (x, y) = (6, 5) and (2, -3) | |
0il7 | Problem:
Eight celebrities meet at a party. It so happens that each celebrity shakes hands with exactly two others. A fan makes a list of all unordered pairs of celebrities who shook hands with each other. If order does not matter, how many different lists are possible? | [
"Solution:\n\nLet the celebrities get into one or more circles so that each circle has at least three celebrities, and each celebrity shook hands precisely with his or her neighbors in the circle.\n\nLet's consider the possible circle sizes:\n\n- There's one big circle with all $8$ celebrities. Depending on the ord... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 3507 | |
05j6 | Problem:
Soit $\left(a_{n}\right)_{n \in \mathbb{N}}$ une suite croissante et non constante d'entiers strictement positifs tels que $a_{n}$ divise $n^{2}$ pour tout $n \geqslant 1$. Prouver que l'une des affirmations suivantes est vraie:
a) Il existe un entier $n_{1}>0$ tel que $a_{n}=n$ pour tout $n \geqslant n_{1}$... | [
"Solution:\n\nTout d'abord, puisque pour tout entier $n$, on a $a_{n} \\in \\mathbb{N}^{*}$, et que la suite $\\left(a_{n}\\right)$ est croissante et non constante, il existe un entier $n_{0}$ tel que $a_{n} \\geqslant 2$ pour tout $n \\geqslant n_{0}$. Par conséquent, pour tout nombre premier $p>n_{0}$, on a $a_{p... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
063a | Problem:
In der Ebene liegen zwei konzentrische Kreise mit den Radien $r_{1}=13$ und $r_{2}=8$.
Es sei $AB$ ein Durchmesser des größeren Kreises und $BC$ eine seiner Sehnen, die den kleineren Kreis im Punkt $D$ berührt.
Man berechne die Länge der Strecke $AD$. | [
"Solution:\n\nDie beiden möglichen Lagen von $D$ sind symmetrisch zur Geraden $(AB)$, so dass es ausreicht, den Fall zu betrachten, bei dem das Dreieck $ABD$ gegen den Uhrzeigersinn orientiert ist (siehe Figur). Der gemeinsame Mittelpunkt der beiden Kreise sei mit $M$ bezeichnet. Weil der Berührradius $MD$ auf der ... | Germany | 1. Auswahlklausur | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 19 | |
0bys | Let $A = \{z \in \mathbb{C} \mid |z| = 1\}$.
a) Prove that $(|z+1| - \sqrt{2})(|z-1| - \sqrt{2}) \le 0$, for all $z \in A$.
b) Prove that, for any $z_1, z_2, \dots, z_{12} \in A$, one can choose the signs "\pm" such that
$$
\sum_{k=1}^{12} |z_k \pm 1| < 17.
$$ | [
"a) Observe that\n$$\n|z+1|^2 + |z-1|^2 = (z+1)(\\bar{z}+1) + (z-1)(\\bar{z}-1) = 2|z|^2 + 2 = 4,\n$$\nhence $|z+1|^2 - 2 = 2 - |z-1|^2$, that is,\n$$\n(|z+1| - \\sqrt{2})(|z+1| + \\sqrt{2}) = -(|z-1| - \\sqrt{2})(|z-1| + \\sqrt{2}).\n$$\nClearly, this implies that $|z+1| - \\sqrt{2}$ and $|z-1| - \\sqrt{2}$ have o... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
0fjc | Problem:
Hallar todos los polinomios $P(t)$ de una variable, que cumplen
$$
P\left(x^{2}-y^{2}\right)=P(x+y) P(x-y)
$$
para todos los números reales $x$ e $y$. | [
"Solution:\nLa ecuación funcional dada\n$$\nP\\left(x^{2}-y^{2}\\right)=P(x+y) P(x-y)\n$$\nes equivalente a la ecuación funcional\n$$\nP(u v)=P(u) P(v)\n$$\ncon el cambio de variables $u=x+y$ y $v=x-y$, para todo $u, v \\in \\mathbb{R}$.\n\nPoniendo $u=v=0$ en $(**)$ se obtiene $P(0)=(P(0))^{2}$, de donde $P(0)=1$ ... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | All solutions are the zero polynomial, the constant one polynomial, and the monomials x to the n for positive integers n. | |
07vz | We are given a triangle *ABC* such that $\angle BAC < 90^\circ$. The point $D$ is on the opposite side of the line $AB$ to $C$ such that $|AD| = |BD|$ and $\angle ADB = 90^\circ$. Similarly, the point $E$ is on the opposite side of $AC$ to $B$ such that $|AE| = |CE|$ and $\angle AEC = 90^\circ$. The point $X$ is such t... | [
"Since $ADXE$ is a parallelogram, we have $\\angle ADX = \\angle AEX$. This implies that $\\angle XDB = 90^\\circ - \\angle ADX = 90^\\circ - \\angle AEX = \\angle CEX$. Since triangle $ADB$ is isosceles and $ADXE$ is a parallelogram, we have $|DB| = |DA| = |XE|$. Similarly, $|EC| = |EA| = |XD|$. We conclude that t... | Ireland | IRL_ABooklet_2023 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals"
] | English | proof only | null | |
0j5r | Problem:
Let
$$
F(x) = \frac{1}{\left(2 - x - x^{5}\right)^{2011}},
$$
and note that $F$ may be expanded as a power series so that $F(x) = \sum_{n=0}^{\infty} a_n x^n$. Find an ordered pair of positive real numbers $(c, d)$ such that
$$
\lim_{n \rightarrow \infty} \frac{a_n}{n^d} = c.
$$ | [
"Solution:\nAnswer: $\\left(\\frac{1}{6^{2011} 2010!},\\ 2010\\right)$\n\nFirst notice that all the roots of $2 - x - x^{5}$ that are not $1$ lie strictly outside the unit circle. As such, we may write\n$$\n2 - x - x^{5} = 2(1 - x)(1 - r_1 x)(1 - r_2 x)(1 - r_3 x)(1 - r_4 x)\n$$\nwhere $|r_i| < 1$, and let\n$$\n\\f... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof and answer | (1/(6^{2011} 2010!), 2010) | |
0eq4 | Mollie, Alfred and four other people want to be in a group photograph. In how many different ways can they be arranged in a row with Mollie and Alfred together in the middle?
