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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0768 | Problem:
In an acute-angled triangle $A B C$, a point $D$ lies on the segment $B C$. Let $O_{1}, O_{2}$ denote the circumcentres of triangles $A B D$ and $A C D$, respectively. Prove that the line joining the circumcentre of triangle $A B C$ and the orthocentre of triangle $O_{1} O_{2} D$ is parallel to $B C$. | [
"Solution:\n\nWithout loss of generality assume that $\\angle A D C \\geq 90^{\\circ}$. Let $O$ denote the circumcenter of triangle $A B C$ and $K$ the orthocentre of triangle $O_{1} O_{2} D$. We shall first show that the points $O$ and $K$ lie on the circumcircle of triangle $A O_{1} O_{2}$. Note that circumcircle... | India | Indian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jzu | Problem:
On a blackboard a stranger writes the values of $s_{7}(n)^{2}$ for $n=0,1, \ldots, 7^{20}-1$, where $s_{7}(n)$ denotes the sum of digits of $n$ in base $7$. Compute the average value of all the numbers on the board. | [
"Solution:\n\nWe solve for $0$ to $b^{n}-1$ and $s_{b}(n)^{2}$ (i.e. base $b$).\nLet $n=d_{1} \\ldots d_{n}$ in base $b$, where there may be leading zeros. Then $s_{b}(n)=d_{1}+\\cdots+d_{n}$, regardless of the leading zeros.\n\n$$\n\\mathbb{E}\\left[s_{d}(n)^{2}\\right]=\\mathbb{E}\\left[\\left(d_{1}+\\cdots+d_{n}... | United States | HMMT November 2017 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 3680 | |
0k0d | Problem:
A malfunctioning digital clock shows the time $9:57\ \mathrm{AM}$; however, the correct time is $10:10\ \mathrm{AM}$. There are two buttons on the clock, one of which increases the time displayed by 9 minutes, and another which decreases the time by 20 minutes. What is the minimum number of button presses nece... | [
"Solution:\nWe need to increase the time by 13 minutes. If we click the 9 minute button $a$ times and the 20 minute button $b$ times, then we must have $9a - 20b = 13$. Note that if this equation is satisfied, then $b$ increases as $a$ increases, so it suffices to minimize $a$. This means that $a$ must end in a 7. ... | United States | HMMT November | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 24 | |
02ke | Problem:
Quantos dentre os números $-5,-4,-3,-2,-1,0,1,2,3$ satisfazem a desigualdade $-3 x^{2}<-14$ ?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 | [
"Solution:\nSe $-3 x^{2}<-14$ então $3 x^{2}>14$ ou $x^{2} > \\frac{14}{3} = 4 \\frac{2}{3}$. Como estamos olhando apenas para valores inteiros de $x$, então $x^{2}$ também é inteiro. Sendo $x^{2} > 4 \\frac{2}{3}$, concluímos que $x^{2}$ é no mínimo $5$. Os números acima que satisfazem essa condição são $-5$, $-4$... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
0gl4 | Determine all primes $p$ such that $2p^2 - 3p - 1$ is a cube of a positive integer. | [
"Let $p$ be a prime and $n$ be a positive integer such that\n$$\n2p^2 - 3p - 1 = n^3. \\qquad (1)\n$$\nSince\n$$\nn^3 = 2p^2 - 3p - 1 < 2p^2 \\le p^3,\n$$\nit follows that $n < p$, so $n + 1 \\le p$.\n\nConsider the case $p = n + 1$. We then get by (1) that\n$$\nn^3 - 2n^2 - n + 2 = 0\n$$\nwhich implies\n$$\n(n - 2... | Thailand | Tajland 2014 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2, 3 | |
0e22 | Problem:
Naj bosta $p$ in $q$ polinoma stopnje 3 s celoštevilskimi koeficienti, katerih vodilna koeficienta sta si tuja. Naj bo $a$ tako racionalno število, da sta $p(a)$ in $q(a)$ celi števili. Dokaži, da je tedaj tudi $a$ celo število. | [
"Solution:\n\nNaj bo $p(x) = b_1 x^3 + c_1 x^2 + d_1 x + e_1$ in $q(x) = b_2 x^3 + c_2 x^2 + d_2 x + e_2$. Zapišimo $a = \\frac{r}{s}$ kot okrajšan ulomek, kjer sta $r$ in $s$ celi števili in je $s > 0$. Označimo $p(a) = m$ in $q(a) = n$. Tedaj velja\n$$\nb_1 \\cdot \\frac{r^3}{s^3} + c_1 \\cdot \\frac{r^2}{s^2} + ... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0929 | Problem:
Let $n \geqslant 3$ be an integer. An inner diagonal of a simple $n$-gon is a diagonal that is contained in the $n$-gon. Denote by $D(P)$ the number of all inner diagonals of a simple $n$-gon $P$ and by $D(n)$ the least possible value of $D(Q)$, where $Q$ is a simple $n$-gon. Prove that no two inner diagonals... | [
"Solution:\n\nFirst we prove that for every $n$-gon $P$ with $n \\geqslant 4$ we have $D(P) \\geqslant 1$. Let $A$ be one of the vertices of $P$ with inner angle less than $180^{\\circ}$. Denote the two vertices adjacent to $A$ by $B$ and $C$. The segment $BC$ is a diagonal of $P$, since $n \\geqslant 4$. If it lie... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0hqa | Problem:
Given a triangle $A B C$, a circle $k$ is tangent to the lines $A B$ and $A C$ at $B$ and $P$. Let $H$ be the foot of perpendicular from the center $O$ of $k$ to $B C$, and let $T$ be the intersection point of $O H$ and $B P$. Prove that $A T$ bisects the segment $B C$. | [
"Solution:\n\nLet $X$ be the intersection of $A T$ and $B C$, $S$ the intersection of $B C$ and $A O$, and $Y$ the intersection of $B P$ and $A O$. Since $B Y \\perp S O$ and $O H \\perp B S$, $T$ is the orthocenter of $\\triangle B O S$ and $S T \\perp B O$, hence $S T \\parallel A B$. Thus\n$$\nS T : A B = T X : ... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ers | All four-digit positive integers which are rearrangements of the number $2316$ are written in increasing order. What is the largest possible difference between two adjacent numbers in this list? | [
"$6123$ is the smallest number starting with a $6$ and $3621$ is the largest number starting with a $3$. The difference between these is $6123 - 3621 = 2502$. Adjacent numbers between $1236$ and $3621$ will be less than $2500$ apart, while adjacent numbers larger than $6123$ will be less than $1000$ apart. Thus $25... | South Africa | South African Mathematics Olympiad Third Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Other"
] | English | final answer only | 2502 | |
06ic | A number is *good* if the sum of its digits is $18$ (for example $130518$ is a good number). Find the number of good $6$-digit numbers. | [
"The answer is $21087$.\nWe use generating function. Since the first digit is nonzero, it corresponds to a factor $x + x^2 + \\cdots + x^9$. All other digits can be one of $0, 1, \\ldots, 9$, and so each of them corresponds to a factor $1 + x + \\cdots + x^9$. Thus, the answer is the coefficient of $x^{18}$ in\n$$\... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 21087 | |
07rg | The game of *Greed* starts with an initial configuration of one or more piles of stones. Player 1 and Player 2 take turns to remove stones, beginning with Player 1. At each turn, a player has two choices:
* take one stone from any one of the piles (a simple *move*);
* take one stone from each of the remaining piles (a ... | [
"The winning strategy in (a) is easily found: all piles are even in number so Player 2 maintains this all-even status by parroting Player 1's every move. This eventually guarantees a win for Player 2.\n\nConfiguration (b) is trickier. We will see that there is a guaranteed win for Player 1. We first mathematically ... | Ireland | Irish | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | (a) Player 2 wins by mirroring moves to keep all piles even. (b) Player 1 wins; begin by removing one stone from an even pile and then follow the W2 strategy. | |
0169 | The equation $x^3 - a x^2 - b = 0$ has 3 integer roots. Prove that $b = d k^2$, where $d$ and $k$ are integers and $d$ divides $a$. | [
"It is sufficient to prove for each prime $p$ that if $b$ is divisible by $p^{2k-1}$ but not divisible by $p^k$ for some positive integer $k$, then $p$ divides $a$.\nLet $u$, $v$, $w$ be the integer roots of the equation. Then by Viète's formulas $u + v + w = a$, $uv + uw + vw = 0$, $uvw = b$. Let $u$ be divisible ... | Baltic Way | Baltic Way SHL | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0epq | Let $ABC$ be an acute-angled triangle with $AB < AC$, and let points $D$ and $E$ be chosen on the sides $AC$ and $BC$ respectively in such a way that $AD = AE = AB$. The circumcircle of $ABE$ intersects the line $AC$ at $A$ and $F$ and the line $DE$ at $E$ and $P$. Prove that $P$ is the circumcentre of $BDF$. | [
"Since $AD = AE$, $AED$ is isosceles, so we also have $\\angle ADE = \\angle AED$. Combining this with the fact that $ABPE$ and $ABPF$ are cyclic, we obtain\n\n$$\n\\angle ABP = 180^{\\circ} - \\angle AEP = \\angle AED = \\angle ADE = \\angle ADP\n$$\nas well as\n$$\n\\angle ABP = 180^{\\circ} - \\angle AFP = \\ang... | South Africa | South African Mathematics Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
08ob | Problem:
Let $a$, $b$, $c$ be positive real numbers such that $a^{2} + b^{2} + c^{2} = 48$. Prove
$$
a^{2} \sqrt{2 b^{3} + 16} + b^{2} \sqrt{2 c^{3} + 16} + c^{2} \sqrt{2 a^{3} + 16} \leq 24^{2}
$$
When does equality hold? | [
"Solution:\nObserve that $2x^{2} + 16 = 2(x^{2} + 8) = 2(x + 2)(x^{3} = 2x + 4)$. From AM-GM:\n$$\n\\sqrt{2x^{3} + 16} = \\sqrt{(2x + 4)(x^{3} - 2x + 4)} \\leq \\frac{2x + 4 + x^{3} - 2x + 4}{2} = \\frac{x^{2} + 8}{2}\n$$\nBy adding the inequality (1) obtained for $x = a$, $x = b$ and $x = c$ it suffices to prove:\... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | Equality holds when a = b = c = 4. | |
0hac | All the diagonals of the lateral faces of an $n$-sided prism are colored in yellow and blue colors, so that single-colored diagonals do not have common points. Prove that the sum of the squares of lengths of all yellow diagonals is equal to the sum of the squares of lengths of the blue ones.
