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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
05h8 | Problem:
Soit $A$, $B$, $C$ et $D$ quatre points sur un cercle dans cet ordre. Soit $U$ le point d'intersection des droites $(AB)$ et $(CD)$, et $V$ le point d'intersection des droites $(BC)$ et $(DA)$. Soit $K$ le point d'intersection de la bissectrice issue de $U$ dans le triangle $AUC$ et de la bissectrice issue de... | [
"Solution:\n\nLe point $L$ est le centre du cercle circonscrit au triangle $UKV$. Pour montrer que les points $U$, $L$ et $V$ sont alignés, il suffit de montrer que $\\widehat{ULV} = 180^{\\circ}$. Par le théorème de l'angle au centre, $\\widehat{UKV} = \\frac{1}{2} \\widehat{ULV}$. Il suffit donc de montrer que $\... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"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"
] | null | proof only | null | |
08rg | Answer the maximum value of $A$ for which, for every positive $x_1, x_2, x_3, y_1, y_2, y_3, z_1, z_2$ and $z_3$, the inequality
$$
(x_1^3 + x_2^3 + x_3^3 + 1)(y_1^3 + y_2^3 + y_3^3 + 1)(z_1^3 + z_2^3 + z_3^3 + 1)
\geq A(x_1 + y_1 + z_1)(x_2 + y_2 + z_2)(x_3 + y_3 + z_3)
$$
holds.
For the maximum value of $A$, establi... | [
"First we prove that, for any positive real numbers $p_1, \\dots, p_n, q_1, \\dots, q_n, r_1, \\dots, r_n$, the following inequality holds:\n$$\n(p_1^3 + \\cdots + p_n^3)(q_1^3 + \\cdots + q_n^3)(r_1^3 + \\cdots + r_n^3) \\ge (p_1q_1r_1 + \\cdots + p_nq_nr_n)^3.\n$$\nIndeed, by using Cauchy-Schwarz inequality repea... | Japan | The 16th Japanese Mathematical Olympiad - The Final Round | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | A = 3/4; equality iff x1 = x2 = x3 = y1 = y2 = y3 = z1 = z2 = z3 = 6^(-1/3). | |
04zs | $$
\frac{a^2 + bc}{b+c} + \frac{b^2 + ca}{c+a} + \frac{c^2 + ab}{a+b} \geq a+b+c
$$
for all positive real numbers $a$, $b$, $c$. | [
"W.l.o.g., assume $a \\ge b \\ge c$. Then\n$$\n\\begin{aligned}\n\\frac{a^2 + bc}{b+c} + \\frac{b^2 + ca}{c+a} + \\frac{c^2 + ab}{a+b} &= \\\\\n&= \\frac{a^2 + (b+c)c - c^2}{b+c} + \\frac{b^2 + (c+a)a - a^2}{c+a} + \\frac{c^2 + (a+b)b - b^2}{a+b} = \\\\\n&= \\frac{a^2 - c^2}{b+c} + c + \\frac{b^2 - a^2}{c+a} + a + ... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0595 | Juku has four cans of juice: a $1$-litre can containing $\frac{1}{2}$ litres of juice, a $\frac{1}{2}$-litre can containing $\frac{1}{3}$ litres of juice, a $\frac{1}{3}$-litre can containing $\frac{1}{4}$ litres of juice and a $\frac{1}{4}$-litre can containing $\frac{1}{5}$ litres of juice. There are no volume markin... | [
"*Answer:* The second, the third, the fourth.\n\nThe second can contains $\\frac{1}{6}$ litres of free space. Pouring juice from the fourth can over to the second can until the second can becomes full leaves $\\frac{1}{30}$ litres of juice in the fourth can. If, after that, one pours all juice from either the secon... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | The second, the third, and the fourth cans. | |
0axx | Problem:
A geometric sequence has a nonzero first term, distinct terms, and a positive common ratio. If the second, fourth, and fifth terms form an arithmetic sequence, find the common ratio of the geometric sequence. | [
"Solution:\n\nLet $a_{1}$ be the first term and $r$ the common ratio of the geometric sequence. Since the second, fourth, and fifth terms form an arithmetic sequence,\n$$\n\\begin{aligned}\na_{1} r^{3} - a_{1} r &= a_{1} r^{4} - a_{1} r^{3} \\\\\nr^{3} - r &= r^{4} - r^{3} \\\\\n0 &= r^{4} - 2 r^{3} + r \\\\\n0 &= ... | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (1+sqrt(5))/2 | |
09wz | Problem:
Voor een positief geheel getal $n$ bekijken we een $n \times n$-bord en tegels met afmetingen $1 \times 1, 1 \times 2, \ldots, 1 \times n$. Op hoeveel manieren kunnen er precies $\frac{1}{2} n(n+1)$ vakjes van het bord rood worden gekleurd, zodat de rode vakjes allemaal bedekt kunnen worden door de $n$ tegels... | [
"Solution:\n\nOmdat de horizontale bedekking een $n$-tegel bevat, bevat elke kolom minstens één rood vakje. In de verticale bedekking moet daarom in elke kolom minstens één tegel liggen; omdat er precies $n$ tegels zijn, betekent dat dat er in elke kolom precies één tegel moet liggen. Net zo moet er in de horizonta... | Netherlands | IMO-selectietoets I | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^{2n-2} | |
0dqe | A finite set of distinct positive integers is called a $T$-set if each of its members divides the sum of them all. Prove that every finite set of positive integers is a subset of some $T$-set. | [
"Clearly, any set containing only one element is a $T$-set. Also, since $\\{1, 2, 3\\}$ is a $T$-set, any of its subsets is certainly contained in a $T$-set.\n\nNow let $S$ be a finite set of positive integers with at least two elements, and let $n\\ (> 3)$ be the largest element in $S$. Let $\\sigma(S)$ denote the... | Singapore | Singapore International Mathematical Olympiad Committee National Team Selection Test | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
09ty | A *complete number* is a 9 digit number that contains each of the digits 1 to 9 exactly once. The *difference number* of a number $N$ is the number you get by taking the differences of consecutive digits in $N$ and then stringing these digits together. For instance, the difference number of 25143 is equal to 3431. The ... | [
"For $a = 4$, an example of such a number is 126734895. For $a = 5$, an example is the number 549832761. (There are other solutions as well.)\n\nWe will show that for $a = 3, 6, 7, 8, 9$ there is no complete number with a difference number equal to $1a1a1a1a$. It then immediately follows that there is also no compl... | Netherlands | Final Round, September 2019 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a = 4, 5 | |
0fol | Se tienen $60$ puntos en el interior del círculo unidad. Demostrar que existe un punto $V$ de la frontera del círculo, tal que la suma de las distancias de $V$ a los $60$ puntos es menor o igual que $80$. | [] | Spain | L Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | Spanish | proof only | null | |
097e | Problem:
Fie $I_{n} = \int_{1}^{n} \frac{[x]}{x^{2}+1} dx$, $n \in \mathbb{N}$, $n \geq 2$. Calculați: $\lim_{n \rightarrow \infty} \frac{I_{n}}{\ln n}$. | [