(A) 8 (B) 16 (C) 24 (D) 48 (E) 80 | [
"The person on the extreme left can be any one of the four people that is neither Alfred nor Mollie; the second left can be any one of the remaining three; the first person on the right of centre... and so on. For every arrangement of the people around them, Alfred and Mollie can swap places to make a new arrangeme... | South Africa | South African Mathematics Olympiad | [
"Statistics > Probability > Counting Methods > Permutations"
] | English | MCQ | D | |
0kd7 | Problem:
Find all real numbers $x$ that satisfy the equation
$$
\frac{x-2020}{1}+\frac{x-2019}{2}+\cdots+\frac{x-2000}{21}=\frac{x-1}{2020}+\frac{x-2}{2019}+\cdots+\frac{x-21}{2000},
$$
and simplify your answer(s) as much as possible. Justify your solution. | [
"Solution:\nThe number $x=2021$ works. Indeed, for $x=2021$, the left-hand side of the equation equals\n$$\n\\frac{2021-2020}{1}+\\frac{2021-2019}{2}+\\cdots+\\frac{2021-2000}{21}=\\frac{1}{1}+\\frac{2}{2}+\\cdots+\\frac{21}{21}=\\underbrace{1+1+\\cdots+1}_{21}=21,\n$$\nand the right-hand side of the equation equal... | United States | Bay Area Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2021 | |
05do | Problem:
We denote the number of positive divisors of a positive integer $m$ by $d(m)$ and the number of distinct prime divisors of $m$ by $\omega(m)$. Let $k$ be a positive integer. Prove that there exist infinitely many positive integers $n$ such that $\omega(n)=k$ and $d(n)$ does not divide $d\left(a^{2}+b^{2}\righ... | [
"Solution:\n\nWe will show that any number of the form $n=2^{p-1} m$ where $m$ is a positive integer that has exactly $k-1$ prime factors all of which are greater than $3$ and $p$ is a prime number such that $(5 / 4)^{(p-1) / 2}>m$ satisfies the given condition.\n\nSuppose that $a$ and $b$ are positive integers suc... | European Girls' Mathematical Olympiad (EGMO) | European Girls' Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0k4v | Problem:
A tourist is learning an incorrect way to sort a permutation $\left(p_{1}, \ldots, p_{n}\right)$ of the integers $(1, \ldots, n)$. We define a fix on two adjacent elements $p_{i}$ and $p_{i+1}$, to be an operation which swaps the two elements if $p_{i}>p_{i+1}$, and does nothing otherwise. The tourist performs... | [
"Solution:\nNote that the given algorithm is very similar to the well-known Bubble Sort algorithm for sorting an array. The exception is that in the $i$-th round through the array, the first $i-1$ pairs are not checked.\n\nWe claim a necessary and sufficient condition for the array to be sorted after the tourist's ... | United States | HMMT February 2018 | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 1010! * 1009! | |
034y | Problem:
Let $c$ be a positive integer and let $\{a_{n}\}_{n=1}^{\infty}$ be a sequence of positive integers such that $a_{n} < a_{n+1} < a_{n} + c$ for every $n \geq 1$. The terms of the sequence are written one after another and in this way one obtains an infinite sequence of digits. Prove that for every positive int... | [
"Solution:\nLet $M$ be an arbitrary positive integer. We shall prove that there exists a term of the sequence $\\{a_{n}\\}_{n=1}^{\\infty}$, whose decimal representation is obtained from that of $M$ by adding several digits from the right, i.e. the number $M$ is a \"beginning\" of that member.\nLet $k$ be an index ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
05jv | Problem:
1) Soient $a$ et $b$ deux nombres réels tels que $a^{n}+b^{n}$ est un entier pour $n=1,2,3,4$. Montrer que $a^{n}+b^{n}$ est un entier pour tout $n \in \mathbb{N}^{*}$.
