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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0g4a | Problem:
There are 924 fans of the Liechtenstein football team from either Liechtenstein or Switzerland who have gathered to get the autographs of their favourite players. There are 11 players on the team, and every fan has exactly 6 favourite players. No two people from a given country share the same group of favouri... | [
"Solution:\n\nWe begin by observing that the number of possibilities for choosing a group of 6 favourite players is exactly\n$$\n\\binom{11}{6} = \\frac{11 \\cdot 10 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6}{6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1} = 462\n$$\nwhich is precisely half the number of fans. This impli... | Switzerland | Second round 2022 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0c8l | Michael is a good chess player. He took part in a competition where all the best chess players of the city were invited. The competition had two stages. After the first stage, looking at the partial results, Michael found that the number of players ranked higher than him was a half of the number of players ranked lower... | [
"If we denote by $x$ the number of players ranked higher than Michael after the first stage, it turns out that the number of players ranked lower than him will be $2x$.\n\nBecause in the second stage Michael surpassed 4 players that were before him and he was surpassed by 2 players that were after him, he \"advance... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 4th place | |
02mw | Problem:
Quantos são os pares de números inteiros positivos $(x, y)$ tais que
$$
\frac{x y}{x+y}=144 ?
$$ | [
"Solution:\n\nTemos a equação:\n$$\n\\frac{x y}{x+y} = 144\n$$\nMultiplicando ambos os lados por $x + y$:\n$$\nx y = 144(x + y)\n$$\nRearranjando:\n$$\nx y - 144x - 144y = 0\n$$\n$$\nx y - 144x - 144y + 144^2 = 144^2\n$$\n$$\n(x - 144)(y - 144) = 144^2\n$$\n\nAgora, como $x$ e $y$ são inteiros positivos, $x - 144$ ... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 45 | |
0a0r | In how many ways can you divide the numbers $1$ up to $10$ into pairs such that, for each pair, the largest number is at least twice the smallest number? | [
"$12$"
] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | final answer only | 12 | |
0a57 | Problem:
A school offers three subjects: Mathematics, Art and Science. At least $80\%$ of students study both Mathematics and Art. At least $80\%$ of students study both Mathematics and Science. Prove that at least $80\%$ of students who study both Art and Science, also study Mathematics. | [
"Solution:\n\nLet $n$ be the total number of students. Let $x$ be the number of students that study all three subjects. Let $a$ be the number of students that study Maths and Art but not Science. Let $b$ be the number of students that study Maths and Science but not Art. Let $c$ be the number of students that study... | New Zealand | NZMO Round One | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof only | null | |
0a4c | Problem:
Een groep van 4050 vrienden speelt een videospel-toernooi. Daarvoor staan er 2025 computers in een zaal gelabeld $a_1, \dots, a_{2025}$ en 2025 computers in een andere zaal gelabeld $b_1, \dots, b_{2025}$. De speler op computer $a_i$ speelt altijd tegen de spelers $b_i$, $b_{i+2}$, $b_{i+3}$ en $b_{i+4}$ (in ... | [
"Solution:\n\nVoor de tegenstandercomputers van $a_i$ bekijken we tegen welke $a_j$ zij spelen in de volgende tabel.\n\n| | | | | |\n|---------|---------|---------|---------|---------|\n| $b_i$ | $a_{i-4}$ | $a_{i-3}$ | $a_{i-2}$ | $a_i$ |\n\n| | |... | Netherlands | Maarttoets | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
09z6 | When you add the digits of the number $2022$, you get $6$.
How many $4$-digit numbers are there (including $2022$) such that, when you add the digits, you get $6$? The numbers are not allowed to start with the digit $0$.
A) $40$ B) $45$ C) $50$ D) $55$ E) $56$ | [] | Netherlands | First Round | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | MCQ | E | |
0e7r | Problem:
Meta je včeraj kupila 5 litrov mleka in 4 hlebce kruha ter plačala 9,10 evra. Danes je mleko cenejše za $10\%$, kruh pa za $15\%$. Za enako količino mleka in kruha bi danes Meta plačala 7,94 evra. Koliko sta stala liter mleka in hlebec kruha včeraj? | [] | Slovenia | Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | final answer only | Milk: 0.82 € per liter; Bread: 1.25 € per loaf | |
0k9g | Problem:
A point $P$ lies at the center of square $ABCD$. A sequence of points $\{P_n\}$ is determined by $P_0 = P$, and given point $P_i$, point $P_{i+1}$ is obtained by reflecting $P_i$ over one of the four lines $AB$, $BC$, $CD$, $DA$, chosen uniformly at random and independently for each $i$. What is the probabilit... | [
"Solution:\nWLOG, $AB$ and $CD$ are horizontal line segments and $BC$ and $DA$ are vertical. Then observe that we can consider the reflections over vertical lines separately from those over horizontal lines, as each reflection over a vertical line moves $P_i$ horizontally to point $P_{i+1}$, and vice versa. Now con... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 1225/16384 | |
0han | For natural number $n$ let $S(n)$ denote sum of its digits. From all pairs of natural numbers $(n, m)$, that satisfy equality $S(n) \cdot S(n+1) \cdots S(n+m) = 2018$, find such, for which sum $n+m$ takes the least possible value. | [
"Numbers $S(n)$ and $S(n+1)$ have the following ratios: $S(n+1) = S(n)+1$, or $S(n) > S(n+1)$, and the last inequality can not be true for two numbers in a row. As $2018 = 2018 \\cdot 1 = 1009 \\cdot 2 = 1009 \\cdot 1 \\cdot 2$, and there are no other factor decompositions, that satisfy the condition, so the follow... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | n = 1 followed by 112 nines; m = 1 | |
0jt5 | Problem:
Let $ABC$ be a triangle with $AB = 13$, $AC = 14$, and $BC = 15$. Let $G$ be the point on $AC$ such that the reflection of $BG$ over the angle bisector of $\angle B$ passes through the midpoint of $AC$. Let $Y$ be the midpoint of $GC$ and $X$ be a point on segment $AG$ such that $\frac{AX}{XG} = 3$. Construct ... | [
"Solution:\nObserve that $BG$ is the $B$-symmedian, and thus $\\frac{AG}{GC} = \\frac{c^2}{a^2}$. Stewart's theorem gives us\n$$\nBG = \\sqrt{\\frac{2a^2c^2b}{b(a^2 + c^2)} - \\frac{a^2b^2c^2}{a^2 + c^2}} = \\frac{ac}{a^2 + c^2} \\sqrt{2(a^2 + c^2) - b^2} = \\frac{390\\sqrt{37}}{197}\n$$\nThen by similar triangles,... | United States | HMMT February | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem"
] | null | proof and answer | 1170√37/1379 | |
0blw | Four families have two children each and all eight children are born after the year 1989. All four youngest siblings are born in the same year, and the sum of the digits of the year is equal to the product of the non zero digits. The differences of the ages between the siblings of each family is a perfect square. Find ... | [
"There exists no digit $a$ for which $1 + 9 + 9 + a = 1 \\cdot 9 \\cdot 9 \\cdot a$, hence the youngest are born after $2000$, say in $200a$ or $201a$. In the first case, the condition $2 + a = 2a$ gives $a = 2$. The second case leads to $a = 3$. The years are $2002$ or $2013$.\n\nThe age differences between the si... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 2012, 2009, 2004, 1997 | |
01dk | Prove that
$$
\sqrt{1 + \frac{1}{n^2} + \frac{1}{(n+1)^2}}
$$
is rational for every positive integer $n$. | [
"As\n$$\n\\sqrt{1 + \\frac{1}{n^2} + \\frac{1}{(n+1)^2}} = \\frac{\\sqrt{n^2(n+1)^2 + (n+1)^2 + n^2}}{n(n+1)}\n$$\nit suffices to show that $n^2(n + 1)^2 + (n + 1)^2 + n^2$ is a perfect square. This follows from\n$$\n\\begin{aligned}\nn^2(n+1)^2 + (n+1)^2 + n^2 &= n^2((n+1)^2 + 1) + (n+1)^2 \\\\\n&= n^4 + 2n^2(n+1)... | Baltic Way | Baltic Way 2016 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
0isk | Problem:
The function $f$ satisfies
$$
f(x)+f(2x+y)+5xy = f(3x-y)+2x^{2}+1
$$
for all real numbers $x, y$. Determine the value of $f(10)$. | [] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | -49 | |
0h6n | Vertices of a regular $(6n+3)$-gon, which are renumbered clockwise by numbers $1; 2; \ldots; 6n+3$, are the game field. Vertices with numbers $2n+1; 4n+2; 6n+3$ are called holes. At the start of the game there are $3$ chips on the field. $2$ players in turn choose any one of the $3$ chips and move it clockwise to a nei... | [