"Solution:\n\nConform lemei Stolz-Cesàro obținem\n$$\n\\begin{gathered}\n\\lim_{n \\rightarrow \\infty} \\frac{I_{n}}{\\ln n} = \\lim_{n \\rightarrow \\infty} \\frac{I_{n+1} - I_{n}}{\\ln (n+1) - \\ln n} \\\\\nI_{n+1} - I_{n} = \\int_{n}^{n+1} \\frac{[x]}{x^{2}+1} dx = n \\int_{n}^{n+1} \\frac{dx}{x^{2}+1} = \\\\\n... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 1 | |
00fv | Let $S$ be a set of $2n+1$ points in the plane such that no three are collinear and no four concyclic. A circle will be called good if it has 3 points of $S$ on its circumference, $n-1$ points in its interior and $n-1$ in its exterior. Prove that the number of good circles has the same parity as $n$. | [
"Lemma 1. Let $P$ and $Q$ be two points of $S$. The number of good circles that contain $P$ and $Q$ on their circumference is odd.\n\n\n\nLet $N$ be the number of good circles that pass through $P$ and $Q$. Number the points on one side of the line $PQ$ by $A_{1}, A_{2}, \\ldots, A_{k}$ and... | Asia Pacific Mathematics Olympiad (APMO) | XI APMO | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
06of | The sequence $c_{0}, c_{1}, \ldots, c_{n}, \ldots$ is defined by $c_{0}=1$, $c_{1}=0$ and $c_{n+2}=c_{n+1}+c_{n}$ for $n \geq 0$. Consider the set $S$ of ordered pairs $(x, y)$ for which there is a finite set $J$ of positive integers such that $x=\sum_{j \in J} c_{j}$, $y=\sum_{j \in J} c_{j-1}$. Prove that there exist... | [
"Let $\\varphi=(1+\\sqrt{5}) / 2$ and $\\psi=(1-\\sqrt{5}) / 2$ be the roots of the quadratic equation $t^{2}-t-1=0$. So $\\varphi \\psi=-1$, $\\varphi+\\psi=1$ and $1+\\psi=\\psi^{2}$. An easy induction shows that the general term $c_{n}$ of the given sequence satisfies\n$$\nc_{n}=\\frac{\\varphi^{n-1}-\\psi^{n-1}... | IMO | IMO 2006 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0jtw | Problem:
Complex number $\omega$ satisfies $\omega^{5}=2$. Find the sum of all possible values of
$$
\omega^{4}+\omega^{3}+\omega^{2}+\omega+1
$$ | [
"Solution:\n\nThe value of $\\omega^{4}+\\omega^{3}+\\omega^{2}+\\omega+1=\\frac{\\omega^{5}-1}{\\omega-1}=\\frac{1}{\\omega-1}$. The sum of these values is therefore the sum of $\\frac{1}{\\omega-1}$ over the five roots $\\omega$. Substituting $z=\\omega-1$, we have that $(z+1)^{5}=2$, so $z^{5}+5 z^{4}+10 z^{3}+1... | United States | HMMT November 2016 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 5 | |
0746 | Problem:
Define a sequence $\left\langle a_{n}\right\rangle_{n=1}^{\infty}$ as follows:
$$
a_{n}= \begin{cases}0, & \text{ if the number of positive divisors of } n \text{ is odd } \\ 1, & \text{ if the number of positive divisors of } n \text{ is even }\end{cases}
$$
(The positive divisors of $n$ include 1 as well ... | [
"Solution:\n\nWe show that $x$ is irrational. Suppose that $x$ is rational. Then the sequence $\\left\\langle a_{n}\\right\\rangle_{n=1}^{\\infty}$ is periodic after some stage; there exist natural numbers $k, l$ such that $a_{n}=a_{n+l}$ for all $n \\geq k$. Choose $m$ such that $m l \\geq k$ and $m l$ is a perfec... | India | Indian National Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof and answer | x is irrational | |
0415 | Given any $n$ ($> 1$) coprime positive integers $a_1, a_2, \dots, a_n$, denote $A = a_1 + a_2 + \dots + a_n$. Let $d_i = (A, a_i)$ (the greatest common divisor), $i = 1, 2, \dots, n$.
Let $D_i$ be the greatest common divisor of $\{a_1, a_2, \dots, a_n\} \setminus \{a_i\}$, $i = 1, 2, \dots, n$. Find the minimum of $\pr... | [
"Consider\n$$\nD_1 = (a_2, a_3, \\dots, a_n) \\text{ and } d_2 = (a_2, A) = (a_2, a_1 + a_2 + \\dots + a_n).\n$$\nLet $(D_1, d_2) = d$. Then $d \\mid a_2, d \\mid a_3, \\dots, d \\mid a_n, d \\mid a_1 + a_2 + \\dots + a_n$. Thus, $d \\mid a_1$. Consequently,\n$$\nd \\mid (a_1, a_2, \\dots, a_n).\n$$\nSince $a_1, a_... | China | China National Team Selection Test | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | (n-1)^n | |
00i4 | Let $\alpha$ and $\beta$ be positive real numbers. Emerald makes a trip in the coordinate plane, starting off from the origin $(0,0)$. Each minute she moves one unit up or one unit to the right, restricting herself to the region $|x-y|<2025$, in the coordinate plane. By the time she visits a point ( $x, y$ ) she writes... | [
"Let $(x_{n}, y_{n})$ be the point that Emerald visits after $n$ minutes. Then $(x_{n+1}, y_{n+1}) \\in \\{(x_{n}+1, y_{n}), (x_{n}, y_{n}+1)\\}$. Either way, $x_{n+1}+y_{n+1}=x_{n}+y_{n}+1$, and since $x_{0}+y_{0}=0+0=0$, $x_{n}+y_{n}=n$.\n\nThe $n$-th number would be then\n$$\nz_{n}=\\left\\lfloor x_{n} \\alpha+\... | Asia Pacific Mathematics Olympiad (APMO) | APMO 2025 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | All positive real pairs with α + β = 2 | |
0a29 | We define a sequence by $a_1 = 850$ and
$$
a_{n+1} = \frac{a_n^2}{a_n - 1}
$$
for $n \ge 1$. Determine all values of $n$ for which $\lfloor a_n \rfloor = 2024$. | [
"The only value that satisfies is $n = 1175$.\n\nFirst, we note that we can rewrite the recursion as\n$$\na_{n+1} = \\frac{a_n^2 - 1 + 1}{a_n - 1} = \\frac{a_n^2 - 1}{a_n - 1} + \\frac{1}{a_n - 1} = a_n + 1 + \\frac{1}{a_n - 1}. \\quad (1)\n$$\nSince the difference of $a_{n+1} - a_n > 1$, there is at most one $n$ t... | Netherlands | BxMO/EGMO Team Selection Test | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 1175 | |
0j78 | Problem:
Let $ABC$ be a triangle with $AB = 5$, $BC = 8$, and $CA = 7$. Let $\Gamma$ be a circle internally tangent to the circumcircle of $ABC$ at $A$ which is also tangent to segment $BC$. $\Gamma$ intersects $AB$ and $AC$ at points $D$ and $E$, respectively. Determine the length of segment $DE$. | [
"Solution:\n\nAnswer: $\\frac{40}{9}$\n\n\n\nFirst, note that a homothety $h$ centered at $A$ takes $\\Gamma$ to the circumcircle of $ABC$, $D$ to $B$ and $E$ to $C$, since the two circles are tangent. As a result, we have $DE \\parallel BC$. Now, let $P$ be the center of $\\Gamma$ and $O$ ... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, ince... | null | proof and answer | 40/9 | |
09nn | Two cars start simultaneously toward each other, one from city $A$ and the other from city $B$. Each time they reach the other city, they immediately return. The first car is $1.25$ times faster than the second. If the distance between their $2$nd and $3$rd meeting points is $56$ km, find the distance between cities $A... | [] | Mongolia | MMO2025 Round 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 126 km | |
0ffp | Problem:
Hallar un par de enteros positivos $a$ y $b$ tales que
1) $a b(a+b)$ no es divisible por $7$ ;
2) $(a+b)^{7}-a^{7}-b^{7}$ es divisible por $7^{7}$.