(Oleg Kryzhanovsky.) | [
"\n**Fig. 50**\nAccording to the condition, single-colored diagonals don't have common points. That's why without loss of generality let's consider that the following diagonals are colored in blue: $B_1A_2, B_2A_3, \\ldots, B_{n-1}A_n, B_nA_1$, the others are colored in yellow. As $\\overri... | Ukraine | The Problems of Ukrainian Authors | [
"Geometry > Solid Geometry > 3D Shapes",
"Algebra > Linear Algebra > Vectors"
] | English | proof only | null | |
0hsl | Problem:
Draw a rectangle. Connect the midpoints of the opposite sides to get 4 congruent rectangles. Connect the midpoints of the lower right rectangle for a total of 7 rectangles. Repeat this process infinitely. Let $n$ be the minimum number of colors we can assign to the rectangles so that no two rectangles sharing... | [
"Solution:\n\n$(3,4)$."
] | United States | null | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | final answer only | (3, 4) | |
065n | Let $x, y, z$ be positive real numbers. Prove that
$$
\sum_{cyclic} \frac{xy}{xy + x^2 + y^2} \le \sum_{cyclic} \frac{x}{2x + z}.
$$ | [
"Given inequality is equivalent to:\n$$\n\\frac{1}{1+\\frac{x}{y}+\\frac{y}{x}}+\\frac{1}{1+\\frac{y}{z}+\\frac{z}{y}}+\\frac{1}{1+\\frac{z}{x}+\\frac{x}{z}} \\le \\frac{1}{2+\\frac{z}{x}}+\\frac{1}{2+\\frac{x}{y}}+\\frac{1}{2+\\frac{y}{z}}.\n$$\nNow we can take substitution $\\frac{x}{y} = a$, $\\frac{y}{z} = b$, ... | Greece | Mediterranean Mathematical Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
04wc | An online vote is being held between options $A$ and $B$. Before Paul voted, the percentage of votes for option $A$ was a positive integer. Paul's vote increased this number by exactly one. Prove that Paul's vote was the nineteenth vote for option $A$. (Josef Tkadlec) | [
"Let the total number of votes before Paul voted be $n$, and the number of votes for $A$ before Paul voted be $k$. The percentage of votes for $A$ before Paul voted is $\\frac{100k}{n}$, which is a positive integer. After Paul votes for $A$, the number of votes for $A$ becomes $k+1$, and the total number of votes b... | Czech Republic | District Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 19 | |
0axc | Problem:
Define a sequence of integers as follows: $a_{1}=1$, $a_{2}=2$, and for $k \in \mathbb{N}$, $a_{k+2}=a_{k+1}+a_{k}$. How many different ways are there to write $2017$ as a sum of distinct elements of this sequence? | [
"Solution:\n\nNote that these $a_{k}$'s are in fact the Fibonacci numbers. Denote by $f(n)$ the number of distinct ways to express a number as a sum of $a_{k}$. Note that $2017=1597+377+34+8+1=a_{15}+a_{12}+a_{8}+a_{5}+a_{1}$.\n\nWe prove the following lemma:\n$$\na_{1}+a_{2}+\\cdots+a_{k}=a_{k+2}-2\n$$\nThis follo... | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 24 | |
0igl | Problem:
A triangular piece of paper of area $1$ is folded along a line parallel to one of the sides and pressed flat. What is the minimum possible area of the resulting figure? | [
"Solution:\nLet the triangle be denoted $ABC$, and suppose we fold parallel to $BC$. Let the distance from $A$ to $BC$ be $h$, and suppose we fold along a line at a distance of $c h$ from $A$. We will assume that neither angle $B$ nor $C$ is obtuse, for the area of overlap will only be smaller if either is obtuse.\... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 2/3 | |
0h20 | In triangle $ABC$ $AL$, $BM$, $CN$ are medians. Prove, that $\angle ANC = \angle ALB$ if and only if $\angle ABM = \angle LAC$. | [
"Lines $LN$ and $AC$ are parallel, hence $\\angle NLA = \\angle LAC$. We have to show the following implication (fig. 7):\n$$\n\\angle ANC = \\angle ALB \\Leftrightarrow \\angle ABM = \\angle NLA.\n$$\nLet $G$ be a centroid, then we have the following equivalences. $\\angle ANC = \\angle ALB \\Leftrightarrow \\angl... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"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"
] | English | proof only | null | |
06k7 | Suppose all of the $200$ integers lying in between (and including) $1$ and $200$ are written on a blackboard. Suppose we choose exactly $100$ of these numbers and circle each one of them. By the *score* of such a choice, we mean the square of the difference between the sum of the circled numbers and the sum of the non-... | [
"The average score is $670000$.\nLet $n = 100$, and let $S$ be any subset of $\\{1, 2, \\dots, 2n\\}$ such that $|S| = n$. There are $\\binom{2n}{n}$ such sets $S$. The score of $S$ is\n$$\n\\begin{aligned}\n\\left( \\sum_{a \\in S} a - \\sum_{b \\notin S} b \\right)^2 &= \\left( \\sum_{k=1}^{2n} k - 2 \\sum_{a \\i... | Hong Kong | HKG TST | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 670000 | |
046d | Given an integer $n \ge 2$. Find the minimum real number $\lambda$ such that for any real numbers $a_1, a_2, \dots, a_n$, and $b$, the following inequality holds:
$$
\lambda \sum_{i=1}^{n} \sqrt{|a_i - b|} + \sqrt{n \left| \sum_{i=1}^{n} a_i \right|} \ge \sum_{i=1}^{n} \sqrt{|a_i|}.
$$ | [
"We will prove that $\\lambda_n := \\frac{n-1+\\sqrt{n-1}}{\\sqrt{n}}$ is the desired minimum value. Let's first prove a few lemmas.\n\n**Lemma 1:** The sequence $\\{\\lambda_n\\}$ is strictly increasing.\n\n**Proof:** Note that\n$$\n\\lambda_{n+1}^2 - \\lambda_n^2 = \\frac{n+1+2n^{\\frac{5}{2}}-2(n+1)(n-1)^{\\frac... | China | China-TST-2023B | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | (n - 1 + sqrt(n - 1)) / sqrt(n) | |
00kz | Let $x, y, z$ be positive real numbers with $x + y + z \ge 3$. Prove that
$$
\frac{1}{x+y+z^2} + \frac{1}{y+z+x^2} + \frac{1}{z+x+y^2} \le 1
$$ | [
"By Cauchy's inequality, we have\n$$\n(x + y + z^2)(x + y + 1) \\ge (x + y + z)^2, \\qquad (6)\n$$\nhence\n$$\n\\frac{1}{x+y+z^2} \\le \\frac{x+y+1}{(x+y+z)^2}.\n$$\nThus it suffices to show that\n$$\n\\sum_{cyc} \\frac{x+y+1}{(x+y+z)^2} = \\frac{2(x+y+z)+3}{(x+y+z)^2} \\le 1.\n$$\nThis is equivalent to the inequal... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
014x | Problem:
Some $1 \times 2$ dominoes, each covering two adjacent unit squares, are placed on a board of size $n \times n$ so that no two of them touch (not even at a corner). Given that the total area covered by the dominoes is $2008$, find the least possible value of $n$. | [
"Solution:\n\nFollowing the pattern from the figure, we have space for\n$$\n6+18+30+\\ldots+150=\\frac{156 \\cdot 13}{2}=1014\n$$\ndominoes, giving the area $2028 > 2008$.\n\n\n\nThe square $76 \\times 76$ is not enough. If it was, consider the \"circumferences\" of the $1004$ dominoes of s... | Baltic Way | Baltic Way 2008 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 77 | |
02vz | Problem:
Encontre todas as soluções inteiras do sistema
$$
\left\{\begin{array}{l}
x z-2 y t=3 \\
x t+y z=1
\end{array}\right.