"Let's call a number of steps that each chip needs to reach the nearest hole, moving clockwise, if there are no other chips in the way, the distance to the hole. For example, for a chip that is placed on the vertex $4n$ the distance to the hole is $2$, if positions $4n+1$ and $4n+2$ are free from other chips; if it... | Ukraine | UkraineMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0gn7 | In a university entrance examination with $2006000$ student participants, each student makes a list of $12$ colleges from among a total of $2006$ colleges. It turns out that, for any $6$ students, there exist two colleges such that each of these $6$ students has included at least one of these colleges into his (her) li... | [] | Turkey | Team Selection Examination for the International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0fp1 | Con baldosas cuadradas de lado un número exacto de unidades se ha podido embaldosar una habitación de superficie $18144$ unidades cuadradas de la siguiente manera: el primer día se puso una baldosa, el segundo dos baldosas, el tercero tres, etc. ¿Cuántas baldosas fueron necesarias? | [
"Supongamos que fueron necesarias $n$ baldosas y que su tamaño es $k \\times k$. Entonces $nk^2 = 18144 = 2^5 \\times 3^4 \\times 7$. Hay nueve casos posibles para $n$, a saber, $2 \\times 7$, $2^3 \\times 7$, $2^5 \\times 7$, $2 \\times 3^2 \\times 7$, $2^3 \\times 3^2 \\times 7$, $2^5 \\times 3^2 \\times 7$, $2 \... | Spain | LII Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Spanish | proof and answer | 2016 | |
0d3n | Let $p$ be a prime number and $n \geq 2$ a positive integer, such that $p \mid (n^{6}-1)$. Prove that $n > \sqrt{p} - 1$. | [
"Because $p$ is prime and divides\n$$\nn^{6}-1 = (n-1)(n+1)(n^{2}-n+1)(n^{2}+n+1),\n$$\nit divides at least one of these positive factors. The prime number $p$ is therefore less or equal to at least one of these factors. Because\n$$\nn-1 < n+1 \\leq (n-1)n+1 = n^{2}-n+1 < n^{2}+n+1 < (n+1)^{2},\n$$\nwe have $p < (n... | Saudi Arabia | SAMC | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English, Arabic | proof only | null | |
0hjo | Problem:
A fair coin is flipped nine times. Which is more likely, having exactly four heads or having exactly five heads? | [
"Solution:\nBoth outcomes are equally likely! Notice that for every sequence with four heads (such as HTTHHTHTT) there is a corresponding sequence with exactly five heads formed by reversing all the coin flips (in this case $T H H T H T H H$). Thus the number of sequences of nine flips with exactly four heads is eq... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | Both are equally likely; each has probability 126/512. | |
036w | Problem:
A circle $k$ is tangent to the arms of an acute angle $A O B$ at points $A$ and $B$. Let $A D$ be the diameter of $k$ through $A$ and $B P \perp A D$, $P \in A D$. The line $O D$ meets $B P$ at point $M$. Find the ratio $\frac{B M}{B P}$. | [
"Solution:\nLet the tangent line to $k$ at $D$ meet the ray $\\overrightarrow{O B}$ at point $S$. Then the lines $S D$, $B P$ and $O A$ are parallel and therefore $\\triangle O B M \\sim \\triangle O S D$ and $\\triangle D P M \\sim \\triangle D A O$. It follows that $\\frac{B M}{S D} = \\frac{O B}{O S}$ and $\\fra... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 1/2 | |
083n | Problem:
Un cubetto di sughero di spigolo $10$ ha un peso attaccato ad un vertice e galleggia in un secchio d'acqua in modo che una diagonale del cubetto sia in posizione verticale. Quanto misura, in $\mathrm{cm}^2$, la superficie del cubetto a contatto con l'acqua, sapendo che il volume immerso è pari a $5$ volte il ... | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area"
] | null | proof and answer | 450 | |
0ee6 | Let $a$ be a real number for which the quadratic equations $x^2 + a x + 2 = 0$ and $x^2 + 2x + a = 0$ have real roots. The sum of the squares of the roots of the first equation equals the sum of the squares of the roots of the second equation. What is the value of $a$?
(A) $-4$
(B) $-2$
(C) $0$
(D) $4$
(E) None of the... | [
"The discriminant of the equation $x^2 + a x + 2 = 0$ is $D = a^2 - 8 \\ge 0$. Its roots $x_{1,2} = \\frac{-a \\pm \\sqrt{a^2-8}}{2}$ satisfy $x_1^2 + x_2^2 = a^2 - 4$.\n\nThe discriminant of the equation $x^2 + 2x + a = 0$ is $D = 4 - 4a \\ge 0$ and the roots $x_{1,2} = \\frac{-2 \\pm \\sqrt{4-4a}}{2}$ satisfy $x_... | Slovenia | Slovenija 2016 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | A | |
0fhx | Problem:
Un polígono convexo de $n$ lados se descompone en $m$ triángulos con interiores disjuntos, de modo que cada lado de esos $m$ triángulos, lo es también de otro triángulo contiguo o del polígono dado. Demostrar que $m+n$ es par. Conocidos $m$ y $n$, hallar el número de lados distintos que quedan en el interior ... | [
"Solution:\n\nComo hay $m$ triángulos, hay $3m$ lados; de ellos $3m-n$ son interiores, y como un lado interior pertenece a dos triángulos, hay $\\frac{3m-n}{2}$ lados interiores distintos. En particular $3m-n$ es par, luego $m$ y $n$ tienen la misma paridad y $m+n$ es par.\n\nSupongamos que el número de vértices $v... | Spain | OME 30 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | m + n is even; the number of distinct interior sides is (3m − n)/2; the number of distinct interior vertices is (m − n + 2)/2. | |
09e1 | A class has $n$ students. Every day class teacher chooses 3 students to form a team to study together. Each student works in a team with every other student not more than once. If $k$ is the number of maximum possible days then prove the inequality
$$
\frac{n(n-3)}{6} \le k \le \frac{n(n-1)}{6} \text{ holds.}
$$ | [
"Let's form from every triple $\\{a, b, c\\}$ a pair, namely $\\{a, b\\}$. By the given condition, different triples must correspond to different pairs. The pair $\\{a, b\\}$ represents pairs $\\{a, c\\}$, $\\{b, c\\}$. Since total number of pairs is $\\frac{n(n-1)}{2}$, number of representing pairs not greater tha... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof only | null | |
0dl2 | The monic polynomial $P(x)$ is called “nice” if its coefficients are in $\{-1, 0, 1\}$. Given a nice polynomial $P(x)$ of degree $2025$ and it is divisible by $x^7 - 1$, what is the maximum number of non-zero coefficients in $P(x)$? | [
"Let $f(x) = x^7 - 1$. We have $x^{7k+r} \\equiv x^r \\pmod{f(x)}$ for all $r = 0, 1, \\ldots, 6$ and $k$ is a positive integer. We will group together all exponents divided by $7$ with remainder $r$. We have\n$$\n\\begin{aligned}\nP(x) &= a_{2025}x^{2025} + \\cdots + a_2x^2 + a_1x + a_0 \\\\\n&\\equiv (a_0 + a_7 +... | Saudi Arabia | Saudi Booklet | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2022 | |
0anj | Problem:
Let $r = \log 50$ and $s = \log 80$. Express $7 \log 20$ in terms of $r$ and $s$.
(a) $2r + s$
(b) $2r + 3s$
(c) $r + 2s$
(d) $3r + 2s$ | [] | Philippines | Qualifying Round | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | MCQ | b | |
03pd | Let $M_n = \{0, a_1a_2\cdots a_n \mid a_i = 0 \text{ or } 1, 1 \le i \le n-1, a_n = 1\}$ be a set of decimal fractions, $T_n$ and $S_n$ be the number and the sum of the elements in $M_n$ respectively. Then
$$
\lim_{n \to \infty} \frac{S_n}{T_n} = \underline{\hspace{2cm}}.
$$ | [
"Since $a_1, a_2, \\dots, a_{n-1}$ all have exactly two possible values, so $T_n = 2^{n-1}$. Meanwhile, the frequency of $a_i = 1$ is the same as that of $a_i = 0$ for $1 \\le i \\le n-1$, and $a_n = 1$. Then\n$$\n\\begin{aligned}\nS_n &= \\frac{1}{2} \\times 2^{n-1} \\times \\left( \\frac{1}{10} + \\frac{1}{10^2} ... | China | China Mathematical Competition (Shaanxi) | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 1/18 | |
003q | Denotamos $29!$ al producto de los 29 enteros positivos desde 1 hasta 29, es decir,
$$29! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 \cdot 11 \cdot 12 \cdot 13 \cdot 14 \cdot 15 \cdot 16 \cdot 17 \cdot 18 \cdot 19 \cdot 20 \cdot 21 \cdot 22 \cdot 23 \cdot 24 \cdot 25 \cdot 26 \cdot 27 ... | [] | Argentina | Argentina 2006 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Español | final answer only | a=3, b=7, c=6, d=0 | |
0763 | Problem:
Let $a, b, c, x, y, z$ be positive real numbers such that $a + b + c = x + y + z$ and $a b c = x y z$. Further, suppose that $a \leq x < y < z \leq c$ and $a < b < c$. Prove that $a = x$, $b = y$ and $c = z$. | [
"Solution:\nLet\n$$\nf(t) = (t - x)(t - y)(t - z) - (t - a)(t - b)(t - c)\n$$\nThen $f(t) = k t$ for some constant $k$. Note that $k a = f(a) = (a - x)(a - y)(a - z) \\leq 0$ and hence $k \\leq 0$. Similarly, $k c = f(c) = (c - x)(c - y)(c - z) \\geq 0$ and hence $k \\geq 0$. Combining the two, it follows that $k =... | India | INMO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
06oh | Let $a, b, c$ be the sides of a triangle. Prove that
$$
\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}} \leq 3 .