2) Est-il vrai que si $a$ et $b$ sont deux nombres réels tels que $a^{n}+b^{n}$ est un entier pour $n=1,2,3$ alors $a^{n}+b^{n}$ est un entie... | [
"Solution:\n1) Notons $s=a+b$ et $p=a b$. Soit $S_{n}=a^{n}+b^{n}$. Par hypothèse, $S_{n}$ est un entier pour $1 \\leqslant n \\leqslant 4$. Comme pour tout $n \\geqslant 1$ on a\n$$\nS_{n+1}=a^{n+1}+b^{n+1}=(a+b)\\left(a^{n}+b^{n}\\right)-a b\\left(a^{n-1}+b^{n-1}\\right)=S_{1} S_{n}-p S_{n-1},\n$$\nsi on montre q... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Part 1: True; the sums are integers for all positive exponents. Part 2: False; for example, taking 1 plus and minus one over the square root of two gives integer sums for the first three exponents but not for the fourth. | |
08su | Suppose a positive integer has the property that the sum of the remainders when its factors are divided by $4$ equals $1000$. Determine all positive integers having this property. | [
"For a positive integer $n$, let us denote by $S(n)$ the sum of all the positive factors of $n$ whose remainder when divided by $4$ is not equal to $2$. Let us first determine $S(n)$.\nSuppose the prime factorization of $n$ is given by\n$$\n2^m p_1^{m_1} \\cdots p_k^{m_k} \\quad (p_1, \\ldots, p_k \\text{ are disti... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 448, 796 | |
06i9 | In a school there are $2013$ boys and $2013$ girls. For each pair of a boy and a girl, together they have to choose one (and only one) of $25$ different clubs to join. Determine the maximum possible value of the integer $k$, such that no matter what the choices of the students are, there is a club with $k$ or more memb... | [
"The answer is $806$.\n\nBy the pigeonhole principle, there is a club with $n \\ge \\frac{2013^2}{25}$ pairs. Suppose there are $a$ boys and $b$ girls in this club. Then the number of pairs is at most $ab$. By the AM-GM inequality, we have\n$$\n\\frac{a+b}{2} \\ge \\sqrt{ab} \\ge \\sqrt{n} \\ge \\frac{2013}{5}.\n$$... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 806 | |
0d8o | Let $\mathbb{R}$ be the set of real numbers. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the condition
$$
f(x f(y)-y)+f(x y-x)+f(x+y)=2 x y
$$
for all $x, y \in \mathbb{R}$. | [
"Let denote by $P(x, y)$ the equation\n$$\nf(x f(y)-y)+f(x y-x)+f(x+y)=2 x y.\n$$\n$P(0, y)$ gives us $f(-y)+f(y)=0, \\forall y$. Thus $f$ is an odd function.\n\n$P(-1, y)$ follows\n$$\nf(-f(y)-y)+f(-y+1)+f(-1+y)=-2 y.\n$$\nFrom this, since $f$ is odd, we have $f(-f(y)-y)=-2 y$ and thus $f(f(y)+ y)=2 y$. So $f$ is ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x)=x or f(x)=-2x | |
0f8h | Problem:
Given a sequence of $19$ positive integers not exceeding $88$ and another sequence of $88$ positive integers not exceeding $19$. Show that we can find two subsequences of consecutive terms, one from each sequence, with the same sum. | [
"Solution:\n\nWe prove the general case. Let the first sequence be $a_1, a_2, \\ldots, a_m$ and the second sequence be $b_1, b_2, \\ldots, b_n$, where $0 < a_i \\leq n$ and $0 < b_j \\leq m$. Put $s_k = a_1 + a_2 + \\ldots + a_k$, $t_k = b_1 + b_2 + \\ldots + b_k$. Assume $s_m > t_n$ (if they are equal, then we are... | Soviet Union | 22nd ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
00di | Let $m, n \ge 2$. You want to completely cover an $m \times n$ board without any gaps or overlaps, using only pieces of the following two types:
Type A
Type B
Each type A piece must cover exactly 4 squares on the board, and each type B piece must cover exactly 5 squa... | [
"We will prove that the only boards that can be covered with the given pieces are the following:\n* Those with both sides even.\n* Those with both sides divisible by 3.\n* Those with at least one side divisible by 6.\n\nIf both $m$ and $n$ are even, then the $m \\times n$ board can be divided into $2 \\times 2$ squ... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | All boards where both sides are even, or both sides are divisible by three, or at least one side is divisible by six. | |
0do8 | Problem:
Дат је $\triangle ABC$. Нека је $A_{1}$ централносиметрична слика пресечне тачке симетрале $\measuredangle BAC$ и странице $BC$, где је центар симетрије средина странице $BC$. Аналогно дефинишемо тачке $B_{1}$ (на страници $CA$) и $C_{1}$ (на страници $AB$). Пресек кружнице описане око $\triangle A_{1}B_{1}C_... | [
"Solution:\n\nПодсетимо се да тачке $P$ и $Q$ унутар $\\triangle ABC$ зовемо изогонално спрегнутим ако је $\\varangle PAB = \\varangle QAC$ и $\\varangle PBC = \\varangle QBA$. Тада такође важи $\\varangle PCA = \\varangle QCB$.\n\nЛема. Подножја нормала из тачака $P$ и $Q$ на праве $BC, CA$ и $AB$ леже на истом кр... | Serbia | 13. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incen... | null | proof only | null | |
0cs3 | Initially, the blackboard contains two polynomials $x^3-3x^2+5$ and $x^2-4x$. If the polynomials $f(x)$ and $g(x)$ are written on the blackboard, it is permitted to write onto the board any polynomial of the form $f(x) \pm g(x)$, $f(x)g(x)$, $f(g(x))$, or $cf(x)$, where $c$ may be any (not necessarily integer) constant... | [
"**Ответ.** Не может.\nПусть $f(x)$ и $g(x)$ — два многочлена, и для некоторой точки $x_0$ выполняются равенства $f'(x_0) = 0$ и $g'(x_0) = 0$. Тогда, очевидно, $(f \\pm g)'(x_0) = 0$ и $cf'(x_0) = 0$. Также $(fg)'(x_0) = f(x_0)g'(x_0) + f'(x_0)g(x_0) = 0$. Наконец, если $h(x)$ — многочлен, то $(h(g(x_0)))' = h'(g(... | Russia | XL Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0205 | Problem:
Yesterday, $n \geq 4$ people sat around a round table. Each participant remembers only who his two neighbours were, but not which one sat on his left and which one sat on his right. Today, you would like the same people to sit around the same round table so that each participant has the same two neighbours as ... | [
"Solution:\n\na. $f(n) = n-3$.\n\n- Asking $n-4$ questions is not enough since the $n-4$ people queried might be sitting in a consecutive string, in which case the $n-4$ answers allow one to sit $n-2$ people in the same positions as yesterday, but there is still an ambiguity among the two remaining ones.\n\n- Let u... | Benelux Mathematical Olympiad | 4th Benelux Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | f(n) = n - 3; g(n) = n - 1 - ceil(n/3) | |
0jaw | Problem:
Let $S_{7}$ denote all the permutations of $1,2, \ldots, 7$. For any $\pi \in S_{7}$, let $f(\pi)$ be the smallest positive integer $i$ such that $\pi(1), \pi(2), \ldots, \pi(i)$ is a permutation of $1,2, \ldots, i$. Compute $\sum_{\pi \in S_{7}} f(\pi)$. | [
"Solution:\nExtend the definition of $f$ to apply for any permutation of $1,2, \\ldots, n$, for any positive integer $n$. For positive integer $n$, let $g(n)$ denote the number of permutations $\\pi$ of $1,2, \\ldots, n$ such that $f(\\pi)=n$. We have $g(1)=1$. For fixed $n, k$ (with $k \\leq n$), the number of per... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 29093 | |
09ur | In a table with two rows and five columns, each of the squares is coloured black or white according to the following rules:
* Two adjacent columns may never have the same number of black squares.