$$ | [
"Solution:\nUma boa estratégia será aplicar alguma manipulação algébrica, como somar as equações, multiplicá-las, somar um fator de correção, entre outras para obtermos alguma fatoração envolvendo esses números. Elevando ambas as equações ao quadrado, temos:\n$$\n\\left\\{\\begin{aligned}\nx^{2} z^{2}-4 x y z t+4 y... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | proof and answer | (3, 1, 1, 0), (-3, -1, -1, 0), (1, 0, 3, 1), (-1, 0, -3, -1) | |
01dy | Let $a_1, a_2, \dots, a_{100}$ be a permutation of numbers $1, 2, \dots, 100$. Denote by $N$ the number of different values of the sums
$$
\sum_{i=u}^{v} a_i, \quad \text{where} \quad 1 \le u \le v \le 100.
$$
Is it possible that $N \ge 2500$? | [
"Answer: yes.\nFor example consider a permutation $1, 100, 2, 99, 3, 98, \\ldots$ For odd $i$ we have $a_i + a_{i+1} = 101$. It is not difficult to check that if $u$ and $v$ have the same parity (and therefore the number of summands is odd) then for all choices of $u$ and $v = u + 2\\ell$ all the sums\n$$\n\\sum_{i... | Baltic Way | Baltic Way shortlist | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | yes | |
08sy | Suppose we denote by $n_{(10)}$ the decimal representation for a positive integer $n$. Suppose three distinct positive integers $a, b, c$ satisfy all of the following conditions:
* $c_{(10)}$ coincides with the number obtained by removing a 6 from $a_{(10)}$.
* $c_{(10)}$ coincides with the number obtained by removing ... | [
"We note that under the hypothesis the top digits of $a_{(10)}$ and $b_{(10)}$ must be different, since if they are the same, then we get $\\frac{a}{b} < 2$ which violates the condition that $a$ is a multiple of $b$. Therefore, we must have the top digit of $a_{(10)}$ or $b_{(10)}$ to be 6, but if the top digit of ... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 15384 | |
074e | Problem:
Let $a, b, c$ be positive real numbers such that $a^{3}+b^{3}=c^{3}$. Prove that
$$
a^{2}+b^{2}-c^{2}>6(c-a)(c-b)
$$ | [
"Solution:\nThe given inequality may be written in the form\n\n$$\n7 c^{2}-6(a+b) c-\\left(a^{2}+b^{2}-6 a b\\right)<0\n$$\n\nPutting $x=7 c^{2}$, $y=-6(a+b) c$, $z=-\\left(a^{2}+b^{2}-6 a b\\right)$, we have to prove that $x+y+z<0$. Observe that $x, y, z$ are not all equal ($x>0, y<0$). Using the identity\n\n$$\nx... | India | Indian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0had | It is known, that for some value $a$ the equality: $a^4 - \frac{1}{a^2} = 4$ is true. Is it possible, that number $x = a^4 + \frac{1}{a^2}$ is integer? | [
"If we add these two equalities, we have equality $4 + x = 2a^4$, if we subtract, $x - 4 = \\frac{2}{a^2}$. Then we have, that $(x+4)(x-4)^2 = \\frac{4}{a^4} \\cdot 2a^4 = 8$. As we are interested only in integer $x$, each of expressions $x+4$ and $x-4$ has to be integer. As these two numbers are of the same parity... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | No | |
0gj1 | Find all positive integers $n$ and sequences of integers $a_0, a_1, \dots, a_n$ such that $a_n \neq 0$ and
$$
f(a_{i-1}) = a_i
$$
for all $i = 1, 2, \dots, n$, where $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$.
試決定所有的正整數 $n$ 及整數數列 $a_0, a_1, \dots, a_n$ 滿足 $a_n \neq 0$ 且
$$
f(a_{i-1}) = a_i
$$
對於 $i = 1, ... | [
"$n = 1$ with $(a_0, a_1) = (2, -2)$; $n = 2$ with $(a_0, a_1, a_2) = (-1, 1, 3)$; or $n$ even with $a_0 = \\dots = a_n = -1$.\n\nIt is clear that $a_0 \\neq 0$ as otherwise $a_n = 0$. For any $k = 1, \\dots, n+1$, let $I_k$ be the convex hull of $0, a_0, \\dots, a_{k-1}$. We will define $a_{-1} = 0$ for convenienc... | Taiwan | IMO 3J, Mock Exam 2 | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Chinese; English | proof and answer | All solutions are: (1) n equals 1 with sequence (2, −2); (2) n equals 2 with sequence (−1, 1, 3); (3) n is even and all terms are −1. | |
08b4 | Problem:
Sia $ABC$ un triangolo, sia $K$ il piede della bisettrice relativa a $BC$ e sia $J$ il piede della trisettrice relativa a $BC$ più vicina al lato $AC$ (ossia $J$ è il punto su $BC$ tale che $3 \cdot \angle CAJ = \angle CAB$). Siano poi $C'$ e $B'$ due punti sulla retta $AJ$, dalla parte di $J$ rispetto ad $A$... | [
"Solution:\n\nPoniamo anzitutto $\\theta = \\angle CAJ$, cosicché $\\angle JAB = 2\\theta$ e $\\angle JAK = \\theta / 2$.\n\nSe il quadrilatero $ABB'C$ è ciclico, allora $\\angle CBB' = \\theta$, poiché insiste sullo stesso arco dell'angolo $\\angle CAB'$. D'altra parte, poiché $\\angle AB'B = \\angle ABB' = (180^\... | Italy | XXXI Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0cox | Given a circle $\omega$ with center $I$. Two lines touching $\omega$ at points $A$ and $B$ intersect at point $O$. A point $C$ is chosen on the smaller arc $AB$ so that $AC \neq CB$. Lines $AC$ and $OB$ intersect at point $D$, while lines $BC$ and $OA$ intersect at point $E$. Prove that the circumcenters of triangles $... | [
"Пусть $M$ — вторая точка пересечения описанных окружностей треугольников $ACE$ и $BCD$ (она есть, так как в случае касания прямые $AE$ и $BD$ были бы параллельны). Нам достаточно показать, что описанная окружность треугольника $OCI$ также проходит через точку $M$, так как в этом случае центры всех трех окружностей... | Russia | Final round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasi... | English; Russian | proof only | null | |
0dzb | Problem:
Na višini na stranico $AC$ enakokrakega trikotnika $ABC$ z vrhom $B$ izberemo točko $D$ tako, da je premica $AC$ tangenta na očrtano krožnico $\mathcal{K}$ trikotnika $ABD$. Naj bo $E$ taka točka na krožnici $\mathcal{K}$, da je tetiva $DE$ pravokotna na tetivo $AB$. Dokaži, da sta trikotnika $ABE$ in $ABC$ s... | [
"Solution:\n\nOznačimo s $T$ presečišče tetiv $DE$ in $AB$. Vemo, da je trikotnik $DTB$ pravokotni. Označimo $\\angle CAB = \\angle ACB = \\alpha$. Ker je $AC$ tangenta, je torej kot $\\angle CAB$ enak nepriležnemu kotu $\\angle AEB$ nad tetivo $AB$. Zato je $\\angle AEB = \\alpha$. Velja tudi, da je $\\angle ABD =... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0jle | Let $k$ be a positive integer. Two players $A$ and $B$ play a game on an infinite grid of regular hexagons. Initially all the grid cells are empty. Then the players alternately take turns with $A$ moving first. In his move, $A$ may choose two adjacent hexagons in the grid which are empty and place a counter in both of ... | [
"** **The answer is $k = 6$. First we show that $A$ cannot win for $k \\ge 6$. Color the grid in three colors so that no two adjacent spaces have the same color, and arbitrarily pick one color $C$. $B$ will play by always removing a counter from a space colored $C$ that $A$ just played. If there is no such counter,... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 6 | |
02ry | Problem:
Seja $a$ um número inteiro positivo tal que há exatamente 10 quadrados perfeitos maiores que $a$ e menores que $2a$.
a) Encontre o menor valor possível de $a$.
b) Encontre o maior valor possível de $a$. | [
"Solution:\n\na) Como existem exatamente dez quadrados perfeitos maiores do que $a$ e menores do que $2a$, então esses dez quadrados perfeitos têm que ser consecutivos. Logo deve existir um número inteiro positivo $x$ tal que\n$$\n(x-1)^2 \\leq a < x^2 < (x+1)^2 < \\cdots < (x+9)^2 < 2a \\leq (x+10)^2\n$$\nEm parti... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | min a = 481, max a = 684 | |
05dq | Problem:
Let $n$ be a positive integer. We have $n$ boxes where each box contains a nonnegative number of pebbles. In each move we are allowed to take two pebbles from a box we choose, throw away one of the pebbles and put the other pebble in another box we choose. An initial configuration of pebbles is called solvabl... | [
"Solution:\n\nNumber the boxes from $1$ through $n$ and denote a configuration by $x = (x_{1}, x_{2}, \\ldots, x_{n})$ where $x_{i}$ is the number of pebbles in the $i$th box. Let\n$$\nD(x) = \\sum_{i=1}^{n} \\left\\lfloor \\frac{x_{i} - 1}{2} \\right\\rfloor\n$$\nfor a configuration $x$. We can rewrite this in the... | European Girls' Mathematical Olympiad (EGMO) | European Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Exactly those configurations in which every box contains an even number of pebbles and the total number of pebbles is 2n − 2. | |
02bv | Problem:
Dado um quadrilátero convexo, se as quatro bissetrizes de seus ângulos formam um novo quadrilátero $H I J E$, calcule a soma dos ângulos opostos $\angle H I J + \angle J E H$.