$$ | [
"Note first that the denominators are all positive, e.g. $\\sqrt{a}+\\sqrt{b}>\\sqrt{a+b}>\\sqrt{c}$.\nLet $x=\\sqrt{b}+\\sqrt{c}-\\sqrt{a}$, $y=\\sqrt{c}+\\sqrt{a}-\\sqrt{b}$ and $z=\\sqrt{a}+\\sqrt{b}-\\sqrt{c}$. Then\n$b+c-a=\\left(\\frac{z+x}{2}\\right)^{2}+\\left(\\frac{x+y}{2}\\right)^{2}-\\left(\\frac{y+z}{2... | IMO | IMO 2006 Shortlisted Problems | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
04mv | Determine all positive integers $n$ for which the quadratic equation
$$
x^2 - 3nx + n + 3 = 0
$$
has integer solutions. | [] | Croatia | Croatia_2018 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | n = 2 | |
0ez0 | Problem:
Given positive numbers $a$, $b$, $c$, $d$ prove that at least one of the inequalities does not hold:
$a + b < c + d$ ;
$(a + b)(c + d) < ab + cd$ ;
$(a + b)cd < ab(c + d)$ . | [
"Solution:\nFrom the first and second inequalities we have $ab + cd > a(c + d) + b(a + b)$, so $cd > ad$, and hence $c > a$.\n\nWe also have $ab + cd > a(a + b) + b(c + d)$, so $cd > bc$, and hence $d > b$.\n\nSo $1/a + 1/b > 1/c + 1/d$, which contradicts the third inequality."
] | Soviet Union | 3rd ASU | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0l3p | Problem:
Let triangle $ABC$ have $AB = 5$, $BC = 8$, and $\angle ABC = 60^{\circ}$. A circle $\omega$ tangent to segments $\overline{AB}$ and $\overline{BC}$ intersects segment $\overline{CA}$ at points $X$ and $Y$ such that points $C, Y, X$, and $A$ lie along $\overline{CA}$ in this order. If $\omega$ is tangent to $\... | [
"Solution:\n\nLet $\\omega$ be tangent to $BC$ at $T$. Observe that $BT = BZ$ and $\\angle ABC = 60^{\\circ}$, so $\\triangle TBZ$ is equilateral. Moreover, the tangent to $\\omega$ at $T$ is parallel to $BC$, so $TY = TZ$. Combining this with $\\angle TZY = \\angle ZTB = 60^{\\circ}$, it f... | United States | HMMT November | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 40/(13*sqrt(3)) | |
0cbq | Let $n \in \mathbb{N}$, $n \ge 4$ and $a_1, a_2, \dots, a_n$ be real numbers such that
$$
a_k^3 = a_{k+1}^2 + a_{k+2}^2 + a_{k+3}^2
$$
for all $k \in \{1, 2, \dots, n\}$ – indices are considered modulo $n$. Show that $a_1 = a_2 = \dots = a_n$. | [
"Suppose there exists $i \\in \\{1, 2, \\dots, n\\}$ for which $a_i = 0$. Clearly $a_{i+1} = 0$, and consequently $a_1 = a_2 = \\dots = a_n = 0$.\n\nSuppose now that $a_i \\ne 0$ for all $i \\in \\{1, 2, \\dots, n\\}$. Choose $p, q \\in \\{1, 2, \\dots, n\\}$ such that $a_q \\le a_i \\le a_p$ for all $i \\in \\{1, ... | Romania | THE Sixteenth STARS OF MATHEMATICS Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
00b1 | A four digit number not ending with $0$ is written on the blackboard.
Carlos has multiplied the number on the blackboard by $4$, added $30$ to the result and written the obtained number in his notebook.
Dora has written in her notebook the number which is obtained when reading the digits of the number on the blackboard... | [
"Let $N$ be the number on the blackboard. Then, the number written by Carlos in his notebook is $C = 4N + 30$. Call $D$ the number written by Dora.\nSince $D = C \\ge N$ is a four digit number, the same holds for $C$; then, the first digit of $N$ is not greater than $2$, since, otherwise, $C \\ge 4 \\cdot 3000 = 12... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Other"
] | English | proof and answer | 2018 | |
05xj | Problem:
Soient $ABC$ un triangle et $O$ le centre de son cercle circonscrit. Soit $d$ la parallèle à $(BC)$ passant par $O$. Soit $A'$ le symétrique de $A$ par rapport à $(BC)$. La parallèle à $(A'B)$ passant par $C$ coupe $d$ en $C_1$, et les droites $(A'C)$ et $(BC_1)$ s'intersectent en $C_2$. La parallèle à $(A'C)... | [
"Solution:\n\n\n\nOn pose $P$ le point d'intersection des droites $(CB_1)$ et $(BC_1)$, et $Q$ le point d'intersection de $(BB_1)$ et $(CC_1)$. On pose également $T$ l'intersection de $(AO)$ et $(BC)$. On peut alors remarquer plusieurs propriétés sur la figure, qu'on va montrer.\n\n- **Lemm... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constru... | null | proof only | null | |
0gqi | $3m$ balls numbered $1, 1, 1, 2, 2, 2, 3, 3, 3, \dots, m, m, m$ are distributed into $8$ boxes so that any two boxes contain identical balls. Find the minimal possible value of $m$. | [
"If some box contains a ball numbered $i$ then at most two other boxes can also contain ball numbered $i$. Therefore, each box must contain at least $4$ balls and consequently the total number of balls is at least $32$. Thus, $m \\ge 11$. The example for $m = 11$: $(1,2,3,4)$, $(1,4,5,6)$, $(1,7,8,9)$, $(2,5,7,9)$,... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 11 | |
05wa | Problem:
Soit $\Gamma$ un cercle et $P$ un point à l'extérieur de $\Gamma$. Les tangentes à $\Gamma$ issues de $P$ touchent $\Gamma$ en $A$ et $B$. Soit $K$ un point distinct de $A$ et $B$ sur le segment $[AB]$. Le cercle circonscrit au triangle $PBK$ recoupe le cercle $\Gamma$ au point $T$. Soit $P'$ le symétrique du... | [
"Solution:\n\n\n\nD'après le théorème de l'angle inscrit, $\\widehat{\\mathrm{PBT}} = \\widehat{\\mathrm{PKT}}$. Ainsi, il suffit de montrer que $\\widehat{\\mathrm{PKT}} = \\widehat{\\mathrm{P}'\\mathrm{KA}}$. La figure semble suggérer que les triangles $PKT$ et $P'KA$ sont semblables, nou... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03ui | Given a unit cube $ABCD-A_1B_1C_1D_1$, construct a ball with point $A$ as the center and of radius $\frac{2\sqrt{3}}{3}$. Then the length of the curves resulting from the intersection between the surfaces of the ball and cube is ______. | [
"As shown in the figure, the surface of the ball intersects all of the six surfaces of the cube. The intersection curves are divided into two kinds: One kind lies on the three surfaces including vertex $A$ respectively, that is $AA_1B_1B$, $ABCD$, and $AA_1D_1D$; while the other lies on the three surfaces not inclu... | China | China Mathematical Competition | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | English | final answer only | 5*sqrt(3)*pi/6 | |
04ha | On a $5 \times 7$ board all squares are coloured white. Exactly $17$ squares need to be coloured black, so that the new arrangement of black and white squares is centrally symmetric, i.e. so that it does not change if it is rotated by $180^\circ$ around the centre of the board. In how many ways is it possible to do tha... | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof and answer | 24310 | |
0246 | Problem:
Seja $n$ um número inteiro positivo maior ou igual a 5. Para números $a_{i}$ escolhidos no conjunto $\{-1,1\}$, calcula-se o número
$$
S_{n}=a_{1} a_{2} a_{3} a_{4}+\ldots+a_{n} a_{1} a_{2} a_{3}
$$
que soma os produtos de cada quatro termos $a_{i}$ de índices consecutivos, inclusive os que começam em $a_{n-2... | [
"Solution:\n\na.\nCom os valores dados, tem-se:\n$$\nS_{8}=1 \\cdot 1 \\cdot 1 \\cdot 1+1 \\cdot 1 \\cdot 1 \\cdot 1+\\ldots+1 \\cdot 1 \\cdot 1 \\cdot 1=8\n$$\nÉ a soma de oito parcelas iguais a 1. Veja que ao trocar o $a_{4}$ de 1 para -1, os quatro produtos em que ele aparece mudam de sinal. Então a soma perde q... | Brazil | NÍVEL 3 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) 8; after flipping the fourth entry: 0; after also flipping the fifth entry: 4.