* Two $2 \times 2$-squares that overlap in one column may never have the same number of black squares.
How many possible col... | [
"D) 20"
] | Netherlands | First Round, January 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | MCQ | D) 20 | |
08am | Problem:
Alessandro, Daniele e Manuela discutono di un numero naturale $n$ di due cifre. Ognuno di loro fa due affermazioni, ma siccome sono tutti un po' scarsi in matematica ognuno di loro fa un'affermazione vera ed una falsa.
Alessandro dice: "$n$ è pari. Inoltre è un multiplo di 3.";
Daniele risponde: "Sì, $n$ è ... | [
"Solution:\n\nLa risposta è (D). Supponiamo prima che $n$ sia pari: allora non è un multiplo di 3, perché una delle due affermazioni di Alessandro deve essere falsa, e quindi (usando quello che dice Daniele) la cifra delle unità di $n$ deve essere 5, ma questo è impossibile per un numero pari. Il numero $n$ (se esi... | Italy | Progetto Olimpiadi della Matematica - Gara di Febbraio | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Logic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
0291 | Problem:
Um número perfeito - Um número natural $n$ é dito perfeito se a soma de todos os seus divisores próprios, isto é, diferentes de $n$, é igual a $n$. Por exemplo, $6$ e $28$ são perfeitos, pois: $6=1+2+3$ e $28=1+2+4+7+14$. Sabendo que $2^{31}-1$ é um número primo, mostre que $2^{30}\left(2^{31}-1\right)$ é um ... | [
"Solution:\n\nSe $2^{31}-1$ é um número primo, seu único divisor próprio é o número $1$. Então os divisores próprios de $2^{30}\\left(2^{31}-1\\right)$ são:\n$$\n1, 2, 2^{2}, 2^{3}, \\ldots, 2^{29}, 2^{30}, \\left(2^{31}-1\\right), 2\\left(2^{31}-1\\right), 2^{2}\\left(2^{31}-1\\right), \\ldots, 2^{29}\\left(2^{31}... | Brazil | Nível 3 | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0ifl | Problem:
Working together, Jack and Jill can paint a house in 3 days; Jill and Joe can paint the same house in 4 days; or Joe and Jack can paint the house in 6 days. If Jill, Joe, and Jack all work together, how many days will it take them? | [
"Solution:\n\nSuppose that Jack paints $x$ houses per day, Jill paints $y$ houses per day, and Joe paints $z$ houses per day. Together, Jack and Jill paint $1 / 3$ of a house in a day - that is,\n$$\nx+y=1 / 3 .\n$$\nSimilarly,\n$$\ny+z=1 / 4\n$$\nand\n$$\nz+x=1 / 6\n$$\nAdding all three equations and dividing by 2... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 8/3 days | |
0aa4 | Problem:
King George has decided to connect the 1680 islands in his kingdom by bridges. Unfortunately the rebel movement will destroy two bridges after all the bridges have been built, but not two bridges from the same island.
What is the minimal number of bridges the King has to build in order to make sure that it is... | [
"Solution:\n\nAn island cannot be connected with just one bridge, since this bridge could be destroyed. Consider the case of two islands, each with only two bridges, connected by a bridge. (It is not possible that they are connected with two bridges, since then they would be isolated from the other islands no matte... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | 2016 | |
0kqq | Problem:
Find all solutions to $m^{4} = n^{3} + 137$ over the positive integers. | [
"Solution:\nThe fourth powers mod $13$ are $0, 1, 3, 9$ and the cubes mod $13$ are $0, 1, 5, 8, 12$. Therefore, $m^{4} - n^{3} \\equiv 7 \\pmod{13}$ is impossible, meaning that there are no solutions."