 | [
"Solution:\n\n\nComo a soma dos ângulos internos de um quadrilátero é $360^\\circ$, temos:\n$$\n\\begin{aligned}\n\\alpha + \\beta & = 360^\\circ - \\angle I H E - \\angle I J E \\\\\n& = 360^\\circ - \\angle D H A - \\angle C J B \\\\\n& = 360^\\circ - \\left(180^\\circ - \\angle A D H - \... | Brazil | null | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 180° | |
06lw | For a positive integer $n$, let $d(n)$ be the number of positive divisors of $n$, and let $\varphi(n)$ be the number of positive integers not exceeding $n$ which are coprime to $n$. Does there exist a constant $C$ such that
$$ \frac{\varphi(d(n))}{d(\varphi(n))} \le C $$
for all $n \ge 1$? | [] | Hong Kong | Year 2021 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | Yes; C = 1. | |
01r7 | Find all integers $a$ and $b$ satisfying the equality
$$
3^a - 5^b = 2.
$$ | [
"(Solution of D. Babrou.) If $a \\le 3$ or $b \\le 2$, then we see that only the pairs $(a; b) = (1; 0)$, $(a; b) = (3; 2)$ satisfy the equation $3^a - 5^b = 2$.\n\nLet now $a \\ge 4$ and $b \\ge 3$. We rewrite the equation in the form $3^3(3^{a-3} - 1) = 5^2(5^{b-2} - 1)$. Setting $x = a - 3$, $y = b - 2$ ($x, y >... | Belarus | Selection and Training Session | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Number-Theoretic Functions > φ (Euler's to... | English | proof and answer | (a, b) = (1, 0) and (a, b) = (3, 2) | |
04ql | Let $ABC$ be an acute triangle with the perimeter of $2s$. We are given three pairwise disjoint circles with pairwise disjoint interiors with the centres $A$, $B$ and $C$, respectively. Prove that there exists a circle with the radius of $s$ which contains all the three circles. | [
"To simplify the formulations, we say that a point lies inside of the circle if it lies on that circle or in its interior. Assume we are given a circle $\\omega$ with the radius of $r$ and the centre $O$. A circle $\\omega'$ with the centre $O'$ contains the circle $\\omega$ if and only if its radius is at least $O... | Czech Republic | null | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0b0g | Problem:
Points $A$, $B$, $C$, and $D$ lie on a line $\ell$ in that order, with $AB = CD = 4$ and $BC = 8$. Circles $\Omega_1$, $\Omega_2$, and $\Omega_3$ with diameters $AB$, $BC$, and $CD$, respectively, are drawn. A line through $A$ and tangent to $\Omega_3$ intersects $\Omega_2$ at the two points $X$ and $Y$. Find... | [] | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 24√5/7 | |
0025 | El trapezio $ABCD$ de bases $AB$ y $CD$, y lados no paralelos $BC$ y $DA$, tiene $\angle A=90^\circ$, $AB=6$, $CD=3$ y $AD=4$. Sean $E$, $G$, $H$ los circuncentros de los triángulos $ABC$, $ACD$, $ABD$, respectivamente. Hallar el área del triángulo $EGH$.
**ACLARACIÓN:** El circuncentro de un triángulo es el punto de ... | [] | Argentina | XX Olimpiada Matemática Argentina | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | español | proof and answer | 27/32 | |
06qq | Let $ABC$ be a triangle with incenter $I$ and let $X$, $Y$ and $Z$ be the incenters of the triangles $BIC$, $CIA$ and $AIB$, respectively. Let the triangle $XYZ$ be equilateral. Prove that $ABC$ is equilateral too. | [
"$AZ$, $AI$ and $AY$ divide $\\angle BAC$ into four equal angles; denote them by $\\alpha$. In the same way we have four equal angles $\\beta$ at $B$ and four equal angles $\\gamma$ at $C$. Obviously $\\alpha + \\beta + \\gamma = \\frac{180^\\circ}{4} = 45^\\circ$; and $0^\\circ < \\alpha, \\beta, \\gamma < 45^\\ci... | IMO | IMO Problem Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triang... | English | proof only | null | |
0lep | For each integer $n \ge 2$, let $s(n)$ denote the sum of all positive integers that are at most $n$ and not relatively prime to $n$.
a) Prove that $s(n) = \frac{n}{2}(n+1-\varphi(n))$, where $\varphi(n)$ is the number of positive integers that are at most $n$ and are relatively prime to $n$.
b) Prove that there does no... | [
"a) Notice that if $k$ is a positive integer such that $\\gcd(k, n) = 1$ and $k < n$ then $n-k$ and $n$ are relatively prime. It follows that\n$$\n\\sum_{k \\le n, \\gcd(k,n)=1} k = \\sum_{k \\le n, \\gcd(k,n)=1} (n-k)\n$$\nLet $A = \\{k \\in \\mathbb{N} \\mid 1 \\le k \\le n, \\gcd(k, n) = 1 = \\{k_1, k_2, \\dots,... | Vietnam | Vietnamese MO 2021 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
08jq | Problem:
Fie $ABC$ un triunghi isoscel cu $AC = BC$, $M$ mijlocul segmentului $AC$ şi $\ell$ dreapta ce trece prin $C$ şi este perpendiculară pe $AB$. Cercul ce trece prin punctele $B$, $C$ şi $M$ intersectează dreapta $\ell$ în punctele $C$ şi $Q$. Să se afle raza cercului circumscris triunghiului $ABC$ în funcţie de... | [] | JBMO | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | R = m - h/2 | |
04i2 | Let $I$ be the incentre of the acute triangle $ABC$ and let $|AC| > |BC|$. The angle bisector and the altitude from vertex $C$ close an angle of $10^\circ$. If $\angle AIB = 120^\circ$, determine the angles of the triangle $ABC$. (Ilko Brnetić) | [] | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | A = 50°, B = 70°, C = 60° | |
03hr | Problem:
Suppose
$$
n(n+1) a_{n+1} = n(n-1) a_n - (n-2) a_{n-1}
$$
for every positive integer $n \geq 1$.
Given that $a_0 = 1$, $a_1 = 2$, find
$$
\frac{a_0}{a_1} + \frac{a_1}{a_2} + \frac{a_2}{a_3} + \cdots + \frac{a_{50}}{a_{51}}.
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2655/2 | |
0eau | Let $a$ and $b$ be two different real numbers. For which $x$ does the equality $\frac{x-a}{x-b} = \frac{x-b}{x-a}$ hold?
(A) $\frac{a-b}{2}$
(B) $\frac{a^2+b^2}{a+b}$
(C) $\frac{a^2+b^2}{2(a+b)}$
(D) $a+b$
(E) $\frac{a+b}{2}$ | [
"Multiplying the given equality by $(x-a)(x-b)$, we get $(x-a)^2 = (x-b)^2$. Squaring both sides and subtracting $x^2$ we get $-2a x + a^2 = -2b x + b^2$, which can be further rearranged into $2(b-a)x = (b-a)(b+a)$. Since $b-a \\neq 0$, we can divide both sides of the equality by $b-a$ to get $2x = b+a$. From here ... | Slovenia | National Math Olympiad in Slovenia | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | E | |
0ddu | Each of $N$ people have chosen some $5$ elements from a $23$-element set so that any two people share at most $3$ chosen elements. Does this mean that $N \le 2020$? Answer the same question with $25$ instead of $23$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | For 23: Yes (in fact N ≤ 1771). For 25: No (there exist families with more than 2020, e.g., at least 2126). | |
0gc8 | 給定一平面上 100 個半徑為 1 的圓, 使得任三個圓心所構成的三角形面積至多為 100. 試證: 存在一條直線至少與 10 個圓相交。 | [
"- 令 $S$ 為這 $n$ 個圓心所成集合。我們先證明：存在一條直線 $L$ 使得 $S$ 中的點投影在 $L$ 上會落在一個長度為 $\\sqrt{4n}$ 的區間內。\n\n證明：設 $A, B$ 為 $S$ 中相距最遠的兩個點，令其距離為 $d$。\n\n* 任取 $S$ 中異於 $A, B$ 得一點 $C$，因為 $|\\triangle ABC| \\le n$，故 $C$ 到直線 $AB$ 的距離至多為 $2n/d$。\n* 從而若 $L$ 垂直直線 $AB$ 於 $D$，則 $S$ 中任一點在 $L$ 上的投影點，將落於以 $D$ 為中心，長度為 $4n/d$ 的區間內。\n* 又因 $S$ 中兩點距離最大... | Taiwan | 二〇一八數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0bpn | Problem:
1) Să se rezolve sistemul:
$$
\left\{\begin{array}{l}
x + [y] + \{z\} = 1,2 \\
y + [z] + \{x\} = 2,3 \\
z + [x] + \{y\} = 3,5
\end{array} \quad \text{unde } x, y, z \in \mathbf{R}\right.