Justificar la respuesta. | [] | Spain | International Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | a = 324, b = 1 | |
0lbz | Define a sequence $(a_n)$ as follows
$$
\begin{cases} a_1 = 1, \\ a_{n+1} = 3 - \frac{a_n + 2}{2^{a_n}} \quad \text{for } n \ge 1. \end{cases}
$$
Prove that this sequence has a finite limit and find this limit. | [
"Firstly, by induction, one can prove that\n$$\na_n > 1, \\quad \\forall n > 1\n$$\nbecause the function $u(x) = 2^{x+1} - x - 2$ is increasing on $(1, +\\infty)$. It implies that\n$$\na_{n+1} = 3 - \\frac{a_n + 2}{2^{a_n}} < 3, \\quad \\forall n \\ge 1.\n$$\nNext, we will prove that $(a_n)$ is an increasing sequen... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 2 | |
0ba7 | Given a positive integer number $k$, define the function $f$ on the set of all positive integer numbers to itself by
$$
f(n) = \begin{cases} 1, & \text{if } n \le k+1, \\ f(f(n-1)) + f(n - f(n-1)), & \text{if } n > k+1. \end{cases}
$$
Show that preimage of every positive integer number under $f$ is a finite non-empty... | [
"It is sufficient to show that the difference $\\Delta(n) = f(n) - f(n-1)$ is 0 or 1 for all integer numbers $n \\ge 2$, and $f$ is unbounded.\n\nClearly, $\\Delta(n) = 0$, $2 \\le n \\le k+1$, provides the basis for an inductive proof. If $n > k+1$, apply the recurrence. By the induction hypothesis, $\\Delta(n-1)$... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0bhg | Let $n \ge 2$ be an integer and $z$ be a complex number so that $z^n = 1$. Prove that:
$$ \prod_{k=1}^{n} (1 - z^k - z^{2k}) = \left\lfloor \left( \frac{1 + \sqrt{5}}{2} \right)^n \right\rfloor - \frac{1 + (-1)^n}{2} $$ | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
078c | In triangle $ABC$ with $CA = CB$, point $E$ lies on the circumcircle of $ABC$ such that $\angle ECB = 90^\circ$. The line through $E$ parallel to $CB$ intersects $CA$ in $F$ and $AB$ in $G$. Prove that the centre of the circumcircle of triangle $EGB$ lies on the circumcircle of triangle $ECF$. | [
"\n\nWe have $FG = FA$ since $FG$ is parallel to $BC$. But also $\\triangle GAE$ is a right angle triangle. Thus, if $F'$ is the midpoint of $GE$, then $\\angle GAF = \\angle FGA = \\angle F'GA = \\angle GAF'$ which implies $F \\equiv F'$. Thus, $F$ is the midpoint of $GE$.\n\nIf $O$ is the... | India | INMO | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06rr | Several positive integers are written in a row. Iteratively, Alice chooses two adjacent numbers $x$ and $y$ such that $x > y$ and $x$ is to the left of $y$, and replaces the pair $(x, y)$ by either $(y+1, x)$ or $(x-1, x)$. Prove that she can perform only finitely many such iterations. | [
"Note first that the allowed operation does not change the maximum $M$ of the initial sequence. Let $a_{1}, a_{2}, \\ldots, a_{n}$ be the numbers obtained at some point of the process. Consider the sum\n$$\nS = a_{1} + 2 a_{2} + \\cdots + n a_{n}\n$$\nWe claim that $S$ increases by a positive integer amount with ev... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0fs8 | Problem:
Find all the positive integers $a, b, c$ such that
$$
a!\cdot b! = a! + b! + c!
$$ | [
"Solution:\n(Valentin and Tanish) Without loss of generality we assume $a \\leq b$. We now divide the entire expression by $b!$ and get\n$$\na! = \\frac{a!}{b!} + \\frac{c!}{b!} + 1\n$$\nIf $a < b$ then also $c < b$ since the right hand side should be an integer. But then\n$$\n\\frac{a!}{b!} + \\frac{c!}{b!} + 1 < ... | Switzerland | null | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (3, 3, 4) | |
0jxf | Problem:
Horizontal parallel segments $AB = 10$ and $CD = 15$ are the bases of trapezoid $ABCD$. Circle $\gamma$ of radius $6$ has center within the trapezoid and is tangent to sides $AB$, $BC$, and $DA$. If side $CD$ cuts out an arc of $\gamma$ measuring $120^{\circ}$, find the area of $ABCD$. | [
"Solution:\n\nSuppose that the center of the circle is $O$ and the circle intersects $CD$ at $X$ and $Y$. Since $\\angle XOY = 120^{\\circ}$ and triangle $XOY$ is isosceles, the distance from $O$ to $XY$ is $6 \\cdot \\sin(30^{\\circ}) = 3$. On the other hand, the distance from $O$ to $AB$ is $6$ as the circle is t... | United States | HMMT November 2017 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 225/2 | |
07ov | Let $a_1 = 1$, $b_1 = 0$, $c_1 = 1$ and define for all $n \ge 1$
$$
a_{n+1} = a_n + 2b_n, \quad b_{n+1} = b_n + 2c_n, \quad c_{n+1} = a_n + c_n.
$$
Find integers $A$, $B$, $C$ such that $a_{n+3} = Aa_{n+2} + Ba_{n+1} + Ca_n$ for all $n \ge 1$. | [
"**Solution 1.** In order to write the required equations for $n = 1, 2, 3$ we calculate the following values:\n\n| n | 1 | 2 | 3 | 4 | 5 | 6 |\n|---|---|---|---|---|---|---|\n| $a_n$ | 1 | 1 | 5 | 17 | 41 | 97 |\n| $b_n$ | 0 | 2 | 6 | 12 | 28 | 78 |\n| $c_n$ | 1 | 2 | 3 | 8 | 25 | 66 |\n\nThe integers $A$, $B$, $C... | Ireland | Irska 2014 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Algebraic Number Theory > Algebraic numbers"
] | null | proof and answer | A = 3, B = -3, C = 5 | |
0ew6 | Problem:
The triangle $ABC$ satisfies $0 \leq AB \leq 1 \leq BC \leq 2 \leq CA \leq 3$. What is the maximum area it can have? | [
"Solution:\nIf we ignore the restrictions of $CA$, then the maximum area is $1$, achieved when $AB$ is perpendicular to $BC$. But in this case $CA$ satisfies the restrictions."
] | Soviet Union | 2nd ASU | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 1 | |
0gnf | The number of unordered edge pairs without common vertex in a graph $G$ with $30$ vertices and $105$ edges is equal to $4822$. What is the maximal possible value of the differences between degrees of vertices? (Azer Kerimov). | [
"Suppose that $G$ has $n$ vertices and $k$ edges and $i$-th vertex has a degree $d_i$. The number of unordered edge pairs without common vertex is equal to\n$$\n\\left(\\binom{k}{2} - \\sum_{i=0}^{n} \\binom{d_i}{2}\\right) = \\frac{k(k-1)}{2} - \\frac{1}{2} \\sum_{i=0}^{n} d_i^2 + k = \\frac{k(k+1)}{2} - \\frac{1}... | Turkey | Team Selection Test for IMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 5 | |
022y | Problem:
Uma circunferência de raio $r$ está inscrita em um setor circular de raio $R$. O comprimento da corda $AB$ é igual a $2a$.
Prove que
$$
\frac{1}{r} = \frac{1}{R} + \frac{1}{a}
$$ | [
"Solution:\n\nDenotemos por $D$ o ponto de tangência de $AO$ com a circunferência. Então $ODO_1 = 90^{\\circ}$. Observe também que $AC = AB / 2 = a$.\n\nPor outro lado, $OCA = 90^{\\circ}$. Os triângulos $ODO_1$ e $OCA$ são semelhantes pois possuem um ângulo comum e um ângulo reto. Portanto,\n$$\n\\frac{OO_1}{OA} =... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08ee | Problem:
Un intero positivo $m$ si dice portatore di zeri se esiste un intero positivo $k$ tale che
- $k$ è un quadrato perfetto,
- $k$ è multiplo di $m$,
- l'espressione decimale di $k$ contiene almeno 2021 cifre 0, ma l'ultima cifra (quella più a destra) è diversa da 0 .
Determinare tutti gli interi portatori di zer... | [
"Solution:\n\nUn intero positivo $m$ è portatore di zeri se e solo se $m$ non è divisibile per $10$.\n\nCondizione necessaria\nSe $m$ è multiplo di $10$, allora $m$ non può essere portatore di zeri, in quanto ogni multiplo di $m^{2}$ terminerà necessariamente con almeno due cifre $0$.\n\nCondizione sufficiente\nDim... | Italy | XXXVII Olimpiade Italiana di Matematica | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All positive integers not divisible by ten | |
0jcx | Problem:
Given $\triangle ABC$ with $AB < AC$, the altitude $AD$, angle bisector $AE$, and median $AF$ are drawn from $A$, with $D, E, F$ all lying on $\overline{BC}$. If $\measuredangle BAD = 2 \measuredangle DAE = 2 \measuredangle EAF = \measuredangle FAC$, what are all possible values of $\measuredangle ACB$? | [
"Solution:\n\n$30^{\\circ}$ or $\\pi / 6$ radians\n\nLet $H$ and $O$ be the orthocenter and circumcenter of $ABC$, respectively: it is well-known (and not difficult to check) that $\\measuredangle BAH = \\measuredangle CAO$. However, note that $\\measuredangle BAH = \\measuredangle BAD = \\measuredangle CAF$, so $\... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 30° | |
0koe | Problem:
Let $S$ be a set of size $11$. A random $12$-tuple $(s_{1}, s_{2}, \ldots, s_{12})$ of elements of $S$ is chosen uniformly at random. Moreover, let $\pi: S \rightarrow S$ be a permutation of $S$ chosen uniformly at random. The probability that $s_{i+1} \neq \pi(s_{i})$ for all $1 \leq i \leq 12$ (where $s_{13}... | [
"Solution:\nGiven a permutation $\\pi$, let $\\nu(\\pi)$ be the number of fixed points of $\\pi$. We claim that if we fix $\\pi$, then the probability that the condition holds, over the randomness of $s_{i}$, is $\\frac{10^{12}+\\nu\\left(\\pi^{12}\\right)-1}{11^{12}}$. Note that a point in $S$ is a fixed point of ... | United States | HMMT February 2022 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 1000000000004 | |
0co3 | Let $AD$, $BE$ and $CF$ be the angle bisectors in a triangle $ABC$, and let $I$ be their intersection point. The perpendicular bisector of segment $AD$ intersects lines $BE$ and $CF$ at points $M$ and $N$, respectively. Prove that the points $A$, $I$, $M$, and $N$ are concyclic. (D. Prokopenko)
В треугольнике $ABC$ пр... | [
"Для решения задачи достаточно установить, что $\\angle MAI = \\angle MNI$ (см. рис. 5).\n\nПусть $K$ — середина отрезка $AD$. Заметим, что $\\angle MNI = \\angle KNI = 90^\\circ - \\angle KIN = 90^\\circ - (\\angle ACI + \\angle CAI) = \\frac{1}{2}(180^\\circ - (\\angle ACB + \\angle BAC)) = \\frac{1}{2}\\angle AB... | Russia | Regional 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; Russian | proof only | null | |
0239 | Problem:
Nessa exercício, as letras representam algarismos. Determine cada uma das parcelas da soma abaixo.