b) Four terms change; the possible changes in the sum are −8, −4, 0, 4, 8.
c) For any eight choices, the sum is always a multiple of 4.
d) The length n must be a multiple of 4. | |
03u6 | Let $f(x): f(x + 1) - f(x) = 2x + 1$ ($x \in \mathbb{R}$), and $|f(x)| \le 1$ when $x \in [0, 1]$. Prove:
$$
|f(x)| \le 2 + x^2 \quad (x \in \mathbb{R}).
$$ | [
"Let $g(x) = f(x) - x^2$, then\n$$\n\\begin{align*}\ng(x+1) - g(x) &= f(x+1) - f(x) - (x+1)^2 + x^2 \\\\\n&= (2x + 1) - (x^2 + 2x + 1 - x^2) \\\\\n&= 2x + 1 - (2x + 1) \\\\\n&= 0.\n\\end{align*}\n$$\nThus, $g(x)$ is a periodic function with 1 as its period. On the other hand, as is given $|f(x)| \\le 1$ when $x \\i... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof only | null | |
05v7 | Problem:
Soit $n$ un entier strictement positif et soient $a, a_{1}, \ldots, a_{n}$ des entiers strictement positifs. On suppose que pour tout entier $k$ pour lequel l'entier $a k+1$ est un carré parfait, au moins l'un des entiers $a_{1} k+1, \ldots, a_{n} k+1$ est également un carré parfait.
Montrer qu'il existe un ... | [
"Solution:\n\nCommençons par deux lemmes sur les résidus quadratiques.\n\nLemme 1 : Si $x$ est résidu quadratique modulo $p$ avec $p$ premier impair, il l'est aussi modulo $p^{2}$ (et même $p^{m}$, mais on n'en a pas besoin ici).\n\nDémonstration: En effet, soit $a$ tel que $a^{2} \\equiv x \\pmod{p}$, ce qu'on réé... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Other"
] | null | proof only | null | |
09o9 | Let $S(N)$ denote the sum of the digits of a positive integer $N$. Find all positive integers $n$ such that
$$
S(n) + S(2n) + S(3n) + \dots + S(n^2) = \frac{4n^2}{15} + 9.
$$
(Nursoltan Khavalbolot) | [] | Mongolia | MMO2025 Round 3 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | no solutions | |
0hcl | The convex polygon $M$ is given. There is a square $K$, that contains $M$ inside and has the minimum possible area. Is it obligatory for at least one of the sides of a square $K$ to contain one of the sides of the polygon $M$?
(Bogdan Rublyov) | [
"Let's consider the smallest tangential square, that is circumscribed around the equilateral triangle $\\triangle ABC$ (Fig. 46). And the side of the square contains the side of the triangle $AC$. It's clear, that it will be a square $ADEC$ and the edge $B$ is located inside the square. Then let's make a small turn... | Ukraine | Ukrainian Mathematical Competitions | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Tri... | English | proof and answer | No | |
094f | Problem:
Let $n$ and $m$ be positive integers. We call a set $S$ of positive integers $(n, m)$-good if it satisfies the following three conditions:
(i) We have $m \in S$.
(ii) For all $a \in S$, all of the positive divisors of $a$ are elements of $S$ too.
(iii) For all mutually different numbers $a, b \in S$, we have ... | [
"Solution:\n\nFor $m=1$ we have that $\\{1\\}$ is $(m, n)$-good. For the rest of the solution we assume $m \\geq 2$.\n\n- $n$ is odd\n\nLet $S$ be $(m, n)$-good set. Since $x+y \\mid x^{n}+y^{n}$, for $x, y \\in S$ with $x \\neq y$ we have $x+y \\in S$. Since $1 \\mid m$, it implies $1 \\in S$ and also $m+1 \\in S$... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0ak6 | A quadrilateral $ABCD$ is inscribed in a circle $k$, where $AB > CD$ and $AB$ is not parallel to $CD$. Point $M$ is the intersection of the diagonals $AC$ and $BD$ and the perpendicular from $M$ to $AB$ intersects the segment $AB$ at the point $E$. If $EM$ bisects the angle $CED$, prove that $AB$ is a diameter of the c... | [
"Solution:\nLet the line through $M$ parallel to $AB$ meet the segments $AD, DH, BC, CH$ at points $K, P, L, Q$ respectively. Triangle $HPQ$ is isosceles, so $MP = MQ$. Now from\n$$\n\\frac{MP}{BH} = \\frac{DM}{DB} = \\frac{KM}{AB} \\quad \\text{and} \\quad \\frac{MQ}{AH} = \\frac{CM}{CA} = \\frac{ML}{AB}\n$$\nwe o... | North Macedonia | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0hy6 | Problem:
Given that three roots of $f(x) = x^{4} + a x^{2} + b x + c$ are $2$, $-3$, and $5$, what is the value of $a + b + c$? | [
"Solution:\nBy definition, the coefficient of $x^{3}$ is negative the sum of the roots. In $f(x)$, the coefficient of $x^{3}$ is $0$. Thus the sum of the roots of $f(x)$ is $0$. Then the fourth root is $-4$. Then $f(x) = (x - 2)(x + 3)(x - 5)(x + 4)$. Notice that $f(1)$ is $1 + a + b + c$. Thus our answer is $f(1) ... | United States | HMMT 1998 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | 79 | |
072v | Problem:
a. Prove that if $n$ is a positive integer such that $n \geq 4011^{2}$, then there exists an integer $l$ such that $n < l^{2} < \left(1 + \frac{1}{2005}\right) n$.
b. Find the smallest positive integer $M$ for which whenever an integer $n$ is such that $n \geq M$, there exists an integer $l$, such that $n < ... | [
"Solution:\n\na. Let $n \\geq 4011^{2}$ and $m \\in \\mathbb{N}$ be such that $m^{2} \\leq n < (m+1)^{2}$. Then\n$$\n\\begin{aligned}\n\\left(1+\\frac{1}{2005}\\right) n - (m+1)^{2} & \\geq \\left(1+\\frac{1}{2005}\\right) m^{2} - (m+1)^{2} \\\\\n& = \\frac{m^{2}}{2005} - 2m - 1 \\\\\n& = \\frac{1}{2005}\\left(m^{2... | India | INMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 4010^2 + 1 | |
0dzh | Two players share a pile of coins, alternating in taking one coin off the pile and placing it on any empty square of a $2008 \times 2008$ chessboard. The player who puts the coin onto the board in such a way that together with three other coins already on the board it forms a non-rectangular isosceles trapezoid with ba... | [
"Let us call a non-rectangular isosceles trapezoid with the bases parallel to two of the edges of the board **regular**. We show that the second player can always win. After the first player's every move the second player should check to see if they can form a regular trapezoid by the placing of a single coin. If n... | Slovenia | Slovenija 2008 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | the second player | |
0ay3 | Problem:
Find the remainder when $30! - 1$ is divided by $930$. | [
"Solution:\nSince $31$ is prime, by Wilson's Theorem, we obtain $30! - 30 \\equiv 0 \\pmod{31}$. Because $30! - 30 \\equiv 0 \\pmod{30}$ and $\\gcd(30, 31) = 1$, we get $30! - 30 \\equiv 0 \\pmod{930}$. Therefore, $30! - 1 \\equiv 29 \\pmod{930}$."