$$ | [] | Romania | OLIMPIADA NATIONALĂ DE MATEMATICĂ ETAPA LOCALĂ | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | x = 1, y = 0.3, z = 2.2 | |
0956 | Problem:
Un elev trebuia să înmulțească un număr de 3 cifre cu un număr de 4 cifre (numerele nu se încep cu 0). În loc să procedeze astfel, el doar a alipit numărul de 4 cifre la dreapta numărului de 3 cifre, obținând în rezultat un număr de 7 cifre, care este de $N$ ori ($N \in \mathbb{N}$) mai mare decât rezultatul ... | [
"Solution:\n\nFie $\\overline{abc}$ - numărul de 3 cifre și $\\overline{defg}$ - numărul de 4 cifre. Atunci numărul de 7 cifre are forma $\\overline{abcdefg}$. Din condiția problemei avem $\\overline{abcdefg} = N \\cdot \\overline{abc} \\cdot \\overline{defg}$, echivalent cu $\\overline{abc} \\cdot 10^{4} = \\overl... | Moldova | A 61-a OLIMPIADA DE MATEMATICA A REPUBLICII MOLDOVA | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) 2 is possible; 10 is impossible. b) 9 | |
039n | Given a $\triangle ABC$. A circle $k$ through $A$ and $B$ intersects the sides $AC$ and $BC$ at points $L$ and $N$, respectively. Let $M$ be the midpoint of the arc $LN$ lying in the triangle. Set $AM \cap BL = D$, $AM \cap BN = F$, $BM \cap AL = G$ and $BM \cap AN = E$. Prove that:
a) $DE\parallel FG$;
b) if $DEFG$ ... | [
"Let $AN \\cap BL = P$.\n\na) Since $\\angle LAM = \\angle MAN = \\angle LBM = \\angle MBN$, the quadrilaterals $ABED$ and $ABFG$ are cyclic. Then $\\angle AED = \\angle ABL = \\angle ANL$ and hence $DE\\parallel LN$. Analogously $FG\\parallel LN$.\n\nb) Since $\\frac{DP}{LP} = \\frac{DE}{LN} = \\frac{GF}{LN} = \\f... | Bulgaria | Spring Mathematical Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03fq | Let $A_0B_0C_0$ be a triangle. For a positive integer $n \ge 1$, we define $A_n$ on the segment $B_{n-1}C_{n-1}$ such that $B_{n-1}A_n : C_{n-1}A_n = 2 : 1$ and $B_n, C_n$ are defined cyclically in a similar manner. Show that there exists an unique point $P$ that lies in the interior of all triangles $A_nB_nC_n$. | [
"We have nested compact sets (closed triangles), so they have non empty intersection. We prove that they intersect in only one point. It's enough to prove that the three points $A_n, B_n, C_n$ converge to a common point $P$. Assume it's false. Then there exists some subsequences of $A_n, B_n, C_n$ (which for simpli... | Bulgaria | 4 Bulgarian National Olympiad - Regional Round | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06wk | Let $A B C D$ be a parallelogram such that $A C = B C$. A point $P$ is chosen on the extension of the segment $A B$ beyond $B$. The circumcircle of the triangle $A C D$ meets the segment $P D$ again at $Q$, and the circumcircle of the triangle $A P Q$ meets the segment $P C$ again at $R$. Prove that the lines $C D$, $A... | [
"Common remarks. The introductory steps presented here are used in all solutions below.\nSince $A C = B C = A D$, we have $\\angle A B C = \\angle B A C = \\angle A C D = \\angle A D C$. Since the quadrilaterals $A P R Q$ and $A Q C D$ are cyclic, we obtain\n$$\n\\angle C R A = 180^{\\circ} - \\angle A R P = 180^{\... | IMO | IMO 2021 Shortlisted Problems | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09p2 | Let us define a binary operation $\star$ for positive numbers $A$ and $B$ by
$$
A \star B = \frac{A}{AB + 1}.
$$
(1) Prove that $(A \star B) \star C = A \star (B + C)$.
(2) Find the value of the expression:
$$
((((\cdots((((1 \star 2) \star 3) \star 4) \cdots) \star 59) \star 60) \star 61).
$$ | [
"1.\nLet us compute $(A \\star B) \\star C$:\n\nFirst, $A \\star B = \\dfrac{A}{AB + 1}$.\n\nNow, $(A \\star B) \\star C = \\dfrac{A \\star B}{(A \\star B) C + 1}$.\n\nSubstitute $A \\star B$:\n$$\n(A \\star B) \\star C = \\frac{\\dfrac{A}{AB + 1}}{\\dfrac{A}{AB + 1} \\cdot C + 1}\n$$\n\nSimplify the denominator:\n... | Mongolia | MMO2025 Round 2 | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 1/1891 | |
0fje | Problem:
La función $g$ se define sobre los números naturales y satisface las condiciones
$$
\begin{aligned}
g(2) & =1 \\
g(2 n) & =g(n) \\
g(2 n+1) & =g(2 n)+1
\end{aligned}
$$
Sea $n$ un número natural tal que $1 \leq n \leq 2002$. Calcula el valor máximo $M$ de $g(n)$. Calcula también cuántos valores de $n$ satisfa... | [
"Solution:\n\nDado cualquier natural $n$, consideramos su representación binaria,\n$$\nn=a_{k} 2^{k}+a_{k-1} 2^{k-1}+\\cdots+a_{1} 2+a_{0}=a_{k} \\ldots a_{1} a_{0}\n$$\ndonde $a_{j}=0$ ó 1.\nProbaremos por inducción sobre $k$ que $g(n)=\\sum_{j=0}^{k} a_{j}$.\nPara $k=0$ es cierto: $g\\left(1_{(2)}\\right)=g(1)=1$... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | M = 10; number of n with g(n) = M: 5 | |
0lda | a) Let $(a_n)$ be a sequence defined by $a_n = \ln(2n^2 + 1) - \ln(n^2 + n + 1)$ for all positive integers $n$. Prove that there are finite values of $n$ such that $\{a_n\} < \frac{1}{2}$.
b) Let $(b_n)$ be a sequence defined by $b_n = \ln(2n^2 + 1) + \ln(n^2 + n + 1)$ for all positive integers $n$. Prove that there a... | [
"a.\nIt is obvious that $1 \\le \\frac{2n^2 + 1}{n^2 - n + 1} \\le 2$ for all $n \\in \\mathbb{Z}^+$. Hence, $0 \\le a_n \\le \\ln 2 < 1$ and $\\lfloor a_n \\rfloor = 0$. It follows that $\\{a_n\\} = a_n$ and\n$$\n\\lim_{n \\to \\infty} \\{a_n\\} = \\lim_{n \\to \\infty} a_n = \\lim_{n \\to \\infty} \\ln \\frac{2n^... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof only | null | |
0dcn | Consider equilateral triangle $ABC$ and suppose that there exist three distinct points $X, Y, Z$ lie inside triangle $ABC$ such that
i) $AX = BY = CZ$.
ii) The triplets of points $(A, X, Z), (B, Y, X), (C, Z, Y)$ are collinear in that order.
Prove that $XYZ$ is an equilateral triangle. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | English | proof only | null | |
0gs1 | Find the minimal possible value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ over all positive real numbers $a, b, c$ satisfying
$$
abc = 1, \quad a+b+c = 5 \text{ and}
$$
$$
(ab+2a+2b-9)(bc+2b+2c-9)(ca+2c+2a-9) \geq 0.
$$ | [
"Answer: 5.\nSince $abc = 1$ we find the minimal value of $ab+bc+ac = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}$. Note that\n$$\nab + 2a + 2b + 2c - 9 = \\frac{1}{c} + 2(5 - c) - 9 = \\frac{1}{c} - 2c + 1 = \\frac{1}{c}(2c + 1)(1 - c).\n$$\nThe similar formulas are held for $bc+2b+2c-9$ and $ca+2c+2a-9$. Therefore... | Turkey | Team Selection Test for EGMO 2019 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 5 | |
06df | Let $n \ge 3$ be an integer. In a conference there are $n$ mathematicians. Every pair of mathematicians communicate in one of the $n$ official languages of the conference. For any three different official languages, there exist three mathematicians who communicate with each other in these three languages. Determine all... | [
"$n$ can be any odd integer larger than $1$.\n\nWe use terminologies in graph theory. We need to colour all edges of the complete graph of $n$ vertices in $n$ colours $C_1, C_2, \\dots, C_n$ such that for any three distinct colours, there exists a triangle whose edges are of these three colours.\n\nNote that there ... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | All odd integers n ≥ 3 | |
0ath | Problem:
Two couples and a single person are seated at random in a row of five chairs. What is the probability that at least one person is not beside his/her partner? | [] | Philippines | Philippine Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | final answer only | 4/5 | |
0jx9 | Problem:
How many sequences of integers $(a_{1}, \ldots, a_{7})$ are there for which $-1 \leq a_{i} \leq 1$ for every $i$, and
$$
a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}+a_{4} a_{5}+a_{5} a_{6}+a_{6} a_{7}=4 ?