$$
\begin{array}{r}
a b c d e f \\
a b c d e f \\
+\quad g h i j \\\hline d e f h j f
\end{array}
$$ | [
"Solution:\n\n$$\n\\begin{array}{r}\n231468 \\\\\n231468 \\\\\n+\\quad 5972 \\\\\\hline 468908\n\\end{array}\n$$\n\n$$\n\\begin{array}{r}\n264538 \\\\\n264538 \\\\\n+\\quad 548698 \\\\\\hline 538178\n\\end{array}\n$$\n\n$$\n\\begin{array}{r}\n273548 \\\\\n273548 \\\\\n+\\quad 548698 \\\\\\hline 548698\n\\end{array}... | Brazil | Desafios | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | abcdef = 247501, ghij = 6389 | |
0ijx | Problem:
Suppose hypothetically that a certain, very corrupt political entity in a universe holds an election with two candidates, say $A$ and $B$. A total of $5,825,043$ votes are cast, but, in a sudden rainstorm, all the ballots get soaked. Undaunted, the election officials decide to guess what the ballots say. Each... | [
"Solution:\n\nLet $N=2912521$, so that the number of ballots cast is $2N+1$. Let $P$ be the probability that $B$ wins, and let $\\alpha=51\\%$ and $\\beta=49\\%$ and $\\gamma=\\beta/\\alpha<1$. We have\n$$\n10^{-X}=P=\\sum_{i=0}^{N}\\binom{2N+1}{N-i} \\alpha^{N-i} \\beta^{N+1+i}=\\alpha^{N} \\beta^{N+1} \\sum_{i=0}... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 510 | |
073s | Let $a$, $b$, $c$ be positive real numbers such that $a^2 + b^2 + c^2 < 2(a + b + c)$. Prove that
$$
3abc < 4(a + b + c).
$$ | [
"The Cauchy-Schwarz inequality gives\n$$\n(a + b + c)^2 \\le 3(a^2 + b^2 + c^2).\n$$\nThus $(a + b + c) < 6$ and hence $\\frac{(a + b + c)^3}{9} < 4(a + b + c)$. Now the AM-GM inequality gives\n$$\n(a + b + c)^3 \\ge 27abc.\n$$\nThus\n$$\n3abc \\le \\frac{(a + b + c)^3}{9} < 4(a + b + c).\n$$\n\nAlternately, we hav... | India | Indija TS 2008 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0ey2 | Problem:
A natural number $k$ has the property that if $k$ divides $n$, then the number obtained from $n$ by reversing the order of its digits is also divisible by $k$. Prove that $k$ is a divisor of $99$. | [
"Solution:\nLet $r(m)$ denote the number obtained from $m$ by reversing the digits.\n\nWe show first that $k$ cannot be divisible by $2$ or $5$. It cannot be divisible by both, for then it ends in a zero and hence $r(k) < k$ and so is not divisible by $k$ (contradiction). So if $5$ divides $k$, then the last digit ... | Soviet Union | 1st ASU | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0271 | Problem:
Num deserto há cobras, ratos e escorpiões. Cada manhã, cada cobra mata um rato. Cada meio-dia, cada escorpião mata uma cobra. Cada noite, cada rato mata um escorpião. Ao final de uma semana, à noite, só restava um rato. Quantos ratos havia na manhã no início da semana? | [
"Solution:\n\n1873"
] | Brazil | null | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 1873 | |
00m8 | a. Man bestimme den größtmöglichen Wert $M$, den $x+y+z$ annehmen kann, wenn $x$, $y$ und $z$ positive reelle Zahlen mit
$$
16xyz = (x + y)^2(x + z)^2
$$
sind.
b. Man zeige, dass es unendlich viele Tripel $(x, y, z)$ positiver rationaler Zahlen gibt, für die
$$
16xyz = (x + y)^2(x + z)^2 \text{ und } x + y + z = M
$$
... | [
"(a) Aufgrund der Nebenbedingung und der arithmetisch-geometrischen Mittelungleichung gilt\n$$\n4\\sqrt{xyz} = (x + y)(x + z) = x(x + y + z) + yz \\ge 2\\sqrt{xyz(x + y + z)}.\n$$\nAlso gilt $2 \\ge \\sqrt{x+y+z}$ und damit $4 \\ge x+y+z$. Da wir im zweiten Teil unendlich viele solche Tripel angeben, für die $x+y+z... | Austria | 48. Österreichische Mathematik-Olympiade | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | German | proof and answer | 4 | |
0bm2 | Problem:
Adott az $f:[0,1] \rightarrow[0,1]$ függvény, amelyre bármely $y \in[0,1]$ és bármely $\varepsilon>0$ esetén létezik $x \in[0,1]$ úgy, hogy $|f(x)-y|<\varepsilon$.
a) Igazold, hogy ha $f$ folytonos a $[0,1]$ intervallumon, akkor $f$ szürjektív!