] | United States | Berkeley Math Circle Monthly Contest 7 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | no solutions | |
07bg | a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$ are $2n$ positive numbers. We know that all the $a_i$'s, $1 \le i \le n$ are not equal, and that they can be separated into two partitions of equal sum. These two properties hold for the $b_i$'s, $1 \le i \le n$, as well. Prove that there exists a simple $2n$-gon with si... | [
"We start with a lemma.\n\n**Lemma 1.** Suppose that are given two sequences $a_1 > a_2 > \\dots > a_m$ and $b_1 < b_2 < \\dots < b_m$ of positive real numbers as lengths of segments. We start from the origin and at the step $i$ ($1 \\le i \\le m$), we go up with a segment of length $a_i$ and then we go right with ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
09tc | Problem:
Bepaal alle functies $f: \mathbb{R} \rightarrow \mathbb{R}$ zodat
$$
(y+1) f(x)+f(x f(y)+f(x+y))=y
$$
voor alle $x, y \in \mathbb{R}$. | [
"Solution:\n\nInvullen van $x=0$ geeft $(y+1) f(0)+f(f(y))=y$, dus $f(f(y))=y \\cdot(1-f(0))-f(0)$. Als $f(0) \\neq 1$, is de rechterkant een bijectieve functie in $y$ en de linkerkant dus ook. Daarmee is in dit geval $f$ bijectief.\n\nWe gaan nu laten zien dat in het geval $f(0)=1$ ook geldt dat $f$ bijectief is. ... | Netherlands | IMO-selectietoets II | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = -x | |
0cdu | Let the numbers $r, s \in [1, \infty)$ with the property that for every positive integers $a, b$, with $a$ dividing $b$, it results that $[ar]$ divides $[bs]$.
a) Prove that $\frac{s}{r}$ is a positive integer.
b) Show that $r$ and $s$ are positive integers.
*Remark.* By $[x]$ we denote the floor of the real number $x$... | [
"a) We suppose that $\\frac{s}{r} \\notin \\mathbb{N}$. Then, there exists $k \\in \\mathbb{N}$ such that $k < \\frac{s}{r} < k+1 \\iff kr < s < (k+1)r$. Choosing $b = a \\in \\mathbb{N}^*$, arbitrary, we obtain $[ar] \\mid [as]$ and thus $[ar] \\mid [as] - k[ar]$. (1)\nFrom $s > kr$, we obtain that there exists $u... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0kyv | Problem:
Isabella the geologist discovers a diamond deep underground via an X-ray machine. The diamond has the shape of a convex cyclic pentagon $P A B C D$ with $A D \| B C$. Soon after the discovery, her X-ray breaks, and she only recovers partial information about its dimensions. She knows that $A D=70$, $B C=55$, ... | [
"Solution:\n\n\nLet $X=P B \\cap A D$ and $Y=P C \\cap A D$. Let $A X=p$, $X Y=q$, and $Y D=r$. From $A B \\| C D$, we get that $A B=C D$, and so $\\angle A P X=\\angle D P Y$. Thus, we may apply Steiner ratio theorem on $\\triangle P A D$ and $\\triangle P X Y$ to get that\n$$\n\\frac{p(p+... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 25 sqrt(6) | |
04wx | Let $a_1 = a_2 = 1$ and $a_{k+2} = a_{k+1} + a_k$ for any $k \in \mathbb{N}$ (the Fibonacci sequence). Prove that for any natural number $m$ there exists an index $k$ such that the number $a_k^4 - a_k - 2$ is divisible by $m$. | [
"All the congruences and remainder classes below are meant mod $m$. We obtain the desired congruence relation $a_k^4 - a_k - 2 \\equiv 0$ as a consequence of the simpler relation $a_k \\equiv -1$.\n\nThe sequence of remainder classes of the numbers $a_k$ has the following property: the remainder classes of any two ... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
01uy | Let $p \ge 2$ be a prime number. Alice and Bob play the following game: they, in turn, select an index $i$ in the set $\{0, 1, 2, \dots, p-1\}$ that was not selected before by either of the two players and then chooses a digit $a_i$. Alice starts. The game ends after all the indices have been selected. The goal of Alic... | [
"2. See IMO-2017 Shortlist, Problem N2."
] | Belarus | Selection and Training Session | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0ezl | Problem:
Prove that we can find a number divisible by $2^{n}$ whose decimal representation uses only the digits $1$ and $2$. | [
"Solution:\nInduction on $n$. We claim that we can find $N$ with $n$ digits, all $1$ or $2$, so that $N$ is divisible by $2^{n}$.\n\nTrue for $n = 1$: take $N = 2$.\n\nSuppose it is true for $n$. If $2^{n + 1}$ divides $N$, then since $2^{n + 1}$ divides $2 \\times 10^{n}$, it also divides $N'$ obtained from $N$ by... | Soviet Union | ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
01xd | The altitudes $CC_1$ and $BB_1$ are drawn in the acute triangle $ABC$. The bisectors of angles $\angle BB_1C$ and $\angle CC_1B$ intersect the line $BC$ at points $D$ and $E$ respectively and meet each other at point $X$.