$$ | [
"Solution:\n\nFor $i=1,2, \\ldots, 6$, let $b_{i}=a_{i} a_{i+1}$. From the problem condition each of $b_{1}, b_{2}, \\ldots, b_{6}$ can only be $-1,0$, or $1$. Since the sum of these six numbers is $4$, either there are five $1$s and a $-1$ or there are four $1$s and two $0$s.\n\nIn the first case, there are $6$ wa... | United States | HMMT November 2017 | [
"Discrete Mathematics > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 38 | |
09a9 | Let $a$, $b$, $c$, $d > 0$. Prove the following equality:
$$
\sqrt{\left(a + \sqrt{\frac{bcd}{a}}\right)\left(b + \sqrt{\frac{acd}{b}}\right)\left(c + \sqrt{\frac{abd}{c}}\right)\left(d + \sqrt{\frac{abc}{d}}\right)} + 2\sqrt{abcd} \ge ab+bc+cd+da+ac+bd.
$$
(proposed by E. Enkhzaya) | [
"Let us consider $f(x) = (x^2 + a^2)(x^2 + b^2)(x^2 + c^2)(x^2 + d^2)$.\n$$\n\\begin{align*}\nf(x) &= \\prod_{cyc} (x + ai) \\prod_{cyc} (x - ai) = \\\\\n&= (((x^4 - x^2(ab+bc+cd+da+ac+bd)+abcd)) + i(x^3(a+b+c+d) - x(abc+bcd+cda+dab))) \\times \\\\\n&\\quad ((x^4 - x^2(ab+bc+cd+da+ac+bd)+abcd)) - i(x^3(a+b+c+d) - x... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0f0v | Problem:
One rat and two cats are placed on a chessboard. The rat is placed first and then the two cats choose positions on the border squares. The rat moves first. Then the cats and the rat move alternately. The rat can move one square to an adjacent square (but not diagonally). If it is on a border square, then it c... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | With two cats, the cats win. With three cats and the rat given an extra initial turn, the rat wins. | |
0a2v | Montasser makes a sequence of numbers. The first two numbers are $6$ and $15$. He always makes the next number in the sequence by dividing the last number by its predecessor and multiplying the result by $2$. Thus, the third number in the sequence is $\frac{15}{6} \cdot 2 = 5$ and the fourth number is $\frac{5}{15} \cd... | [] | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | MCQ | C | |
0c2q | Problem:
Pentru $k \in \mathbb{Z}$ definim polinomul $F_{k}=X^{4}+2(1-k) X^{2}+(1+k)^{2}$. Să se determine toate valorile $k \in \mathbb{Z}$, astfel încât $F_{k}$ să fie ireductibil peste $\mathbb{Z}$ si reductibil peste $\mathbb{Z}_{p}$ pentru orice $p$ prim. | [
"Solution:\n\nVom arăta că numerele care satisfac condiţia cerută sunt toate numerele $k \\in \\mathbb{Z}$ care nu sunt de forma $\\pm l^{2}$, cu $l \\in \\mathbb{Z}$.\n\nArătăm că $F_{k}$ este reductibil peste $\\mathbb{Z}$ dacă şi numai dacă $F_{k}$ se descompune ca produs de două polinoame monice de grad 2.\n\nÎ... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | All integers k that are not of the form ± a perfect square | |
0a9o | Problem:
The number $1$ is written on the blackboard. After that a sequence of numbers is created as follows: at each step each number $a$ on the blackboard is replaced by the numbers $a-1$ and $a+1$; if the number $0$ occurs, it is erased immediately; if a number occurs more than once, all its occurrences are left on ... | [
"Solution:\nLet $S$ be a set of different numbers, all of them less than $2^{n-1}$, and create two new sets as follows: $S_1$, consisting of all the numbers in $S$ except the smallest one, and $S_2$, with elements the smallest element of $S$ and all the numbers we get by adding $2^{n-1}$ to each number in $S$. Note... | Nordic Mathematical Olympiad | The 26th Nordic Mathematical Contest | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | binomial(n, floor(n/2)) | |
07r5 | Suppose $x$, $y$, $z$ are positive numbers that sum to $\pi$. Prove that
$$
\frac{\sin 2x + \sin 2y + \sin 2z}{\sin x + \sin y + \sin z} \le 1,
$$
with equality iff $x = y = z = \pi/3$. | [
"$$\n\\begin{align*}\n\\sin 2x + \\sin 2y + \\sin 2z &= \\sin 2x + \\sin 2y - \\sin 2(x + y) \\\\\n&= \\sin 2x(1 - \\cos 2y) + \\sin 2y(1 - \\cos 2x) \\\\\n&= 2 \\sin 2x \\sin^2 y + 2 \\sin 2y \\sin^2 x \\\\\n&= 4 \\sin x \\sin y (\\cos x \\sin y + \\cos y \\sin x) \\\\\n&= 4 \\sin x \\sin y \\sin z.\n\\end{align*}... | Ireland | Ireland_2017 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Co... | English | proof only | null | |
0a1n | Problem:
Vind alle priemgetallen $p$ waarvoor het natuurlijke getal
$$
3^{p}+4^{p}+5^{p}+9^{p}-98
$$
hoogstens 6 positieve delers heeft. | [
"Solution:\nAntwoord: de enige priemgetallen waarvoor dit geldt zijn $2$, $3$ en $5$.\n\nWe schrijven $f(p) = 3^{p} + 4^{p} + 5^{p} + 9^{p} - 98$. Dan berekenen we de priemfactorisaties $f(2) = 3 \\cdot 11$, $f(3) = 7 \\cdot 11^{2}$ en $f(5) = 7 \\cdot 9049$. Die hebben dus respectievelijk $4$, $6$ en $4$ delers. H... | Netherlands | IMO-selectietoets III | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 2, 3, 5 | |
04eq | Determine all positive integers $n$ such that the product of all positive divisors of $n$ is equal to $n^3$. Represent these numbers as a product of prime powers. | [
"Let $1 = d_1 < d_2 < \\dots < d_k = n$ be all positive divisors of the number $n$. We see that\n$$\nd_1 \\cdot d_k = d_2 \\cdot d_{k-1} = d_3 \\cdot d_{k-2} = n.\n$$\nFrom this we conclude that the product of all positive divisors of $n$ is equal to $n^3$ if and only if $n$ has exactly six divisors.\n\nPositive in... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | n = p^5 or n = p^2 q, where p and q are distinct primes | |
08g3 | Problem:
Per ogni intero positivo $n$, indichiamo con $s(n)$ la somma delle cifre di $n$ (nell'usuale rappresentazione in base 10). Così, per esempio, $s(8)=8, s(2023)=7, s(573)=15$.
a) Determinare se esistono due interi positivi distinti $a$ e $b$ tali che
$$
2023 \cdot a + s(a) = 2023 \cdot b + s(b).
$$
b) Determi... | [
"Solution:\n\na.\nEsistono: basta prendere, per esempio, $a = 10^{2024} - 1$ e $b = 10^{2024} + 8$.\nPer dimostrarlo, osserviamo che la scrittura in base 10 di $a$ è costituita da 2024 cifre 9 consecutive, per cui $s(a) = 9 \\cdot 2024$, mentre $b$ si scrive con una cifra 1 seguita da 2023 cifre 0 ed una cifra 8, p... | Italy | XXXIX Olimpiade Italiana di Matematica | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) Yes. For example, a = 10^2024 − 1 and b = 10^2024 + 8.
b) Yes. For example, a = 793 and b = 19000. | |
05fm | Problem:
Soit $ABC$ un triangle et $\omega$ son cercle inscrit dont les points de tangence sur $[BC]$, $[CA]$ et $[AB]$ sont respectivement $D$, $E$ et $F$. On note $\Omega$ le cercle tangent à $\omega$ et passant par $B$ et $C$ et $T$ le point de tangence de ces deux cercles. On note $X$ et $Y$ les milieux des segmen... | [
"Solution:\n\nPendant toute la preuve on va utiliser les résultats du lemme qui suit. On commence d'abord par une notation.\nSoit $ABC$ un triangle, on note $\\Omega$ le cercle circonscrit à $ABC$, on note également $\\omega$ le cercle tangent à $(AB)$, $(AC)$ et intérieurement à $\\Omega$, on dit que ce cercle est... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced C... | null | proof only | null | |
0fdq | Problem:
¿Podemos trazar $2003$ segmentos en el plano de forma que cada uno de ellos corte exactamente a otros tres? | [
"Solution:\n\nNo es posible:\nLlamamos $N$ al número de cortes. Si sumamos, desde $1$ hasta $2003$, el número de segmentos que se cortan con uno dado, cada corte lo contamos dos veces. Por tanto, obtenemos el número $2N$.\nSi la hipótesis del enunciado se cumple, se tiene $2003 \\cdot 3 = 2N$. Pero $2003 \\cdot 3$ ... | Spain | null | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0i8z | Problem:
The sequence $a_{1}, a_{2}, a_{3}, \ldots$ of real numbers satisfies the recurrence
$$
a_{n+1}=\frac{a_{n}^{2}-a_{n-1}+2 a_{n}}{a_{n-1}+1} .