b) Adj példát olyan $f$ függvényre, ami teljesíti a feladatbeli f... | [
"Solution:\n\na)\nConsiderăm o funcţie continuă $f:[0,1] \\rightarrow[0,1]$ având proprietatea din enunţ. Fie $y \\in[0,1]$. Din ipoteză deducem că există un şir $(x_{n})_{n \\geq 1}$, cu termenii în $[0,1]$, astfel încât $|f(x_{n})-y|<1/n$, $\\forall n \\geq 1$.\n\nŞirul $(x_{n})_{n \\geq 1}$ este mărginit, deci a... | Romania | Olimpiada Naţională de Matematică, Etapa Judeţeană şi a Municipiului Bucureşti | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | a) If f is continuous on the unit interval, then f is surjective onto the unit interval. b) Example: define f on the unit interval by f(x) = x for rational x and f(x) = 0 for irrational x; its image is dense but not all real numbers in the interval, so it is not surjective. | |
0atc | Problem:
Let $r$ be some real constant, and $P(x)$ a polynomial which has remainder $2$ when divided by $x - r$, and remainder $-2x^{2} - 3x + 4$ when divided by $(2x^{2} + 7x - 4)(x - r)$. Find all values of $r$. | [] | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | r = 1/2 or r = -2 | |
09de | Гурвалжны хагас параметр-р, багтаасан тойргийн радиус-$R$, багтсан тойргийн радиус-$r$ бол $p^2 \ge 12Rr + 3r^2$ болохыг батал. | [
"$$\nR = \\frac{abc}{4S}, \\quad r = \\frac{S}{p}, \\quad S = \\sqrt{p(p-a)(p-b)(p-c)}\n$$\n\n$p^2 \\ge 12 \\cdot \\frac{abc}{4S} \\cdot \\frac{S}{p} + 3 \\left(\\frac{S}{p}\\right)^2 = \\frac{3abc}{p} + \\frac{3(p-a)(p-b)(p-c)}{p}$\n\n$= 3r^3 - 3r^2(a + b + c) + 3r(ab + bc + ac)$\n\n$= 3r^2 - 3r(a + b + c) + 3(ab ... | Mongolia | ММО-48 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | Mongolian | proof only | null | |
0g96 | 遊戲開始時有 $2^m$ 張紙，每張上寫有一個 $1$。考慮以下操作：每次我們選兩張紙，假設其上的數字分別為 $a$ 與 $b$。將兩張紙上的數字都擦掉，並在兩張紙上都寫上 $a+b$。
試證：經過 $m2^{m-1}$ 步後，所有紙上的數字總和至少為 $4^m$。 | [
"令 $P_k$ 為第 $k$ 次操作後所有紙張上數字的乘積，而 $S_k$ 為第 $k$ 次操作後所有紙張上數字的總和。顯然 $P_0 = 1$。又基於 $(a+b)^2 \\ge 4ab$，易知 $P_{k+1} \\ge 4P_k$，故 $P_{m2^{m-1}} \\ge 4^{m2^{m-1}} = (2^m)^{2^m}$。最後由算幾不等式，$S_k \\ge 4^m$。證畢。"
] | Taiwan | 2015 Math Olympiad Second Stage Training Camp | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0893 | Problem:
Determinare tutti gli interi positivi che sono uguali a 300 volte la somma delle loro cifre. | [
"Solution:\n\nDimostreremo che c'è un'unica soluzione, ossia $n=2700$.\n\nSia $n$ un intero positivo che soddisfa la condizioni date. Osserviamo immediatamente che, poiché $n$ è un multiplo di $300$, e quindi di $100$, le cifre delle unità e delle decine di $n$ devono essere uguali a zero. Supponiamo dunque che la ... | Italy | Cesenatico | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2700 | |
0jec | Problem:
The real numbers $x, y, z$ satisfy $0 \leq x \leq y \leq z \leq 4$. If their squares form an arithmetic progression with common difference $2$, determine the minimum possible value of $|x-y| + |y-z|$. | [
"Solution:\n\n$|x-y| + |y-z| = z - x = \\frac{z^2 - x^2}{z + x} = \\frac{4}{z + x}$, which is minimized when $z = 4$ and $x = \\sqrt{12}$. Thus, our answer is $4 - \\sqrt{12} = 4 - 2\\sqrt{3}$."
] | United States | HMMT 2013 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof and answer | 4 - 2√3 | |
0bwa | Let $p$ and $q$ be two positive real numbers, $p > q$, and let $C$ be the set of the continuous real functions defined on the interval $[0, 1]$. Find
$$
\max_{f \in C} \int_{0}^{1} \left(x^{p} |f(x)|^{q} - x^{q} |f(x)|^{p}\right) dx
$$
and the functions which yield this maximum. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | English | proof and answer | Maximum value: ((p − q) / (p (p + q + 1))) * (q / p)^{q / (p − q)}. Maximizing functions: f(x) = ± (q / p)^{1 / (p − q)} · x. | |
0ahm | An arbitrary triangle $ABC$ is given together with two lines $p$ and $q$ which are not parallel to each other and are not perpendicular to any of the sides of the triangle. We denote the perpendiculars through $A$, $B$ and $C$ to the line $p$ by $p_A$, $p_B$ and $p_C$ respectively, and the perpendiculars to $q$ by $q_A... | [
"Without loss of generality we can assume that $p_B$ is between $p_A$ and $p_C$. The first case is if $q_A$ is between $q_B$ and $q_C$ as shown in the picture, obviously $KL$ intersects $PN$. Analogously, if $q_C$ is between $q_A$ and $q_B$ the case is symmetrical to the one we are considering. The second case, if ... | North Macedonia | Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0l8z | Let be given two positive integers $m$, $n$ with $m < 2001$, $n < 2002$; and let be given $2001 \times 2002$ distinct real numbers. Put these numbers into the little squares of a rectangular board of size $2001 \times 2002$ (the board consists of $2001$ rows and $2002$ columns) so that each number is putting in a littl... | [
"We enlarge the problem by replacing $2001$ by $p$, $2002$ by $q$ ($m \\le p, n \\le q$) and prove by induction on $p+q$ that $s \\ge (p-m)(q-n)$ (1).\nIt is easily seen that the assertion (1) is true for $p+q=2, 3, 4$. Suppose that it is true for $p+q=k$.\nConsider a $(p, q)$-board with $p+q=k+1$. It is easy to se... | Vietnam | THE 2002 VIETNAMESE MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | (2001 - m)(2002 - n) | |
08fr | Problem:
Sia $ABCD$ un parallelogramma tale che la bisettrice uscente da $B$ interseca il lato $CD$ nel suo punto medio $M$. Il lato $BC$ è lungo $6$, e la diagonale $AC$ è lunga $14$. Determinare la lunghezza di $AM$.
(A) $2 \sqrt{19}$
(B) $14-3 \sqrt{3}$
(C) $4 \sqrt{5}$
(D) $9$
(E) $2 \sqrt{22}$ | [
"Solution:\n\n\n\nOsserviamo che il triangolo $MBC$ è isoscele di base $MB$: infatti $\\widehat{CBM}=\\widehat{MBA}$, per ipotesi, e $\\widehat{ABM}=\\widehat{BMC}$, in quanto sono angoli alterni interni rispetto le parallele $DC \\parallel AB$. Da cui\n$$\n\\overline{AB}=\\overline{DC}=2\\... | Italy | Olimpiadi di Matematica - Febbraio | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | C | |
00cq | Hallar el mayor número entero capicúa de 5 dígitos que es divisible por $101$.
ACLARACIÓN: Un número es capicúa si se lee igual de izquierda a derecha que de derecha a izquierda. | [] | Argentina | Nacional OMA 2019 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Spanish | proof and answer | 49894 | |
02k0 | Problem:
Dois espelhos formam um ângulo de $30^{\circ}$ no ponto $V$. Um raio de luz parte de um ponto $S$ paralelamente a um dos espelhos e é refletido pelo outro espelho no ponto $A$, como mostra a figura. Depois de uma certa quantidade de reflexões, o raio retorna a $S$.
Se $A S$ e $A V$ têm ambos 1 metro, qual o c... | [
"Solution:\n\nVamos acompanhar o trajeto do raio de luz a partir do ponto $S$. Para isso, lembramos a propriedade básica da reflexão de um raio de luz em um espelho: o ângulo de reflexão é igual ao ângulo de incidência. Por exemplo, na figura ao lado, os ângulos $a$ e $b$ são iguais, bem como $d$ e $e$. Notamos que... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | B | |
0dbw | What is the biggest number of queens of two colors (black and white) one can place on a chessboard, such that no two of them beat one another (the queens do not beat through each other)?
Note: The queen is able to move any number of squares vertically, horizontally or diagonally. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 8 | |
0fu8 | Problem:
Bestimme alle endlichen Folgen $\left(x_{0}, x_{1}, \ldots, x_{n}\right)$ reeller Zahlen, sodass die Zahl $k$ in der Folge genau $x_{k}$ mal auftritt. | [
"Solution:\nBeachte zuerst, dass alle Folgeglieder nichtnegative ganze Zahlen $\\leq n+1$ sind, da $x_{k}$ das Auftreten von $k$ in der Folge zählt und die Länge der Folge $n+1$ ist. Ausserdem ist $x_{k}=n+1$ unmöglich, denn sonst wären alle Folgeglieder gleich $k$, also auch $x_{k}$, Widerspruch. Daher gilt $x_{k}... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All such sequences are exactly the following:
- (1, 2, 1, 0)
- (2, 0, 2, 0)
- (2, 1, 2, 0, 0)
- And, for any integer m ≥ 3, the sequence of length m + 4 with entries x0 = m, x1 = 2, x2 = 1, x_m = 1, and all other entries zero; equivalently written as (m, 2, 1, 0, 0, ..., 0, 1, 0, 0, 0) with exactly m − 3 zeros between ... | |
0ged | 對於每個質數 $p$，都有一個名為 $p$-蘭蒂亞的王國，擁有 $p$ 座島嶼，這些島依序編號為 $1, 2, \dots, p$。編號為 $n$ 和 $m$ 的島之間會有橋相連，若且唯若 $p$ 整除 $(n^2 - m + 1)(m^2 - n + 1)$。證明有無窮多個質數 $p$，在 $p$-蘭蒂亞王國裡面會有兩個島無法藉由一連串的橋來連接。
For each prime $p$, there is a kingdom of $p$-Landia consisting of $p$ islands numbered $1, 2, \dots, p$. Two distinct islands numbered $n$ an... | [
"View it as a directed graph with a directed edge $mn$ iff $n = m^2 + 1$ in $\\mathbb{Z}_p$. Easy to see that out degree is at most one for every vertex. If $a, b$ are roots of $x^2 - x + 1$, then we can check that $a, b$ are distinct if $p > 3$ (otherwise $a = b = 2^{-1}$ and $1 = 2^{-2}$ in $\\mathbb{Z}_p$). So t... | Taiwan | 2021 數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Graph Theory",
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
0hm6 | Problem:
Let $m, n$ be positive integers. Suppose that a given rectangle can be tiled (without overlaps) by a combination of horizontal $1 \times m$ strips and vertical $n \times 1$ strips. Show that it can be tiled using just one of the two types. | [
"Solution:\n\nIf the rectangle is $a \\times b$ ($a$ the horizontal dimension), it is clear that $a$ and $b$ are integers since the rectangle can be divided into rectangles of integer side lengths; we want to show that either $a$ is divisible by $m$ or $b$ is divisible by $n$.\n\nLet $\\zeta = \\cos \\frac{2\\pi}{m... | United States | Berkeley Math Circle Take-Home Contest | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0cdy | For every positive integer $n$ define $a_n = n + 1$, $b_n = 2 \cdot n + 3$, $c_n = 3 \cdot n + 5$, $d_n = 4 \cdot n + 7$. Find all positive integers $n$ which have at most two digits and $a_n, b_n, c_n$, and $d_n$ are prime. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 88 | |
0hjn | Problem:
Let $a$ and $b$ be integers such that
$$
|a+b| > |1 + a b|.