] | Philippines | Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 29 | |
0ksy | Problem:
A jar contains 8 red balls and 2 blue balls. Every minute, a ball is randomly removed. The probability that there exists a time during this process where there are more blue balls than red balls in the jar can be expressed as $\frac{a}{b}$ for relatively prime integers $a$ and $b$. Compute $100 a+b$. | [
"Solution:\nOne can show that the condition in the problem is satisfied if and only if the last ball drawn is blue (which happens with probability $\\frac{1}{5}$), or the blue balls are drawn second-to-last and third-to-last (which happens with probability $\\frac{1}{\\binom{10}{2}} = \\frac{1}{45}$). Thus the tota... | United States | HMMT February | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 209 | |
03hc | Problem:
Prove that if $p$ and $p+2$ are both prime integers greater than $3$, then $6$ is a factor of $p+1$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
072c | Problem:
Let $M$ be the midpoint of side $BC$ of a triangle $ABC$. Let the median $AM$ intersect the incircle of $ABC$ at $K$ and $L$, $K$ being nearer to $A$ than $L$. If $AK = KL = LM$, prove that the sides of triangle $ABC$ are in the ratio $5 : 10 : 13$ in some order. | [
"Solution:\n\nLet $I$ be the incentre of triangle $ABC$ and $D$ be its projection on $BC$. Observe that $AB \\neq AC$ as $AB = AC$ implies that $D = L = M$. So assume that $AC > AB$. Let $N$ be the projection of $I$ on $KL$. Then the perpendicular $IN$ from $I$ to $KL$ is a bisector of $KL$ and as $AK = LM$, it is ... | India | INMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance cha... | null | proof and answer | 5:10:13 | |
0dys | Problem:
Dokaži: če za neničelna realna števila $a, b$ in $c$ velja
$$
a(b+c)+b(c+a)+c(a+b)=a b+b c+c a
$$
je vrednost izraza $\frac{a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b)}{a b c}$ celo število. | [
"Solution:\n\nIz dane enačbe sledi $a b+b c+c a=0$. Zato lahko zapišemo\n$$\n\\frac{a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b)}{a b c}=\\frac{a(a b+a c)+b(b a+b c)+c(c a+c b)}{a b c}\n$$\nUpoštevamo, da je $a b+a c=-b c$, $b a+b c=-c a$ in $c a+c b=-a b$ in dobimo\n$$\n\\frac{a(a b+a c)+b(b a+b c)+c(c a+c b)}{a b c}=\\frac{a... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0a7o | Problem:
We call a finite plane set $S$ consisting of points with integer coefficients a two-neighbour set, if for each point $(p, q)$ of $S$ exactly two of the points $(p+1, q)$, $(p, q+1)$, $(p-1, q)$, $(p, q-1)$ belong to $S$. For which integers $n$ does there exist a two-neighbour set which contains exactly $n$ po... | [
"Solution:\n\nThe points $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$ clearly form a two-neighbour set (which we abbreviate as $2\\{NS\\}$).\n\nFor every even number $n=2k \\geq 8$, the set $S=\\{(0,0), \\ldots, (k-2,0), (k-2,1), (k-2,2), \\ldots, (0,2), (0,1)\\}$ is a $2\\{NS\\}$.\n\nWe show that there is no $2\\{NS\\}$ wit... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 8 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All even integers except two and six; equivalently, four and every even integer at least eight. | |
0bf3 | Suppose $(A, +, \cdot)$ is a ring in which $x^2 = 0$ only for $x = 0$. Let $B = \{a \in A \mid a^2 = 1\}$. Show that:
a. $ab - ba = bab - a$, for all $a \in A$ and $b \in B$.
b. $(B, \cdot)$ is a group. | [
"a. Let $a \\in A$ and $b \\in B$. Since $((b-1)a(b+1))^2 = (b-1)a(b+1)(b-1)a(b+1) = (b-1)a(b^2-1)a(b+1) = 0$, we get $(b-1)a(b+1) = 0$, hence $ab - ba = bab - a$.\n\nb. Let $a$ and $b$ be two elements of $B$. Write\n$$\n\\begin{align*}\n(ab - ba)^2 &= abab - ab^2a - ba^2b + baba \\\\\n&= a(bab) + (bab)a - 2 \\\\\n... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0489 | Given a circle $\omega$ and two points $A, B$ outside $\omega$. A quadrilateral inscribed in $\omega$ is called "good" if one pair of opposite sides intersects at $A$ and the other pair intersects at $B$.
Assume that at least one good quadrilateral exists. Prove that there exists a good quadrilateral $\Gamma$ such that... | [
"Let $O$ be the center of $\\omega$. For any good quadrilateral $PQRS$ (where $PQ$ and $RS$ meet at $A$, and $PS$ and $QR$ meet at $B$), let the diagonals $PR$ and $QS$ intersect at $C$. By Brocard's theorem, $A, B, C, O$ form an orthocentric system, so $C$ is a fixed point. The existence of a good quadrilateral im... | China | 2025 International Mathematical Olympiad China National Team Selection Test | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Concurrency and Collinearity > Men... | English | proof only | null | |
021y | Problem:
Numa classe, $40\%$ dos alunos não enxergam bem. Desses, $70\%$ usam óculos e os $30\%$ restantes usam lentes de contato. Sabendo que 21 alunos usam óculos, quantos alunos tem essa classe? | [
"Solution:\n\nSeja $A$ o número total de alunos da sala. Logo, $\\frac{40}{100} \\times A$ não enxergam bem. Portanto, $\\frac{70}{100} \\times \\frac{40}{100} \\times A$ usam óculos. Consequentemente, temos que:\n$$\n\\frac{70}{100} \\times \\frac{40}{100} \\times A = 21 \\Rightarrow A = \\frac{21 \\times 100}{7 \... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 75 | |
0cma | Problem:
Xenia and Sergey play the following game. Xenia thinks of a positive integer $N$ not exceeding $5000$. Then she fixes $20$ distinct positive integers $a_{1}, a_{2}, \ldots, a_{20}$ such that, for each $k=1,2, \ldots, 20$, the numbers $N$ and $a_{k}$ are congruent modulo $k$. By a move, Sergey tells Xenia a se... | [
"Solution:\n\nSergey can determine Xenia's number in $2$ but not fewer moves.\n\nWe first show that $2$ moves are sufficient. Let Sergey provide the set $\\{17,18\\}$ on his first move, and the set $\\{18,19\\}$ on the second move. In Xenia's two responses, exactly one number occurs twice, namely, $a_{18}$. Thus, S... | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 2 | |
09xv | The integers $1$ to $n$ are written on the board. One of the numbers is wiped out. The average of the remaining numbers is $11\frac{1}{4}$.
Which number has been wiped out?
A) $6$ B) $7$ C) $11$ D) $12$ E) $21$ | [
"A) $6$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | A | |
0a47 | Problem:
We noemen een geheel getal $n \ge 3$ polypythagorees als er $n$ verschillende positieve getallen zijn die je een cirkel achter elkaar kan zetten zo dat de som van de kwadraten van elk paar opvolgende getallen een kwadraat is. Zo is $3$ een polypythagorees getal omdat je bijvoorbeeld met $44$, $117$ en $240$ e... | [
"Solution:\n\nWe bewijzen met inductie dat alle gehele getallen groter of gelijk aan $2$ polypythagorees zijn, waarbij we de definitie uitbreiden naar $n = 2$ op de logische manier. Als inductiebasis nemen we $(3, 4)$ voor $n = 2$ en $(44, 117, 240)$ uit het voorbeeld voor $n = 3$.\n\nStel nu dat $n$ polypythagoree... | Netherlands | IMO-selectietoets III | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All integers greater than or equal to 3 | |
06am | In a convex quadrilateral $ABCD$, its diagonals meet at $E$ and $H$, $G$ are the midpoints of the sides $AD$, $BC$, respectively. The circumcircles $c_1$ and $c_2$, of the triangles $AEB$ and $DEC$, respectively, meet at point $F \neq E$. If the parallel from $E$ to the line $HG$ meets the circle $c_2$ at $S$, prove th... | [
"In order to prove $FS \\parallel CD$, it suffices $FC = SD$, or $S\\hat{C}D = F\\hat{E}C$. From the cyclic $SEDC$ and $FS \\parallel CD$ we get\n$$\nS\\hat{C}D = B\\hat{E}S = B\\hat{I}G,\n$$\nwhere $I$ is the point of intersection of $HG$ and $BD$.\nIt suffices to prove that: $B\\hat{I}G = F\\hat{E}C$. Angle chasi... | Greece | Selection Examination | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ir9 | Problem:
In acute triangle $ABC$, the three altitudes meet at $H$. Given that $AH = BC$, calculate at least one of the angles of $\triangle ABC$.