$$
Given that $a_{1}=1$ and $a_{9}=7$, find $a_{5}$. | [
"Solution:\nLet $b_{n}=a_{n}+1$. Then the recurrence becomes $b_{n+1}-1=\\left(b_{n}^{2}-b_{n-1}\\right) / b_{n-1}=b_{n}^{2} / b_{n-1}-1$, so $b_{n+1}=b_{n}^{2} / b_{n-1}$. It follows that the sequence $\\left(b_{n}\\right)$ is a geometric progression, from which $b_{5}^{2}=b_{1} b_{9}=2 \\cdot 8=16 \\Rightarrow b_... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 3 | |
04w1 | Suppose that the sum of 74 real numbers lying in the interval $[4, 10]$ is 356. Find the maximal possible value of the sum of their squares. | [
"Denote the numbers $x_1, \\dots, x_{74}$. The assumption $4 \\le x_i \\le 10$ guarantees that $(x_i-4)(10-x_i) \\ge 0$ holds for all $i$. Expanding this to $x_i^2 \\le 14x_i - 40$ and summing over $i$ gives us an upper bound\n$$\nx_1^2 + x_2^2 + \\dots + x_{74}^2 \\le 14(x_1 + \\dots + x_{74}) - 40 \\cdot 74 = 14 ... | Czech Republic | Second Round of the 73rd Czech and Slovak Mathematical Olympiad (January 16th, 2024) | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 2024 | |
0g6q | 令 $a_1, a_2, \dots, a_n$ 為非負實數使得對任意正整數 $1 \le k \le n$,
$$
a_1 a_2 \cdots a_k \ge \frac{1}{(2k)!}
$$
試證:
$$
a_1 + a_2 + \cdots + a_n \ge \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}.
$$ | [
"將題意之左式改寫如下：\n$$\n\\begin{aligned}\na_1 + a_2 + \\cdots + a_n &= \\left(1 - \\frac{1}{2}\\right)(1 \\cdot 2a_1) + \\left(\\frac{1}{3} - \\frac{1}{4}\\right)(3 \\cdot 4a_2) \\\\\n&\\quad + \\cdots + \\left(\\frac{1}{2n-1} - \\frac{1}{2n}\\right)((2n-1) \\cdot 2na_n) \\\\\n&= \\left(1 - \\frac{1}{2} - \\frac{1}{3} + ... | Taiwan | 二〇一二數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation"
] | null | proof only | null | |
0jcc | Problem:
$M$ is an $8 \times 8$ matrix. For $1 \leq i \leq 8$, all entries in row $i$ are at least $i$, and all entries on column $i$ are at least $i$. What is the minimum possible sum of the entries of $M$? | [
"Solution:\n\nAnswer: $372$\n\nLet $s_{n}$ be the minimum possible sum for an $n \\times n$ matrix. Then, we note that increasing it by adding row $n+1$ and column $n+1$ gives $2n+1$ additional entries, each of which has minimal size at least $n+1$. Consequently, we obtain\n$$\ns_{n+1} = s_{n} + (2n+1)(n+1) = s_{n}... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 372 | |
0b5i | Prove that all positive integers, except the powers of $2$, can be written as the sum of (at least two) consecutive positive integers. | [
"All symbols in the sequel are denoting integer numbers. Let $n = 2^a b$, $a \\ge 0$, $b \\ge 1$, $b$ odd. We want to have $n = (m+1) + (m+2) + \\dots + (m+k)$, with $m \\ge 0$ and $k \\ge 2$, hence $k(2m+k+1) = 2^{a+1}b$.\n\nIf $b=1$, it follows $k=2^\\alpha$, with $1 \\le \\alpha \\le a+1$, but then $2m+k+1 > 1$ ... | Romania | Local Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
06nt | Given $\triangle ABC$ with $AB < AC$, let $AD$ be the bisector of $\angle BAC$ with $D$ on the side $BC$. Let $\Gamma$ be a circle passing through $A$ and $D$ which is tangent to $BC$ at $D$. Suppose $\Gamma$ cuts the side $AB$ again at $E \neq A$. The tangent to the circumcircle of $\triangle BDE$ at $D$ intersects $\... | [
"Let $Q \\neq A$ be the second intersection point of $\\Gamma$ and $AC$. As\n$$\n\\angle EDB = \\angle EAD = \\angle DAQ = \\angle DEQ,\n$$\nwe have $EQ // BC$. Also, we have $AF // BC$ since\n$$\n\\angle EAF = 180^\\circ - \\angle FDE = 180^\\circ - \\angle DBE.\n$$\n\nThis shows $EQ // AF$, and hence $AEQF$ is an... | Hong Kong | Hong Kong Team Selection Test 1 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fma | Problem:
Sean $a, b$ y $c$ tres números reales positivos cuyo producto es $1$. Demuestra que, si la suma de estos números es mayor que la suma de sus inversos, entonces exactamente uno de ellos es mayor que $1$. | [
"Solution:\n\nPuesto que $a b c = 1$ y $a + b + c > \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}$, tenemos que\n$$\n\\begin{aligned}\n(a-1)(b-1)(c-1) & = a b c - a b - b c - c a + a + b + c - 1 \\\\\n& = a + b + c - \\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) > 0\n\\end{aligned}\n$$\nLa desigualdad ante... | Spain | Spain | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0eqr | If the difference between two prime numbers is also prime, what is the smallest value of the sum of those two primes?
(A) 9
(B) 7
(C) 6
(D) 4
(E) 3 | [
"Looking at the difference of successive pairs of small primes, we note that $3 - 2 = 1$ which is not prime, $5 - 2 = 3$ and $5 - 3 = 2$, and then larger numbers differing by $2$ (or more). So the smallest sum comes from $5 + 2 = 7$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | MCQ | B | |
0bct | Prove that, in every scalene triangle,
$$
\frac{h_a - h_b}{r_b - r_a} + \frac{h_b - h_c}{r_c - r_b} + \frac{h_c - h_a}{r_a - r_c} \ge \frac{3}{2}.
$$ | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0ete | Find all polynomials $a(x), b(x), c(x), d(x)$ with real coefficients satisfying the simultaneous equations
$$
\begin{aligned}
b(x)c(x) + a(x)d(x) &= 0 \\
a(x)c(x) + (1 - x^2)b(x)d(x) &= x + 1
\end{aligned}
$$ | [
"We first show that it is not possible for all four polynomials to be non-zero. Suppose they are. Denote the leading coefficients of the polynomials $a(x), b(x), c(x), d(x)$ (which exist, because the polynomials are non-zero) by $A, B, C, D$, respectively. Then the first equation implies $BC = -AD$ and thus $ABCD =... | South Africa | The South African Mathematical Olympiad, Third Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English, Afrikaans | proof and answer | {(a(x), b(x), c(x), d(x)) : b(x) = 0, d(x) = 0, and a(x)c(x) = x + 1} = {(k, 0, (x+1)/k, 0) : k ∈ ℝ \ {0}} ∪ {((x+1)/k, 0, k, 0) : k ∈ ℝ \ {0}} | |
0ima | Problem:
The four sides of quadrilateral $ABCD$ are equal in length. Determine the perimeter of $ABCD$ given that it has area $120$ and $AC = 10$.