$$
Prove that $a b = 0$. | [
"Solution:\nNotice that replacing both $a$ and $b$ by their negatives does not change either side of the given inequality. Therefore, we may assume that $a + b \\geq 0$. We now have $a + b > |1 + a b|$, so\n$$\n\\begin{array}{rll}\na + b > 1 + a b & \\text{ and } & a + b > -1 - a b \\\\\na b - a - b + 1 < 0 & \\tex... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
09yl | Line up the numbers $1$ to $15$ such that if you add any two numbers that are next to each other, you get a square number.
What do you get if you add the first and last number from the line? | [] | Netherlands | First Round | [
"Discrete Mathematics > Other"
] | English | final answer only | 17 | |
0ens | Let $O$ be the intersection point of the diagonals of the convex quadrilateral $ABCD$, with $AO = OC$. Points $P$ and $Q$ are marked on the segments $AO$ and $CO$, respectively, such that $PO = OQ$. Let $N$ be the intersection of $AB$ and $DP$, and $K$ be the intersection of $CD$ and $BQ$.
Prove that the points $N$, $O... | [
"\nDraw $NM \\parallel KL \\parallel AC$ as in the figure. From the similarity of the triangles $BOQ$ and $BLK$ it follows that\n$$\n\\frac{LK}{OQ} = \\frac{BL}{BO} = 1 + \\frac{LO}{BO},\n$$\nFrom the similarity of the triangles $DOC$ and $DLK$ it follows that\n$$\n\\frac{LK}{OC} = \\frac{D... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
076d | Problem:
There are four basket-ball players $A$, $B$, $C$, $D$. Initially, the ball is with $A$. The ball is always passed from one person to a different person. In how many ways can the ball come back to $A$ after seven passes? (For example $A \rightarrow C \rightarrow B \rightarrow D \rightarrow A \rightarrow B \righ... | [
"Solution:\nLet $x_{n}$ be the number of ways in which $A$ can get back the ball after $n$ passes. Let $y_{n}$ be the number of ways in which the ball goes back to a fixed person other than $A$ after $n$ passes. Then\n$$\nx_{n}=3 y_{n-1},\n$$\nand\n$$\ny_{n}=x_{n-1}+2 y_{n-1}\n$$\nWe also have $x_{1}=0$, $x_{2}=3$,... | India | INMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 546 | |
0a3l | Problem:
Gegeven is een natuurlijk getal $n$. Er zijn $n$ eilanden met $n-1$ bruggen ertussen zo dat je van elk eiland bij elk ander eiland kan komen. Op een middag breekt er brand uit op een van de eilanden. Elke morgen verspreidt het vuur zich naar alle naburige eilanden (die eilanden die met een brug zijn verbonden... | [
"Solution:\n\nVoor $n = k^{2} + 1$ beschouwen we de eilandengroep met $k^{2}$ eilanden in een $k \\times k$-grid, waarin de eilanden per rij verbonden worden en alle eilanden in de meest linker kolom nog met het laatste eiland worden verbonden, waar ook de brand start. Dan zijn er na $\\ell < k$ nachten nog minsten... | Netherlands | IMO-selectietoets II | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | ⌊√(n−1)⌋ | |
00d5 | Ana y Beto juegan al siguiente juego. Ana escribe cuatro enteros consecutivos de tres dígitos. Beto elige tres de los cuatro números de Ana y calcula su suma. Si el número que obtiene se puede escribir como producto de tres enteros positivos mayores que 1, gana Beto. En caso contrario, gana Ana. Determinar si Ana puede... | [
"Veamos que es imposible que gane Ana. Dados cuatro enteros consecutivos, Beto elige los dos impares y uno par para que la suma de los tres sea par. Así se asegura que uno de los factores de la suma es el $2$. Vistos módulo $3$, los dos impares pueden tener restos $0$ y $2$, $1$ y $0$ o $2$ y $1$. Por lo tanto siem... | Argentina | Nacional OMA | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Spanish | proof and answer | No—Ana cannot guarantee a win; Beto can always choose two odds and the even between them so the sum is divisible by two and three, hence a product of three integers greater than one. | |
08ia | Problem:
The prime number $p$ has the following property: the remainder $r$ of the division of $p$ by $210$ is a composite number which can be represented as a sum of two perfect squares. Find the number $r$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | proof and answer | 169 | |
070j | Problem:
$\mathrm{ABC}$ is an acute-angled triangle with orthocenter $\mathrm{H}$. $\mathrm{AE}$ and $\mathrm{BF}$ are altitudes. $\mathrm{AE}$ is reflected in the angle bisector of angle $\mathrm{A}$ and $\mathrm{BF}$ is reflected in the angle bisector of angle $\mathrm{B}$. The two reflections intersect at $\mathrm{... | [
"Solution:\n\n\nWe show first that $\\mathrm{O}$ is the circumcenter of $\\mathrm{ABC}$. $\\angle \\mathrm{ABF} = 90^\\circ - \\mathrm{A}$. The line $\\mathrm{BC}$ is the reflection in $BD$ of the line $BA$ and the line $\\mathrm{BF}'$ is the reflection of $\\mathrm{BF}$, so angle $\\mathrm... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0i1n | Problem:
$A$ is the center of a semicircle, with radius $A D$ lying on the base. $B$ lies on the base between $A$ and $D$, and $E$ is on the circular portion of the semicircle such that $E B A$ is a right angle. Extend $E A$ through $A$ to $C$, and put $F$ on line $C D$ such that $E B F$ is a line. Now $E A = 1$, $A C... | [
"Solution:\n\nLet $\\theta = \\angle A E D$ and $x = D E$. By the law of cosines on triangle $A D E$, we have\n$$\n1 = 1 + x^{2} - 2x \\cos \\theta \\implies 2x \\cos \\theta = x^{2}.\n$$\nThen by the law of cosines on triangle $C D E$ (note that $C D = \\sqrt{5}$), we have\n$$\n5 = (1 + \\sqrt{2})^{2} + x^{2} - 2(... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | sqrt(2 - sqrt(2)) | |
09lg | An $n$-digit positive integer written in the digits $0$, $1$, $2$, $3$ is called a *rabbit number* if deleting $n - 4$ digits yields the number $2023$. How many $n$-digit rabbit numbers are there? | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | For n < 4: 0. For n ≥ 4: 3·4^{n−1} − 3^{n−4} [ 2·C(n−1,3) + 9·C(n−1,2) + 27·(n−1) + 81 ]. | |
0idh | Problem:
Find the smallest integer $n$ such that $\sqrt{n+99}-\sqrt{n}<1$. | [
"Solution:\nThis is equivalent to\n$$\n\\begin{aligned}\n\\sqrt{n+99} &< \\sqrt{n} + 1 \\\\\nn+99 &< n + 1 + 2\\sqrt{n} \\\\\n99 &< 1 + 2\\sqrt{n} \\\\\n98 &< 2\\sqrt{n} \\\\\n49 &< \\sqrt{n}\n\\end{aligned}\n$$\nSo the smallest integer $n$ with this property is $49^{2} + 1 = 2402$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2402 | |
02y5 | Problem:
Seja $m=999 \ldots 99$ o número formado por 77 dígitos iguais a 9 e seja $n=777 \ldots 77$ o número formado por 99 dígitos iguais a 7. Qual o número de dígitos de $m \cdot n$ ? | [
"Solution:\n\nComo $m+1=10^{77}$, perceba que:\n$$\n\\begin{aligned}\nm \\cdot n & =(m+1) \\cdot n-n \\\\\n& =\\underbrace{777 \\ldots 77}_{99} \\underbrace{000 \\ldots 00}_{77}-\\underbrace{777 \\ldots 77}_{99} .\n\\end{aligned}\n$$\nComo $\\underbrace{777 \\ldots 77}_{99} \\underbrace{000 \\ldots 00}_{77}$ possui... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | final answer only | 176 | |
00ut | For each positive integer $n$, denote by $\omega(n)$ the number of prime divisors of $n$. Find all polynomials $f(x)$ with integer coefficients, such that if $n$ is a positive integer satisfying $\omega(n) > 2023^{2023}$, then $f(n)$ is also a positive integer with
$$
\omega(f(n)) \le \omega(n).