Answer: $\angle A = 45^{\circ}$ (angles $B$ and $C$ cannot be determined). | [
"Solution:\n\nIn the diagram, $\\angle AFH = \\angle CFB$ (both are right angles) and $\\angle FAH = \\angle FCB$ (both are complementary to $\\angle ABC$). We get $\\triangle AFH \\cong \\triangle CFB$ by AAS. It follows that $AF = FC$, so $\\triangle AFC$ is right isosceles. We find that $\\angle BAC = 45^{\\circ... | United States | Berkeley Math Circle Monthly Contest 2 | [
"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 | Angle A = 45 degrees (angles B and C cannot be determined). | |
03ae | The positive integers $a > b > 1$ are such that the equation
$$
\frac{a^x - 1}{a - 1} = \frac{b^y - 1}{b - 1}
$$
has at least two distinct solutions in positive integers $x > 1$ and $y > 1$. Prove that $a$ and $b$ are co-prime. | [
"Assume that $a$ and $b$ are not co-prime and let the prime $p$ be their common divisor. We denote by $v_p(n)$ the exact degree of $p$ which divides $n$. Note that $(n, \\frac{n^\\ell-1}{n-1}) = 1$ for every positive integer $n > 1$ and $p$ does not divide $n-1$ if it divides $n$.\n\nLet $(x_1, y_1)$ and $(x_2, y_2... | Bulgaria | 58. National mathematical olympiad Final round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
07b4 | Suppose that $I$ is the incenter of triangle $ABC$. The perpendicular to line $AI$ from point $I$ intersects sides $AC$ and $AB$ in points $B'$ and $C'$, respectively. Points $B_1$ and $C_1$ are placed on half-lines $BC$ and $CB$, respectively, in such a way that $AB = BB_1$ and $AC = CC_1$. If $T$ is the second inters... | [
"Throughout the solution, let $\\omega$ and $\\omega_a$ be the circumcircle of triangle $ABC$ and the excircle of $ABC$ tangent to side $BC$, respectively. Furthermore, $O$ and $I_a$ are the centers of $\\omega$ and $\\omega_a$, respectively.\nUnder an inversion with center $A$ and power $AB \\times AC$, and then a... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"G... | English | proof only | null | |
0jig | Problem:
Find the remainder when $1^{2} + 3^{2} + 5^{2} + \cdots + 99^{2}$ is divided by $1000$. | [
"Solution:\nWe have $S = \\sum_{i=0}^{49} (2i+1)^{2} = \\sum_{i=0}^{49} \\left(4i^{2} + 4i + 1\\right) = 4 \\cdot \\sum_{i=0}^{49} i^{2} + 4 \\cdot \\sum_{i=0}^{49} i + 50$.\n\nNow,\n$\\sum_{i=0}^{49} i = \\frac{49 \\cdot 50}{2} = 1225$,\n$\\sum_{i=0}^{49} i^{2} = \\frac{49 \\cdot 50 \\cdot 99}{6} = 40425$.\n\nSo,\... | United States | HMMT November 2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 650 | |
05jq | Problem:
$\mathbb{N}^{*}$ désigne l'ensemble des entiers supérieurs ou égaux à un. Trouver toutes les applications $f: \mathbb{N}^{*} \rightarrow \mathbb{N}^{*}$ vérifiant :
i) quel que soit $n$ appartenant à $\mathbb{N}^{*}$, $f(n+f(n))=f(n)$
ii) $f(2013)=1$. | [
"Solution:\n\nSoit $f$ une solution éventuelle.\nSoit $a \\in \\mathbb{N}^{*}$ tel que $f(a)=1$.\nAlors $f(a+1)=f(a+f(a))=f(a)=1$.\nPuisque $f(2013)=1$, on en déduit par récurrence que $f(n)=1$ pour tout entier $n \\geqslant 2013$.\n\nD'autre part, supposons que $a \\geqslant 2$ soit un entier tel que $f(n)=1$ pour... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | The unique solution is the constant function f(n) = 1 for all positive integers n. | |
01i2 | Let points $A$ and $B$ lie on circle $\omega$ with center $O$. Assume that $O$ does not lie on line $AB$. Let point $C$ lie on segment $AB$ and denote by $M$ and $N$ the midpoints of segments $AC$ and $CB$, respectively. The circumcircle of $AON$ intersects $\omega$ at $A$ and $K$ and the circumcircle of $BOM$ intersec... | [
"**Solution.** Let $K'$ and $H'$ be the projections of $P'$ onto $AB$ and $AC$ respectively, as in figure 14. Now, $HPP'H'$ is a right angled trapezium, and $M$ is the midpoint of $PP'$. If $M'$ is the midpoint of $HH'$, then $M'M||HP$, so $M'M \\perp HH'$. Therefore, $MH = MH'$. Similarly, $MK = MK'$, and so as $M... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral si... | null | proof only | null | |
0cjo | Let $ABCD$ be a parallelogram and let $O$ be the intersection point of the diagonals. Prove that for any point $M \in (AB)$, there exist unique points $N \in (OC)$ and $P \in (OD)$ such that $O$ is the centroid of triangle $MNP$.
Nelu Chichirim | [
"A point $M \\in (AB)$ is uniquely determined by a real number $k \\in (0, \\infty)$ such that $\\overline{AM} = k$, from which we get\n$$\n\\overrightarrow{OM} = \\frac{1}{k+1}\\overrightarrow{OA} + \\frac{k}{k+1}\\overrightarrow{OB}.\n$$\nTo find the points $N$ and $P$ uniquely, we need to find $x, y \\in (0, \\i... | Romania | 75th Romanian Mathematical Olympiad | [
"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 > Vectors"
] | English | proof only | null | |
02z8 | Problem:
Começando com um número inteiro positivo $n$, uma sequência é criada satisfazendo a seguinte regra: cada termo se obtém do anterior subtraindo-se o maior quadrado perfeito que é menor ou igual ao termo anterior, até chegar ao número zero. Por exemplo, se $n=142$, teremos a seguinte sequência de 5 termos:
$$
a... | [
"Solution:\n\na) Um exemplo é a sequência\n$$\na_{1}=23, a_{2}=7=23-16, a_{3}=3=7-4, a_{4}=2=3-1, a_{5}=1=2-1, a_{6}=0=1-1\n$$\n\nb) Como $a_{n+1}=a_{n}-x^{2}$, $\\operatorname{com}~ x^{2} \\leq a_{n} < (x+1)^{2}$, segue que\n$$\n\\begin{aligned}\na_{n+1} & = a_{n} - x^{2} \\\\\n& < (x+1)^{2} - x^{2} \\\\\n& = 2x +... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Example with 6 terms: 23 → 7 → 3 → 2 → 1 → 0. Smallest n for 7 terms: 167. | |
0fix | Problem:
Prueba que la gráfica del polinomio $P$ es simétrica respecto del punto $A(a, b)$ sí y sólo sí existe un polinomio $Q$ tal que: $P(x)=b+(x-a) Q\left((x-a)^{2}\right)$, para todo $x \in \mathbb{R}$. | [
"Solution:\nSupongamos primero que exista el polinomio $P$ que cumple las condiciones requeridas. Sea $x-a=h$ o $x=a+h$. Entonces:\n$$\n\\left\\{\\begin{array}{l}\nP(a-h)=b-h Q\\left(h^{2}\\right) \\\\\nP(a+h)=b+h Q\\left(h^{2}\\right)\n\\end{array}\\right. \\quad \\text{y} \\quad \\frac{P(a-h)+P(a+h)}{2}=b \\text{... | Spain | XXXVII Olimpiada Española de Matemáticas | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
049t | Let $a$, $b$, $c$ be distinct positive integers and let $k$ be a positive integer such that
$$
ab + bc + ca \ge 3k^2 - 1.
$$
Prove that $\frac{1}{3}(a^3 + b^3 + c^3) \ge abc + 3k$. | [
"The desired inequality is equivalent to $a^3 + b^3 + c^3 - 3abc \\ge 9k$. We have\n$$\na^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca).\n$$\nIntegers $|a-b|$, $|b-c|$ and $|c-a|$ can not all be equal $1$ so\n$$\n\\begin{aligned}\na^2 + b^2 + c^2 - ab - bc - ca &= \\frac{1}{2}((a-b)^2 + (b-c)^2... | Croatia | CroatianCompetitions2011 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
091d | Problem:
For each integer $n \geq 2$, determine the largest real constant $C_n$ such that for all positive real numbers $a_1, \ldots, a_n$, we have
$$
\frac{a_1^2+\cdots+a_n^2}{n} \geq\left(\frac{a_1+\cdots+a_n}{n}\right)^2+C_n \cdot\left(a_1-a_n\right)^2
$$ | [
"Solution:\nDefine $x_{ij}=a_i-a_j$ for $1 \\leq i<j \\leq n$. After multiplication with $n^2$, the difference of the squares of quadratic and arithmetic mean equals\n$$\n\\begin{aligned}\nn^2\\left(\\mathrm{QM}^2-\\mathrm{AM}^2\\right) & =n\\left(a_1^2+\\cdots+a_n^2\\right)-\\left(a_1+\\cdots+a_n\\right)^2=(n-1) \... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | C_n = 1/(2n) | |
0bqi | Problem:
a) Fie $A$ o mulţime finită, nevidă şi $f: A \rightarrow A$ o funcţie injectivă. Arătaţi că $f$ este surjectivă.