 | [
"Solution:\n\nLet $M$ be the midpoint of $AC$. Then triangles $AMB$, $BMC$, $CMD$, and $DMA$ are all right triangles having legs $5$ and $h$ for some $h$.\n\nThe area of $ABCD$ is $120$, but also $4 \\cdot \\left(\\frac{1}{2} \\cdot 5 \\cdot h\\right) = 10h$, so $h = 12$.\n\nThen $AB = BC = CD = DA = \\sqrt{12^2 + ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals"
] | null | proof and answer | 52 | |
0jts | Problem:
Let $V$ be a rectangular prism with integer side lengths. The largest face has area $240$ and the smallest face has area $48$. A third face has area $x$, where $x$ is not equal to $48$ or $240$. What is the sum of all possible values of $x$? | [
"Solution:\n\nLet the length, width, and height of the prism be $s_{1}, s_{2}, s_{3}$. Without loss of generality, assume that $s_{1} \\leq s_{2} \\leq s_{3}$. Then, we have that $s_{1} s_{2} = 48$ and $s_{2} s_{3} = 240$. Noting that $s_{1} \\leq s_{2}$, we must have $(s_{1}, s_{2}) = (1,48), (2,24), (3,16), (4,12... | United States | HMMT November 2016 | [
"Geometry > Solid Geometry > 3D Shapes",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 260 | |
0d3g | Tarik and Sultan are playing the following game. Tarik thinks of a number that is greater than $100$. Then Sultan is telling a number greater than $1$. If Tarik's number is divisible by Sultan's number, Sultan wins, otherwise Tarik subtracts Sultan's number from his number and Sultan tells his next number. Sultan is fo... | [
"Yes, Sultan has winning strategies. Here are two examples:\n\nFirst winning strategy. Sultan plays the following numbers in order: $2$, $3$, $4$, $6$, $8$, $20$, $24$ and he wins. Indeed, let $n \\geq 100$ be Tarik's number and assume that Sultan will not win before playing the last number $24$.\n- $n$ is not divi... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic"
] | English, Arabic | proof and answer | Yes | |
03sa | Assume that $\alpha, \beta, \gamma$ satisfy $0 < \alpha < \beta < \gamma < 2\pi$. If
$$
\cos(x + \alpha) + \cos(x + \beta) + \cos(x + \gamma) = 0
$$
for arbitrary $x \in \mathbb{R}$, then $\gamma - \alpha = \underline{\hspace{2cm}}$. | [
"Write $f(x) = \\cos(x+\\alpha) + \\cos(x+\\beta) + \\cos(x+\\gamma)$. Since $f(x) \\equiv 0$ for $x \\in \\mathbb{R}$,\n$$\nf(-\\alpha) = 0,\\quad f(-\\gamma) = 0 \\text{ and } f(-\\beta) = 0.\n$$\nThat is\n$$\n\\begin{align*}\n\\text{That is}\\quad & \\cos(\\beta - \\alpha) + \\cos(\\gamma - \\alpha) = -1, \\\\\n... | China | China Mathematical Competition (Jiangxi) | [
"Precalculus > Trigonometric functions"
] | English | proof and answer | 4π/3 | |
0dm6 | Let $ABC$ be a triangle. Point $D$ lies on side $BC$, such that the incircles of triangles $ABD$ and $ACD$ are congruent. Let $\Omega_B$ be the circle with diameter $AB$, and let $\Omega_C$ be the circle with diameter $AC$. Prove that line $AD$ is perpendicular to one of the common tangents to the circles $\Omega_B$ an... | [
"Denote $AH$ as the altitude of triangle $ABC$ then clearly $H \\in \\Omega_B, \\Omega_C$ so $AH$ is the common chord of the two circles. Let $EF$ be the common tangent near $A$ of the two circles with $E \\in \\Omega_B, F \\in \\Omega_C$. According to the familiar property, $AH$ passes through the midpoint $K$ of ... | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter... | null | proof only | null | |
07s7 | A point $P$, not the centre, lies inside a circle of radius $r$. A point $A$ lies on the circumference of the circle and $E$ is the mid-point of $AP$. Prove that there exists a circle of radius $r/2$ that contains $E$ for all possible positions of $A$, whereby $P$ remains fixed. | [
"Draw the diameter $HK$ that passes through $P$ and let $C$ and $D$ be the mid-points of $PK$ and $PH$, respectively.\n\n\n\nSince $E$ is the mid-point of $AP$, $EC$ is parallel to $AK$ and $ED$ is parallel to $AH$. This implies that $\\angle DEC$ is similar to $\\angle HAK$, and so $\\angl... | Ireland | Irish | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
04cn | Let $p$ and $q$ be real numbers. The graph of the function $f(x) = x^2 + px + q$ intersects with the coordinate axes in three different points $A$, $B$ and $C$. Prove that the circumcircle of triangle $ABC$ intersects with the $y$-axis at a point with ordinate $1$. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
057w | Anna, Anne and Anni seek for real solutions $(x, y)$ to the system of equations
$$
\begin{cases} 4x^3y - x^4 - 3x^2y^2 = 2021, \\ 4y^3x - y^4 - 3y^2x^2 = 2021. \end{cases}
$$
Anna claims that the system of equations has a solution. Anne claims that the system of equations has no solution but at least one of the two equ... | [
"The system does not have a solution since adding the equation gives $-(x - y)^4 = 4042$ whose l.h.s. is non-positive but r.h.s. is positive.\nWe show that the first equation $4x^3y - x^4 - 3x^2y^2 = 2021$ has solutions (the same could be done for the second equation by symmetry). Dividing the equation by $x^4$ and... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | Anne | |
0frr | En un grupo de 2022 estudiantes, algunos son amigos entre sí, y la amistad es siempre recíproca. Sabemos que cualquier subconjunto de esos estudiantes tiene la siguiente propiedad: siempre existe un estudiante del subconjunto que es amigo de, a lo sumo, 100 estudiantes del mismo.
a) Determina el menor entero positivo ... | [
"Para la primera parte, seguimos la siguiente estrategia: tomamos un estudiante con a lo sumo 100 amigos (que existe por hipótesis). Diremos que es el estudiante 1. Sacamos a ese estudiante y en el subconjunto resultante, existirá un estudiante con a lo sumo 100 amigos en dicho subconjunto; este será el estudiante ... | Spain | LVIII Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Spanish | proof and answer | a) N = 101. b) Maximum sum of degrees = 394300. | |
0fo3 | Prove that for every integer $S \ge 100$ there exists an integer $P$ for which the following story could hold true: The mathematician asks the shop owner: "How much are the table, the cabinet and the bookshelf?" The shop owner replies: "Each item costs a (positive) integer amount of Euros. The table is more expensive t... | [
"Write $S$ in the form $S = 6k + r$ for integers $k$ and $r$ with $1 \\le r \\le 6$, and note that $k > 2r$. We claim that the number\n$$\nP = 6k(k - r)(k + r)\n$$\nis an appropriate choice.\nDenote the prices of table, cabinet and shelf by $x$, $y$ and $z$, respectively. Then $x = 3(k+r)$, $y = 2(k-r)$, $z = k$ is... | Spain | Mediterranean Mathematical Competition | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Spanish | proof only | null | |
0itn | Problem:
On an infinite chessboard (whose squares are labeled by $(x, y)$, where $x$ and $y$ range over all integers), a king is placed at $(0,0)$. On each turn, it has probability of $0.1$ of moving to each of the four edge-neighboring squares, and a probability of $0.05$ of moving to each of the four diagonally-neig... | [
"Solution:\n\nSince only the parity of the coordinates are relevant, it is equivalent to consider a situation where the king moves $(1,0)$ with probability $0.2$, moves $(0,1)$ with probability $0.2$, moves $(1,1)$ with probability $0.2$, and stays put with probability $0.4$. This can be analyzed using the generati... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | 1/4 + 3/(4*5^2008) | |
0i4i | Problem:
An omino is a 1-by-1 square or a 1-by-2 horizontal rectangle. An omino tiling of a region of the plane is a way of covering it (and only it) by ominoes. How many omino tilings are there of a 2-by-10 horizontal rectangle? | [
"Solution:\n\nThere are exactly as many omino tilings of a 1-by-$n$ rectangle as there are domino tilings of a 2-by-$n$ rectangle. Since the rows don't interact at all, the number of omino tilings of an $m$-by-$n$ rectangle is the number of omino tilings of a 1-by-$n$ rectangle raised to the $m$th power, $F_{n}^{m}... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 7921 | |
0jsn | Problem:
Let $p(x) = x^{2} - x + 1$. Let $\alpha$ be a root of $p(p(p(p(x))))$. Find the value of
$$
(p(\alpha)-1)\, p(\alpha)\, p(p(\alpha))\, p(p(p(\alpha)))
$$ | [
"Solution:\nSince $(x-1)x = p(x)-1$, we can set\n$$\n\\begin{aligned}\n(p(\\alpha)-1)\\, p(\\alpha)\\, p(p(\\alpha))\\, p(p(p(\\alpha))) &= (p(p(\\alpha))-1)\\, p(p(\\alpha))\\, p(p(p(\\alpha))) \\\\\n&= (p(p(p(\\alpha)))-1)\\, p(p(p(\\alpha))) \\\\\n&= p(p(p(p(\\alpha))))-1 \\\\\n&= -1\n\\end{aligned}\n$$"
] | United States | HMMT November | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | -1 | |
08pj | Problem:
Let $x$, $y$ and $z$ be positive numbers. Prove that
$$
\frac{x}{\sqrt{\sqrt[4]{y}+\sqrt[4]{z}}}+\frac{y}{\sqrt{\sqrt[4]{z}+\sqrt[4]{x}}}+\frac{z}{\sqrt{\sqrt[4]{x}+\sqrt[4]{y}}} \geq \frac{\sqrt[4]{(\sqrt{x}+\sqrt{y}+\sqrt{z})^{7}}}{\sqrt{2 \sqrt{27}}}
$$ | [
"Solution:\nReplacing $x = a^{2}$, $y = b^{2}$, $z = c^{2}$, where $a$, $b$, $c$ are positive numbers, our inequality is equivalent to\n$$\n\\frac{a^{2}}{\\sqrt{\\sqrt{b}+\\sqrt{c}}}+\\frac{b^{2}}{\\sqrt{\\sqrt{c}+\\sqrt{a}}}+\\frac{c^{2}}{\\sqrt{\\sqrt{a}+\\sqrt{b}}} \\geq \\frac{\\sqrt[4]{(a+b+c)^{7}}}{\\sqrt{2 \... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0cpp | Let $O$ be the center of a rectangle $ABCD$. Point $K$ is chosen on the circumcircle of the rectangle. Line $CK$ intersects segment $AD$ at point $M$. Given that $AM : MD = 2$, prove that the meeting point of the medians of triangle $OKD$ lies on the circumcircle of triangle $COD$.
На окружности, описанной около прямо... | [
"Отметим на продолжении отрезка $AD$ такую точку $T$, что $AT = DM$. Тогда прямоугольные треугольники $CDM$ и $BAT$ равны, а значит, $BT \\parallel CM$. Заметим, что $DT = DA + AT = 3DM + DM = 4DM$. По теореме Фалеса, прямая $CM$ пересекает отрезок $BD$ в точке $N$ такой, что $DB = 4DN$. Значит, $DN = NO$, то есть ... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English, Russian | proof only | null |
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