$$ | [
"Answer: All polynomials of the form $f(x) = x^m$ for some $m \\in \\mathbb{Z}^+$ and $f(x) = c$ for some $c \\in \\mathbb{Z}^+$ with $\\omega(c) \\le 2023^{2023} + 1$.\n\nFirst of all we prove the following (well-known) Lemma.\n\n*Lemma.* Let $f(x)$ be a non-constant polynomial with integer coefficients. Then, the... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | All and only polynomials of the form f(x) = x^m with m a positive integer, and constant polynomials f(x) = c with c a positive integer satisfying ω(c) ≤ 2023^{2023} + 1. | |
0gjz | Show that $\sum_{k=0}^{n} (-1)^k \binom{2n+1}{2k+1} 2008^k$ is not divisible by $19$ for every positive integer $n$. | [
"Observe that $-2008 \\equiv 6 \\equiv 5^2 \\pmod{19}$. Thus,\n$$\n\\begin{aligned}\n2 \\sum_{k=0}^{n} \\binom{2n+1}{2k+1} (-2008)^k &\\equiv 2 \\sum_{k=0}^{n} \\binom{2n+1}{2k+1} 5^{2k} \\pmod{19} \\\\\n&\\equiv (1+5)^{2n+1} - (1-5)^{2n+1} \\pmod{19} \\\\\n&\\equiv 6^{2n+1} + 4^{2n+1} \\\\\n&\\equiv 2^{2n+1} (3^{2... | Thailand | Thai Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0b7u | We say that a ring $A$ has property (P) if any non-zero element can be written uniquely as the sum of an invertible element and a non-invertible element.
a) If in $A$, $1 + 1 \neq 0$, prove that $A$ has property (P) if and only if $A$ is a field.
b) Give an example of a ring that is not a field, containing at least t... | [
"a) If $A$ is a field and $x \\in A$, $x \\ne 0$, then $x$ is invertible and $x = x + 0$; this representation is unique, since $0$ is the only noninvertible element.\n\nAssume $A$ is not a field. Let $x \\in A$, $x \\ne 0$, be a noninvertible element. Since $1 + x = (1 + x) + 0$ and $-1 + x = (-1 + x) + 0$, the ele... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Field Theory"
] | English | proof only | null | |
03hs | Problem:
Two grade seven students were allowed to enter a chess tournament otherwise composed of grade eight students. Each contestant played once with each other contestant and received one point for a win, one half point for a tie and zero for a loss. The two grade seven students together gained a total of eight poi... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 7 or 14; the solution is not unique | |
0kss | Problem:
Let $n$ be a nonnegative integer and let $r$ be an odd number. Show that there is some $0 \leq i < 2^{n}$ such that
$$
\binom{2^{n}+i}{i} \equiv r \pmod{2^{n+1}}.
$$ | [
"Solution:\nWe can write $\\binom{2^{n}+i}{i}$ as $\\prod_{k=1}^{i} \\frac{2^{n}+k}{k}$. For any $k \\leq i < 2^{n}$, the number of times $2$ divides $2^{n}+k$ is just the number of times $2$ divides $k$, so this product must have an equal number of factors of $2$ in the numerator and denominator, and therefore mus... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
01lh | A function $f: \mathbb{R} \to \mathbb{R}$ satisfies the equality $f(f(x)) = x f(x) + x - 1$ for all real numbers $x$.
a) Find $f(-1)$.
b) Find all possible values of $f(1)$. | [
"Let the function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfy the equality\n$$\nf(f(x)) = x f(x) + x - 1 \\quad (*)\n$$\nfor all $x \\in \\mathbb{R}$.\n\na) Set $c = f(0)$. We have $f(c) = f(f(0)) = 0 \\cdot f(0) + 0 - 1 = -1$. So\n$$\n-1 = f(c). \\quad (1)\n$$\nTherefore, taking into account $(*)$ for $x = c$ and (1... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(-1) = -1; f(1) ∈ {-1, 1} | |
049u | Determine the minimal value of $\sin(x + 3) - \sin(x + 1) - 2\cos(x + 2)$ if $x \in \mathbb{R}$. | [] | Croatia | CroatianCompetitions2011 | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | 2(sin 1 - 1) | |
04bl | Determine all values of the real parameter $m$ for which the equation
$$
(m - 1)x^2 - 2mx + 2 = 0
$$
has no real solutions. | [
"Let us consider the quadratic equation:\n$$\n(m - 1)x^2 - 2mx + 2 = 0.\n$$\nThis equation has no real solutions if and only if its discriminant is negative.\n\nThe discriminant $D$ is:\n$$\nD = [-2m]^2 - 4(m-1) \\cdot 2 = 4m^2 - 8(m-1) = 4m^2 - 8m + 8.\n$$\nWe require:\n$$\n4m^2 - 8m + 8 < 0.\n$$\nDivide both side... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | no real values of m | |
09ru | Problem:
Vind alle niet-negatieve gehele getallen $n$ waarvoor er gehele getallen $a$ en $b$ bestaan met $n^{2}=a+b$ en $n^{3}=a^{2}+b^{2}$. | [
"Solution:\n\nOplossing I. Vanwege de ongelijkheid van het rekenkundig en meetkundig gemiddelde, toegepast op $a^{2}$ en $b^{2}$, geldt $a^{2}+b^{2} \\geq 2 a b$. Aangezien $2 a b=(a+b)^{2}-\\left(a^{2}+b^{2}\\right)$, volgt hieruit $n^{3} \\geq\\left(n^{2}\\right)^{2}-n^{3}$, oftewel $2 n^{3} \\geq n^{4}$. Dit bet... | Netherlands | Selectietoets | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 0, 1, 2 | |
02hy | Problem:
Sete equipes, divididas em dois grupos, participaram do torneio de futebol do meu bairro.
O grupo 1 foi formado pelas equipes Avaqui, Botágua e Corinense.
O grupo 2 foi formado pelas equipes Dinossauros, Esquisitos, Flurinthians e Guaraná.