b) Fie $f: \{1,2,3, \ldots, 63\} \rightarrow \{1,2,3, \ldots, 63\}$ o funcţie astfel încât $f(1)+f(2)+f(3)+\cdots+f(63)-1=2016$. Arătaţi că $f$ nu este injectivă. | [] | Romania | Olimpiada Națională de Matematică | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
00kq | Let $\triangle ABC$ be an acute-angled triangle with $AC < AB$ and circumradius $R$. Furthermore, let $D$ be the foot of the altitude from $A$ on $BC$ and let $T$ denote the point on the line $AD$ such that $AT = 2R$ holds with $D$ lying between $A$ and $T$. Finally, let $S$ denote the mid-point of the arc $BC$ on the ... | [
"As usual, we denote the angles $\\angle BAC$, $\\angle ABC$ and $\\angle BCA$ by $\\alpha$, $\\beta$ and $\\gamma$, respectively. The center of the circumcircle is denoted by $O$, see Figure 1.\n\nFigure 1: Problem 2\nBy assumption, we have $\\beta < \\gamma$. Let $E$ be the point on the c... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
08tl | Let $M$ be the mid-point of the side $BC$ of a triangle $ABC$. Suppose $AB = 4$ and $AM = 1$, where by $XY$ we denote the length of the line segment $XY$. Determine the smallest possible value the angle $\angle BAC$ can take. | [
"Let $N$ be the point on the line $AM$ which is symmetric to the point $A$ with respect to the point $M$. Then the triangle $ABMN$ is congruent to the triangle $CAM$, since $BM = CM$, $MN = MA$ and $\\angle BMN = \\angle CMA$. Therefore, we have\n$$\n\\angle BAC = \\angle BAM + \\angle CAM = \\angle BAM + \\angle B... | Japan | Japan Mathematical Olympiad First Round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geom... | null | proof and answer | 150° | |
0gzq | Nonnegative real numbers $a$, $b$, $c$ satisfy the equation $a + b + c = 1$. Prove the inequality
$$
(1 - a)^2 + (1 - b)^2 + (1 - c)^2 \geq 6\sqrt{abc}.
$$
Can the case of equality occur? | [
"Removing the brackets we will have $(1 - a)^2 + (1 - b)^2 + (1 - c)^2 = 3 + a^2 + b^2 + c^2 - 2(a + b + c) + 3 = a^2 + b^2 + c^2 + a + b + c \\geq 6\\sqrt{abc}$, equality is achieved when $a = b = c = a^2 = b^2 = c^2$, which is impossible, because $a + b + c = 1$."
] | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | No | |
09t4 | Problem:
De ingeschreven cirkel van een niet-gelijkbenige driehoek $\triangle ABC$ heeft middelpunt $I$ en raakt aan $BC$ en $CA$ in respectievelijk $D$ en $E$. Zij $H$ het hoogtepunt van $\triangle ABI$, zij $K$ het snijpunt van $AI$ en $BH$ en zij $L$ het snijpunt van $BI$ en $AH$. Bewijs dat de omgeschreven cirkels... | [
"Solution:\n\nOplossing I. We bekijken de configuratie in de figuur; andere configuraties gaan analoog. Er geldt $\\angle IDB=90^{\\circ}=\\angle IKB$, dus $BKDI$ is een koordenvierhoek. Verder is $\\angle ALB=90^{\\circ}=\\angle AKB$, dus ook $BKLA$ is een koordenvierhoek. Dus\n$$\n\\angle BKD=180^{\\circ}-\\angle... | Netherlands | IMO-selectietoets II | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0i7h | Problem:
Find the smallest value of $x$ such that $a \geq 14 \sqrt{a} - x$ for all nonnegative $a$. | [
"Solution:\nWe want to find the smallest value of $x$ such that $x \\geq 14 \\sqrt{a} - a$ for all $a$. This is just the maximum possible value of $14 \\sqrt{a} - a = 49 - (\\sqrt{a} - 7)^2$, which is clearly $49$, achieved when $a = 49$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 49 | |
078j | Let $N \ge 3$ be an integer, and $a_0, \dots, a_{N-1}$ be pairwise distinct reals so that $a_i \ge a_{2i}$ for all $i$ (indices are taken mod $N$). Find all possible $N$ for which this is possible. | [
"The only such $N$ are powers of $2$.\n\nIf $N$ is not a power of $2$, let's say $p \\nmid N$ where $p$ is an odd prime. Now, observe that $p \\nmid 1$ but $p \\mid 2^k - 1$ for some $k > 1$. Then, let $\\alpha = \\frac{N}{p}$. Now,\n$$\na_{\\alpha} > a_{2\\alpha} \\ge a_{4\\alpha} \\dots \\ge a_{2^k \\alpha} \\imp... | India | EGMO TST | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | N is a power of 2 | |
0bbv | For every $n \ge 3$, determine all the configurations of $n$ distinct points $X_1, X_2, \dots, X_n$ in the plane, with the property that for any pair of distinct points $X_i, X_j$ there exists a permutation $\sigma$ of the integers $\{1, \dots, n\}$, such that $d(X_i, X_k) = d(X_j, X_{\sigma(k)})$ for all $1 \le k \le ... | [
"Let us first prove that the points must be concyclic. Assign to each point $X_k$ the vector $x_k$ in a system of orthogonal coordinates whose origin is the point of mass of the configuration, thus $\\frac{1}{n}\\sum_{k=1}^{n}x_k = 0$.\nThen $d^2(X_i, X_k) = ||x_i - x_k||^2 = \\langle x_i - x_k, x_i - x_k \\rangle ... | Romania | 2011 Fourth ROMANIAN MASTER OF MATHEMATICS | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | All points lie on a common circle. If the number of points is odd, they form a regular polygon. If the number of points is even, they either form a regular polygon or an equiangular polygon whose side lengths alternate between two equal values. | |
0kvc | Problem:
Let $A X B Y$ be a cyclic quadrilateral, and let line $A B$ and line $X Y$ intersect at $C$. Suppose $A X \cdot A Y = 6$, $B X \cdot B Y = 5$, and $C X \cdot C Y = 4$. Compute $A B^{2}$. | [
"Solution:\nObserve that\n$$\n\\begin{aligned}\n& \\triangle A C X \\sim \\triangle Y C B \\Longrightarrow \\frac{A C}{A X} = \\frac{C Y}{B Y} \\\\\n& \\triangle A C Y \\sim \\triangle X C B \\Longrightarrow \\frac{A C}{A Y} = \\frac{C X}{B X}\n\\end{aligned}\n$$\nMultiplying these two equations together, we get th... | United States | HMMT February 2023 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 242/15 | |
0hb8 | Let $ABCD$ be a parallelogram. The circle through $A$ and $D$ intersects the lines $AB$, $BD$, $AC$ and $CD$ in points $B_1$, $B_2$, $C_1$ and $C_2$ respectively. Let $K$ be the intersection point of the lines $B_1B_2$ and $C_1C_2$. Prove that $K$ is equidistant from the lines $AB$ and $CD$.
... | [
"Let $O$ be the intersection point of the diagonals of the parallelogram. Let's prove that the points $C_1$, $O$, $B_2$ and $K$ lie on the same circle. In fact, with the cyclicity of points $A$, $B_1$, $B_2$, $C_1$, $C_2$ and $D$ and parallelism of $AB$ and $CD$ we have (Fig. 41):\n$$\n\\angle(KC_1, C_1O) = \\angle... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0eor | The sum of the squares of 3 consecutive positive integers is 770. The largest of these integers is
(A) 15 (B) 16 (C) 17 (D) 18 (E) 19 | [
"If we call the three numbers $n-1$, $n$, $n+1$, then $(n-1)^2 + n^2 + (n+1)^2 = 770$, which simplifies to $3n^2 + 2 = 770$. Therefore $n^2 = 768/3 = 256 = 16^2$, so $n = 16$ (since $n > 0$), and the largest number is $n+1 = 17$.\n\nWithout doing any algebra, it is clear that $n^2 \\approx 770/3 = 256\\frac{2}{3}$,... | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | C | |
0gg1 | 給定正整數 $n$。設 $a_1, a_2, \dots, a_n$ 為 $1, 2, \dots, n$ 的排列。試確定 $\sum_{i=1}^{n} \lfloor \frac{a_i}{i} \rfloor$ 的最小值。
註：$\lfloor x \rfloor$ 是不超過實數 $x$ 的最大整數。 | [
"Suppose that $2^k \\le n < 2^{k+1}$ with some non-negative integer $k$. First we show a permutation $(a_1, a_2, \\dots, a_n)$ such that $\\sum_{i=1}^n \\lfloor \\frac{a_i}{i} \\rfloor = k+1$; then we will prove that $\\sum_{i=1}^n \\lfloor \\frac{a_i}{i} \\rfloor \\ge k+1$ for every permutation. Hence, the minimal... | Taiwan | 2022 數學奧林匹亞競賽第一階段選訓營, 國際競賽實作(一) | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | Chinese; English | proof and answer | floor(log2 n) + 1 | |
07vf | Triana writes nine positive numbers, one in each cell of a $3 \times 3$ grid. The row sums and column sums are all equal. The row products and column products are also all equal to each other, but not necessarily equal to the row and column sums.