Prove that the intersection points of circumcircles of the triangles $BEX$ and $CDX$ lie on the li... | [
"Since $\\angle BB_1C = \\angle CC_1B = 90^\\circ$, the points $B, C_1, B_1$ and $C$ lie on the circle $\\omega$ with the diameter $BC$. Hence the bisectors of angles $\\angle BB_1C$ and $\\angle CC_1B$ pass through the midpoint of the arc $BC$ of $\\omega$, so this midpoint is $X$.\n\n\n\n... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
0jxn | Problem:
Kelvin the Frog and 10 of his relatives are at a party. Every pair of frogs is either friendly or unfriendly. When 3 pairwise friendly frogs meet up, they will gossip about one another and end up in a fight (but stay friendly anyway). When 3 pairwise unfriendly frogs meet up, they will also end up in a fight.... | [
"Solution:\n\nConsider a graph $G$ with 11 vertices - one for each of the frogs at the party - where two vertices are connected by an edge if and only if they are friendly. Denote by $d(v)$ the number of edges emanating from $v$; i.e. the number of friends frog $v$ has. Note that $d(1)+d(2)+\\ldots+d(11)=2e$, where... | United States | February 2017 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 28 | |
00x8 | Problem:
All faces of a convex polyhedron are parallelograms. Can the polyhedron have exactly 1992 faces? | [
"Solution:\n\nNo, it cannot. Let us call a series of faces $F_{1}, F_{2}, \\ldots, F_{k}$ a ring if the pairs $(F_{1}, F_{2}),(F_{2}, F_{3}), \\ldots, (F_{k-1}, F_{k}),(F_{k}, F_{1})$ each have a common edge and all these common edges are parallel. It is not difficult to see that any two rings have exactly two comm... | Baltic Way | Baltic Way 1992 | [
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | No | |
0h0r | Olesya writes down numbers $1, 2, 3, 4, 5, 6$ at the vertices of a prism. After this, at each edge Andriy writes down the sum of numbers that are written at the vertices that form this edge. Can Olesya write numbers in such a way that all Andriy's numbers are different? | [
"Yes. See fig. 10."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | English | proof and answer | Yes | |
05n1 | Problem:
Soit $S$ un ensemble d'entiers strictement positifs tel que
$$
\lfloor\sqrt{x}\rfloor=\lfloor\sqrt{y}\rfloor \text{ pour tous } x, y \in S
$$
Prouver que si $x, y, z, t \in S$ avec $(x, y) \neq(z, t)$ et $(x, y) \neq(t, z)$, alors $x y \neq z t$.
( $\lfloor.\rfloor$ désigne la partie entière.) | [
"Solution:\nSupposons tout d'abord qu'il existe des entiers $x_{1}, x_{2}, x_{3}, x_{4}$ dans $S$ tels que $x_{1} x_{2} \\leqslant x_{3} x_{4}$ et $x_{1}+x_{2}>x_{3}+x_{4}$. Puisqu'il s'agit d'entiers, on a donc $x_{1}+x_{2}-x_{3}-x_{4} \\geqslant 1$. Soit $n=\\left\\lfloor\\sqrt{x_{1}}\\right\\rfloor$. Par définit... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0a9s | Problem:
Let $ABC$ be an acute angled triangle, and $H$ a point in its interior. Let the reflections of $H$ through the sides $AB$ and $AC$ be called $H_{c}$ and $H_{b}$, respectively, and let the reflections of $H$ through the midpoints of these same sides be called $H_{c}^{\prime}$ and $H_{b}^{\prime}$, respectively... | [
"Solution:\n\nIf at least two of the four points $H_{b}, H_{b}^{\\prime}, H_{c}$, and $H_{c}^{\\prime}$ coincide, all four are obviously concyclic. Therefore we may assume that these four points are distinct.\n\nLet $P_{b}$ denote the midpoint of segment $H H_{b}$, $P_{b}^{\\prime}$ the midpoint of segment $H H_{b}... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05su | Problem:
Soient $P$, $Q$ des polynômes à coefficients réels tels que $P \circ Q = P^{2019}$. On suppose que toutes les racines de $P$ sont réelles. Montrer qu'elles sont toutes égales. | [
"Solution:\n\nA priori, on sait que le terme de droite va être nul si on l'évalue en $x_{j}$, on regarde donc ce qu'on en déduit pour le terme de gauche. Posons $X = \\{x_{1}, \\ldots, x_{k}\\}$, on a alors nécessairement $Q(x_{j}) \\in \\{x_{1}, \\ldots, x_{k}\\} = X$ pour tout $1 \\leqslant j \\leqslant k$.\n\nDe... | France | Envoi 5: Pot Pourri | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
0ecj | There are $n$ lights in a line, $n \ge 3$, numbered with numbers 1 through $n$. At the beginning each odd light in the line is turned on and each even light is turned off. In each move we may simultaneously change the state of three consecutive lights (turn on or turn off).
a. Prove that the order of the moves we make... | [
"a. Let's look at what happens with the state of one light when executing a move. Each move either changes or preserves the state of the light. Thus the final state of some light only depends on the number of moves made that change the state of that light, and not on the order of the moves made. Therefore the final... | Slovenia | National Math Olympiad 2015 – Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Exactly those n that are divisible by 3 | |
069w | In the table are written the positive integers $1, 2, 3, \ldots, 2018$. John and Mary have the possibility to make the following move:
They select two of the written numbers in the table, let $\alpha, \beta$ and they replay them with the numbers $5\alpha - 2\beta$ and $3\alpha - 4\beta$.
John asserts that after a fini... | [
"We observe that after a move the sum of the numbers in the table have a change equal to the difference:\n$$\n(5\\alpha - 2\\beta) + (3\\alpha - 4\\beta) - (\\alpha + \\beta) = 7(\\alpha - \\beta)\n$$\nTherefore we conclude that after every application of a move the difference of the sum $S_{\\text{new}}$ minus the... | Greece | 36th Hellenic Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | Mary is right. | |
0kfu | Let $\alpha \ge 1$ be a real number. Hephaestus and Poseidon play a turn-based game on an infinite grid of unit squares. Before the game starts, Poseidon chooses a finite number of cells to be *flooded*. Hephaestus is building a *levee*, which is a subset of unit edges of the grid (called *walls*) forming a connected, ... | [
"We show that if $\\alpha > 2$ then Hephaestus wins, but when $\\alpha = 2$ (and hence $\\alpha \\le 2$) Hephaestus cannot contain even a single-cell flood initially.\n\n**Strategy for** $\\alpha > 2$: Impose $\\mathbb{Z}^2$ coordinates on the cells. Adding more flooded cells does not make our task easier, so let u... | United States | USA IMO TST | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | alpha > 2 | |
021j | Problem:
É possível dividir um tabuleiro $8 \times 9$ em retângulos $1 \times 6$ ? | [] | Brazil | Desafios | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | No | |
0b1k | Problem:
What is the probability that a rectangle with perimeter $36~\mathrm{cm}$ has area greater than $36~\mathrm{cm}^2$? | [
"Solution:\nLet $x$ and $y$ be the lengths of the sides of the rectangle. We are looking for the probability that $x y > 36$ given that $2x + 2y = 36$. Equivalently, we compute the probability that $x(18 - x) > 36$ given $0 < x < 18$.\n\nNow, $x(18 - x) > 36 \\Leftrightarrow x^2 - 18x + 36 < 0 \\Leftrightarrow (x -... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | sqrt(5)/3 | |
0ceg | Triangle $BAD$ has $\angle BAD = 45^\circ$ and triangle $BDC$ is on its outside, so that $DC = BA$ and $\angle DCB = \angle CDA = 75^\circ$. Find the measure of $\angle ABD$.
Adrian Bud | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 90 | |
027t | Problem:
Sejam $ABCD$ e $EFGH$ quadrados de lados $33$ e $12$, com $EF$ sobre o lado $DC$ (como mostrado na figura abaixo). Seja $X$ o ponto de interseção dos segmentos $HB$ e $DC$. Suponha que $\overline{DE} = 18$.
a) Calcule o comprimento do segmento $\overline{EX}$.
b) Prove que os pontos... | [
"Solution:\na) Denote $\\overline{EX} = x$. Temos que $|\\overline{CX}| = 33 - 18 - x = 15 - x$.\n\nAgora note que os triângulos $EXH$ e $CXB$ são semelhantes, logo:\n$$\n\\frac{|\\overline{EH}|}{|\\overline{CB}|} = \\frac{|\\overline{EX}|}{|\\overline{CX}|} \\Rightarrow \\frac{12}{33} = \\... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof and answer | EX = 4; points A, X, and G are collinear. | |
0i76 | Problem:
Count the number of triangles with positive area whose vertices are points whose $(x, y)$-coordinates lie in the set $\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\}$. | [
"Solution:\n\nThere are $\\binom{9}{3} = 84$ triples of points. 8 of them form degenerate triangles (the ones that lie on a line), so there are $84 - 8 = 76$ nondegenerate triangles."
] | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 76 | |
0ejp | Problem:
a) Vsota prvih osmih členov aritmetičnega zaporedja je 124, prvi člen pa je enak 5. Izračunaj prve štiri člene aritmetičnega zaporedja.
b) Vsota prvih sedmih členov nekega aritmetičnega zaporedja je enaka 105. Prvi, tretji in sedmi člen danega aritmetičnega zaporedja so zaporedni trije členi nekega geometrijs... | [
"Solution:\n\na)\nZapišimo obrazec za vsoto prvih $n$ členov aritmetičnega zaporedja:\n$$S_n = \\frac{n}{2}\\left(2 a_1 + (n-1) d\\right).$$\nVstavimo podatke za vsoto prvih $8$ členov in dobimo:\n$$S_8 = \\frac{8}{2}(2 \\cdot 5 + 7 d) = 124.$$ \nPoenostavimo:\n$$4(10 + 7d) = 124$$\n$$40 + 28d = 124$$\n$$28d = 84$$... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Odbirno tekmovanje | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) 5, 8, 11, 14; b) 15, 15, 15, 15 or 6, 9, 12, 15 | |
03j5 | Problem:
The lengths of the sides of a triangle are $6$, $8$ and $10$ units. Prove that there is exactly one straight line which simultaneously bisects the area and perimeter of the triangle. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0e5i | Find all natural numbers $n \ge 10$ with non-zero digits that satisfy the following condition: if any of the digits of $n$ is deleted, the obtained number is a divisor of $n$. | [
"Suppose the decimal notation of a natural number $n$ is equal to $\\overline{a_k a_{k-1} \\dots a_2 a_1}$. The main condition of the problem says that the number $\\overline{a_k a_{k-1} \\dots a_2}$ divides the number $\\overline{a_k a_{k-1} \\dots a_2 a_1} = 10 \\cdot \\overline{a_k a_{k-1} \\dots a_2} + a_1$, he... | Slovenia | National Math Olympiad 2012 | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 66, 77, 88, 99 | |
02fh | A regular tetrahedron has side $L$. What is the smallest $x$ such that the tetrahedron can be passed through a loop of twine of length $x$? | [
"The answer is $2L$. Consider the following net of the tetrahedron:\n\n\n\nLet $P$ be a point of one of the edges of the tetrahedron. The loop will pass through $P$ some time. But $P'$ on the net coincides with $P$ on the tetrahedron, so by the triangular inequality the loop must be at leas... | Brazil | XVII OBM | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 2L | |
0duk | Problem:
Najmanjše naravno število, katerega kvadrat se konča s tremi štiricami, je 38, saj je $38^{2}=1444$. Katero je naslednje najmanjše naravno število s to lastnostjo? | [
"Solution:\n\nNaj bo $38+n$ iskano število. Tedaj je $(38+n)^{2}=1444+n(76+n)$, kjer se število $n(76+n)$ konča s tremi ničlami, oziroma je večkratnik števila $1000$. Ker je $1000=5^{3} \\cdot 2^{3}$, mora biti ali $n$ ali $76+n$ deljivo s $5$. Toda $n$ in $76+n$ nista hkrati deljivi s $5$, zato mora biti eno izmed... | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 462 | |
01th | Given the triangle $ABC$ with $AB = 2AC$. If points $M$ and $N$ belong to the sides $BC$ and $AB$, respectively, and the perimeter of the trapezoid $CMNA$ is the sum of the lengths of the sides $AB$ and $AC$, construct $M$ using compasses and ruler.