Show that each row and each column of Triana's matrix contains the same t... | [
"Clearly, one solution is\n$$\n\\begin{bmatrix} p & q & r \\\\ q & r & p \\\\ r & p & q \\end{bmatrix}\n$$\nor a $90^\\circ$ rotation of the same. Each row and column clearly contains the same three numbers $\\{p, q, r\\}$. We now show that this is the only solution. Without loss of generality (scaling all entries ... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0htp | Problem:
In convex hexagon $A X B Y C Z$, sides $A X$, $B Y$ and $C Z$ are parallel to diagonals $B C$, $X C$ and $X Y$, respectively. Prove that $\triangle A B C$ and $\triangle X Y Z$ have the same area. | [
"Solution:\n\nLet $[\\mathcal{P}]$ denote the area of a polygon $\\mathcal{P}$.\n\nThe important claim is that if $\\overline{K L} \\parallel \\overline{M N}$, then $[K L M] = [K L N]$. This is a simple consequence of the formula $A = \\frac{1}{2} b h$.\n\nThen, we find that\n$$\n\\begin{aligned}\n& [A B C] = [X B ... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ikp | Problem:
The train schedule in Hummut is hopelessly unreliable. Train $A$ will enter Intersection $X$ from the west at a random time between 9:00 am and 2:30 pm; each moment in that interval is equally likely. Train $B$ will enter the same intersection from the north at a random time between 9:30 am and 12:30 pm, inde... | [
"Solution:\n\nSuppose we fix the time at which Train $B$ arrives at Intersection $X$; then call the interval during which Train $A$ could arrive (given its schedule) and collide with Train $B$ the \"disaster window.\"\n\nWe consider two cases:\n\ni. Train $B$ enters Intersection $X$ between 9:30 and 9:45. If Train ... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 13/48 | |
01jn | Let $p$ be an odd prime. Let $a_1, a_2, \dots, a_{p-1}$ be integers such that $i \cdot a_i \equiv 1 \pmod{p}$, for all $i = 1, 2, \dots, p-1$. Prove that
$$
2^p - 2 \equiv p(a_1 - a_2 + \dots + a_{p-2} - a_{p-1}) \pmod{p^2}.
$$ | [
"From the binomial formula, we have\n$$\n2^p = C_p^0 + \\dots + C_p^p.\n$$\nSince $C_p^0 = C_p^p = 1$, we also have\n$$\n2^p - 2 = C_p^1 + \\dots + C_p^{p-1}.\n$$\nFor $1 \\le k \\le p-1$, we will find the value of $C_p^k$ modulo $p^2$. We know that\n$$\nC_p^k = \\frac{p!}{k!(p-k)!} = p \\cdot \\frac{(p-1)!}{k!(p-k... | Baltic Way | Baltic Way 2023 Shortlist | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
01nl | Given a polynomial $P(x)$ with positive real coefficients.
Prove that $P(1)P(xy) \ge P(x)P(y)$ for all $x \ge 1, y \ge 1$. | [
"We use weighted Chebyshev's inequality: if $a_1, \\dots, a_n > 0$, and $x_1 \\le x_2 \\le \\dots \\le x_n$, $y_1 \\le y_2 \\le \\dots \\le y_n$, then\n$$\n\\frac{(a_1x_1 + \\dots + a_nx_n)(a_1y_1 + \\dots + a_ny_n)}{a_1 + \\dots + a_n} \\le a_1x_1y_1 + \\dots + a_nx_ny_n. \\quad (*)\n$$\n\nNow let $p(x) = a_1 + a_... | Belarus | 62nd Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
0gvm | Do there exist integers $a, b, c, d, t, x, y, z$ such that each of expressions
$$
|ay - bx|, |az - cx|, |at - dx|, |bz - cy|, |bt - dy|, |ct - dz|
$$
would be valued either 1 or 2005? | [
"Помітимо, що серед чисел $x, y, z, t$ принаймні три є непарними. Аналогічно, непарними будуть щонайменше три з чотирьох чисел $a, b, c, d$. Отже, у принаймні двох з чотирьох пар $(a;x), (b;y), (c;z), (d;t)$ обидві компоненти є непарними числами. Але ж тоді один із заданих шести виразів має набувати парне значення.... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | No | |
0hpm | Problem:
Let $A$ and $B$ be two points on the plane with $AB = 7$. What is the set of points $P$ such that $PA^2 = PB^2 - 7$? | [
"Solution:\n\nIf we let $K$ be the point on $AB$ with $AK = 4$, $BK = 3$, then the answer is the line through $K$ perpendicular to $AB$. To see this, set $A = (0, 0)$ and $B = (7, 0)$. Then the points $P = (x, y)$ are exactly those satisfying\n$$\n(x - 0)^2 + (y - 0)^2 = (x - 7)^2 + (y - 0)^2 - 7\n$$\nwhich rearran... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof and answer | The line perpendicular to AB through the point on AB that is 3 units from A and 4 units from B (in coordinates with A at the origin and B at seven on the horizontal axis, this is the line x = 3). | |
0c1u | Given $ABC$ an acute triangle with $AB < AC$, let $AD$ be the altitude and $AE$ the angle bisector, where $D, E \in (BC)$. Given the acute triangle $A'B'C'$, with $A'B' < A'C'$, let $A'D'$ be the altitude and $A'E'$ the angle bisector, where $D', E' \in (B'C')$.
If $[AB] \equiv [A'B']$, $[AD] \equiv [A'D']$ and $[AE] \... | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
06v3 | Let $u_{1}, u_{2}, \ldots, u_{2019}$ be real numbers satisfying
$$
u_{1}+u_{2}+\cdots+u_{2019}=0 \quad \text{and} \quad u_{1}^{2}+u_{2}^{2}+\cdots+u_{2019}^{2}=1 .$$
Let $a=\min \left(u_{1}, u_{2}, \ldots, u_{2019}\right)$ and $b=\max \left(u_{1}, u_{2}, \ldots, u_{2019}\right)$. Prove that
$$
a b \leqslant-\frac{1}{20... | [
"Let $P=\\left\\{i: u_{i}>0\\right\\}$ and $N=\\left\\{i: u_{i} \\leqslant 0\\right\\}$ be the indices of positive and nonpositive elements in the sequence, and let $p=|P|$ and $n=|N|$ be the sizes of these sets; then $p+n=2019$. By the condition $\\sum_{i=1}^{2019} u_{i}=0$ we have $0=\\sum_{i=1}^{2019} u_{i}=\\su... | IMO | IMO 2019 Shortlisted Problems | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
02du | Find all solutions in positive integers to $(n+1)^k - 1 = n!$. | [
"The solutions are $n = 1, k = 1$; $n = 2, k = 1$; $n = 4, k = 2$.\nIt is easy to check that the solutions above are the only solutions for $n \\le 4$. So assume $n > 4$. So $n! + 1 > n + 1$, so $k > 1$. If $n$ is odd, then $n + 1$ is even, but $n! + 1$ is odd, so there are no solutions. So $n$ is even. Hence $n$ i... | Brazil | VI OBM | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (n, k) = (1, 1), (2, 1), (4, 2) | |
0lbt | A sequence of $n$ real numbers ($n \ge 2$) is called *nearly increasing* if it contains an increasing subsequence of $n-1$ terms. How many nearly increasing sequences are there in the set of all permutations of $\{1,2,3,...,n\}$? | [] | Vietnam | Vietnamese Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2^n - n | |
01r1 | Points $X$, $Y$ and $Z$ are marked on the sides $AD$, $AB$ and $BC$ of the rectangular $ABCD$, respectively.
Given $AX = CZ$ prove that $XY + YZ \ge AC$. | [
"Let $ZM$ be parallel to $AC$ (see the figure).\n\nIt is evident that $MZCA$ is a parallelogram, so $MZ = AC$ and $MA = CZ$. By condition, $AX = CZ$, so $AX = MA$. Therefore, the altitude $YA$ of the triangle $MYX$ is a median, hence this triangle is an isosceles triangle and $MY = YX$.\n\nSince the length of the s... | Belarus | Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0blh | Find all the functions $f : \mathbb{R} \to \mathbb{R}$, which have primitives and there exists a primitive $F$ of $f$, such that $F \circ F$ is also a primitive of $f$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof and answer | f(x) ≡ 0 or f(x) ≡ 1 | |
0k8k | Problem:
To celebrate 2019, Faraz gets four sandwiches shaped in the digits $2$, $0$, $1$, and $9$ at lunch. However, the four digits get reordered (but not flipped or rotated) on his plate and he notices that they form a 4-digit multiple of $7$. What is the greatest possible number that could have been formed?
![](a... | [
"Solution:\n\nNote that $2$ and $9$ are equivalent mod $7$. So we will replace the $9$ with a $2$ for now. Since $7$ is a divisor of $21$, a four-digit multiple of $7$ consisting of $2, 0, 1$, and $2$ cannot have a $2$ followed by a $1$ (otherwise we could subtract a multiple of $21$ to obtain a number of the form ... | United States | HMMT November 2019 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 1092 |
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