(S. Mazanik) | [
"$M$ is the intersection point of the line $\\ell \\parallel AC$ passing through the intersection point of the bisector of the angle $ACB$ and the side $AB$.\n\nLet $M$ be the point we search for and $NM \\parallel AC$ (see the Fig.).\n\nLet $P(CMNA)$ denote the perimeter of the trapezoid $CMNA$. Then $P(CMNA) = AN... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | null | |
0ig7 | Problem:
In a town of $n$ people, a governing council is elected as follows: each person casts one vote for some person in the town, and anyone that receives at least five votes is elected to council. Let $c(n)$ denote the average number of people elected to council if everyone votes randomly. Find $\lim_{n \rightarro... | [
"Solution:\n\n$1 - 65 / 24e$\n\nLet $c_{k}(n)$ denote the expected number of people that will receive exactly $k$ votes. We will show that $\\lim_{n \\rightarrow \\infty} c_{k}(n) / n = 1/(e \\cdot k!)$. The probability that any given person receives exactly $k$ votes, which is the same as the average proportion of... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 1 - 65/(24e) | |
08rt | In a mathematical competition, gold medals are given to $\lfloor \frac{n}{a} \rfloor$ people, silver medals to $\lfloor \frac{n}{b} \rfloor$ and bronze medals to $\lfloor \frac{n}{c} \rfloor$ ($a \ge b \ge c$ are integer constants and $n$ is the number of participants). No one gets two or more medals. Determine all tri... | [
"Let $f(n) = n - \\lfloor \\frac{n}{a} \\rfloor - \\lfloor \\frac{n}{b} \\rfloor - \\lfloor \\frac{n}{c} \\rfloor$ for integer $n$. For $n$ positive, $f(n)$ is equal to the contestants with no medals on an $n$-people contest. Since $x - 1 < [x] \\le x$, it follows that $Sn \\le f(n) < Sn + 3$ where $S = 1 - \\frac{... | Japan | Japan 2007 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (6,6,6), (8,8,4), (10,5,5), (12,6,4) | |
086z | Problem:
Sia $x$ la più piccola delle due soluzioni dell'equazione $x^{2}-4x+2=0$. Quali sono le prime tre cifre dopo la virgola nella scrittura (in base 10) del numero
$$
x+x^{2}+x^{3}+\cdots+x^{2009} ?
$$ | [
"Solution:\n\nLa risposta è 414. Dalla consueta formula risolutiva per le equazioni di secondo grado, si ha $x=2-\\sqrt{2}$. Utilizzando ora la formula per la somma di una progressione geometrica, abbiamo\n$$\n\\begin{aligned}\nx+x^{2}+x^{3}+\\cdots+x^{2009} & =x\\left(1+x+x^{2}+\\cdots+x^{2008}\\right) \\\\\n& =x ... | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 414 | |
0g8v | 甲、乙兩人在實數線上玩以下的著色遊戲。甲有一桶顏料共四單位, 其中 $p$ 單位的顏料剛好可以塗滿一個長度為 $p$ 的閉區間。每回合, 甲先指定一個正整數 $m$, 並給乙 $\frac{1}{2^m}$ 單位的顏料。接著, 乙選一個正整數 $k$, 並將 $\frac{k}{2^m}$ 到 $\frac{k+1}{2^m}$ 塗滿 (此區間可能有一部分在之前的回合中已經被塗過。) 如果桶子空了但 $[0, 1]$ 區間還沒被塗滿, 則甲獲勝。
試問: 甲是否有在有限回合內獲勝的必勝法?
Player A and B play a painful game on the real line. Player A has a pot... | [
"否，乙可以確保在顏料用光時 $[0, 1]$ 區間必被塗滿。在第 $r$ 回合開始時，令 $x_r$ 為滿足 $[0, x_r]$ 皆已被塗滿的最大實數 (令 $x_1 = 0$.) 假設 A 選擇 $m$，令 $y_r$ 為滿足\n$$\n\\frac{y_r}{2^m} \\le x_r < \\frac{y_r+1}{2^m}\n$$\n的整數。注意到 $I_0^r := [y_r/2^m, (y_r+1)/2^m]$ 是本回合可以塗，且尚未被塗滿的區間中最左邊的那一個。\n乙的策略是考慮 **下一個** 區間 $I_1^r := [(y_r + 1)/2^m, (y_r + 2)/2^m]$。若 $I_1^r$ 尚... | Taiwan | 二〇一四年國際數學奧林匹亞競賽第二階段選訓營 模擬競賽(二) | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0kb5 | Suppose $P$ is a polynomial with integer coefficients such that for every positive integer $n$, the sum of the decimal digits of $|P(n)|$ is not a Fibonacci number. Must $P$ be constant?
(A *Fibonacci number* is an element of the sequence $F_0, F_1, \dots$ defined recursively by $F_0 = 0$, $F_1 = 1$, and $F_{k+2} = F_{... | [
"The answer is yes, $P$ must be constant. By $S(n)$ we mean the sum of the decimal digits of $|n|$.\nWe need two claims.\n\n**Claim** — If $P(x) \\in \\mathbb{Z}[x]$ is nonconstant with positive leading coefficient, then there exists an integer polynomial $F(x)$ such that all coefficients of $P \\circ F$ are positi... | United States | USA TSTST | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null |
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