Na primeira rodada do torneio, cada equipe enfrentou cada uma das equi... | [
"Solution:\n\na) Foram disputadas 3 partidas que são: $A \\times B$, $B \\times C$, $C \\times A$.\n\nb) Foram disputadas 6 partidas que são: $D \\times E$, $D \\times F$, $D \\times G$, $E \\times F$, $E \\times G$, $F \\times G$\n\nc) Na segunda rodada, cada equipe do grupo 1 joga 4 partidas; uma com cada equipe ... | Brazil | Brazilian Mathematical Olympiad | [
"Math Word Problems",
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | a) 3; b) 6; c) 12 | |
0a7x | Problem:
The circle whose diameter is the altitude dropped from the vertex $A$ of the triangle $A B C$ intersects the sides $A B$ and $A C$ at $D$ and $E$, respectively $(A \neq D, A \neq E)$. Show that the circumcentre of $A B C$ lies on the altitude dropped from the vertex $A$ of the triangle $A D E$, or on its exte... | [
"Solution:\n\n(See Figure 8.) Let $A F$ be the altitude of $A B C$. We may assume that $\\angle A C B$ is sharp. From the right triangles $A C F$ and $A F E$ we obtain $\\angle A F E = \\angle A C F$. $\\angle A D E$ and $\\angle A F E$ subtend the same arc, so they are equal. Thus $\\angle A C B = \\angle A D E$, ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 10 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and... | null | proof only | null | |
0639 | Problem:
Eine natürliche Zahl $n$ habe die folgende Eigenschaft:
Für beliebige reelle Zahlen $a_{1}, a_{2}, \ldots, a_{d}$, die sowohl $a_{1}+a_{2}+\ldots+a_{d}=2013$ als auch $0 \leq a_{i} \leq 1$ für $i=1,2, \ldots, d$ erfüllen, existiert eine Zerlegung der Menge dieser reeller Zahlen in $n$ paarweise disjunkte Teil... | [
"Solution:\n\nDie kleinste Zahl $n$ mit dieser Eigenschaft ist $4025$.\n\nWir zeigen zunächst $n \\geq 4025$. Dazu wählen wir $d=4025$ sowie $a_{1}=\\ldots=a_{4025}=\\frac{2013}{4025}>\\frac{1}{2}$. Dann ist $a_{1}+\\ldots+a_{4025}=2013$ und wegen $a_{i}+a_{j}=\\frac{4026}{4025}>1$ für alle $1 \\leq i \\neq j \\leq... | Germany | Germany TST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 4025 | |
0jxk | Problem:
Let $P$ and $A$ denote the perimeter and area respectively of a right triangle with relatively prime integer side-lengths. Find the largest possible integral value of $\frac{P^{2}}{A}$ | [
"Solution:\nAssume WLOG that the side lengths of the triangle are pairwise coprime. Then they can be written as $m^{2}-n^{2}$, $2 m n$, $m^{2}+n^{2}$ for some coprime integers $m$ and $n$ where $m>n$ and $m n$ is even. Then we obtain\n$$\n\\frac{P^{2}}{A}=\\frac{4 m(m+n)}{n(m-n)}\n$$\nBut $n$, $m-n$, $m$, $m+n$ are... | United States | February 2017 | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 45 | |
0g5m | 已知 $f(x)$ 為整係數多項式滿足:
對任意正整數 $n$, $f(n)$ 非零且 $f(n)$ 至多有 2011 個質因數不是 $n$ 的質因數。
試證明 $f$ 可以表示成 $f(x) = cx^k$, 其中 $c$ 為整數, $k$ 為非負整數。 | [
"對於題目所述的這樣的 $f$, 若 $f$ 的常數項為 0, 則可考慮 $g(x) = f(x)/x$ 亦滿足題目條件, 因此可假設 $f$ 的常數項非零, 並僅須證明此時 $f$ 是常數多項式。\n令 $P = \\{p$ 是質數 $|$ 存在正整數 $n$ 使得 $f(n)$ 是 $p$ 的倍數, 但 $n$ 不是 $p$ 的倍數 $\\}$。我們宣稱 $P$ 至多包含 2011 個質數。\n若不然, 用反證法, 令 $p_1, \\cdots, p_{2012}$ 為 $P$ 中的相異質數, 令正整數 $n_1, \\cdots, n_{2012}$ 分別滿足 $p_i|f(n_i)$ 且 $p_i \\nmid n... | Taiwan | 二〇一一數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0fvo | Problem:
Bestimme alle Funktionen $f: \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ mit folgenden Eigenschaften:
a) $f(1)=0$,
b) $f(x)>0$ für alle $x>1$,
c) Für alle $x, y \geq 0$ mit $x+y>0$ gilt
$$
f(x f(y)) f(y)=f\left(\frac{x y}{x+y}\right) .
$$ | [
"Solution:\n\nMit $y=1$ folgt aus (a) und (c) für alle $x \\geq 0$\n$$\n0=f(x f(1)) f(1)=f\\left(\\frac{x}{x+1}\\right)\n$$\nDer Ausdruck $\\frac{x}{x+1}$ nimmt alle Werte im Intervall $[0,1[$ an, wenn $x$ alle nichtnegativen reellen Zahlen durchläuft. Somit gilt zusammen mit (a)\n$$\nf(x)=0, \\quad 0 \\leq x \\leq... | Switzerland | SMO Finalrunde | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 0 for 0 ≤ x ≤ 1; f(x) = 1 − 1/x for x > 1 | |
0gqh | In a triangle $ABC$, the external bisector of $\angle BAC$ intersects the ray $[BC$ at $D$. The feet of the perpendiculars from $B$ and $C$ to the line $AD$ are $E$ and $F$ respectively, and the foot of the perpendicular from $D$ to $AC$ is $G$. Show that $\angle DGE + \angle DGF = 180^\circ$. | [
"\n\nLet $GD$ and $EB$ intersect at the point $P$. Since $\\angle PEA = \\angle PGA = 90^\\circ$, the points $P$, $G$, $A$, $E$ are concyclic and hence $\\angle EPA = \\angle EGA$. Since $AD$ is the external bisector of $\\angle BAC$, we have $\\angle GAD = \\angle BAE$ and hence $\\angle E... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0iyj | Problem:
Triangle $A B C$ has side lengths $A B=231$, $B C=160$, and $A C=281$. Point $D$ is constructed on the opposite side of line $A C$ as point $B$ such that $A D=178$ and $C D=153$. Compute the distance from $B$ to the midpoint of segment $A D$. | [
"Solution:\n\nNote that $\\angle A B C$ is right since\n\n$$\nB C^{2}=160^{2}=50 \\cdot 512=(A C-A B) \\cdot(A C+A B)=A C^{2}-A B^{2}\n$$\n\nConstruct point $B'$ such that $A B C B'$ is a rectangle, and construct $D'$ on segment $B' C$ such that $A D=A D'$. Then\n\n$$\nB' D'^{2}=A D'^{2}-A B'^{2}=A D^{2}-B C^{2}=(A... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 208 | |
04nu | Let $\overline{BD}$ and $\overline{CE}$ be the altitudes of an acute-angled triangle $ABC$. The circle with diameter $\overline{AC}$ meets $\overline{BD}$ at $F$. The circle with diameter $\overline{AB}$ meets the line $CE$ at points $G$ and $H$, where $G$ is between $C$ and $E$. If $\angle CHF = 12^\circ$, find the me... | [
"The chord $\\overline{GH}$ is perpendicular to $\\overline{AB}$, so $AB$ is the bisector of the segment $\\overline{GH}$. Hence $|AG| = |AH|$.\n\nSince $AFC$ is a right-angled triangle, Euclid's theorem gives us $|AF|^2 = |AD| \\cdot |AC|$.\n\nAnalogously, since $ABG$ is a right-angled triangle, we have $|AG|^2 = ... | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
... | English | proof and answer | 78° | |
0ezq | Problem:
Let $f(x, y) = x^2 + x y + y^2$. Show that given any real $x$, $y$ one can always find integers $m$, $n$ such that $f(x - m, y - n) \leq 1/3$. What is the corresponding result if $f(x, y) = x^2 + a x y + y^2$ with $0 \leq a \leq 2$? | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis... | null | proof and answer | For f(x, y) = x^2 + x y + y^2, there exist integers m, n such that f(x − m, y − n) ≤ 1/3 for all real x, y. More generally, for f(x, y) = x^2 + a x y + y^2 with 0 ≤ a ≤ 2, there exist integers m, n such that f(x − m, y − n) ≤ 1/(2 + a) for all real x, y. | |
04hk | Let $ABC$ be an acute triangle such that $|AC| > |BC|$. Let $H$ be the orthocentre of that triangle, $N$ the foot of the altitude from $B$, and $P$ the midpoint of the side $AB$. The circumcircles of the triangles $ABC$ and $CHN$ intersect in $C$ and $D$. Prove that the points $B, D, N$ and $P$ lie on the same circle. | [
"Denote $\\angle BAC = \\alpha$.\n\nSince $ABN$ is a right-angled triangle and $P$ is the midpoint of its hypotenuse, we have $\\angle BPN = 2\\alpha$. On the other hand,\n$$\n\\begin{align*}\n\\angle NDB &= \\angle CDB - \\angle CDN \\\\\n&= (180^\\circ - \\angle BAC) - \\angle CHN \\quad ... | Croatia | Mathematica competitions in Croatia | [
"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"
] | English | proof only | null | |
03aq | Let $a$, $b$, $c$ and $d$ be positive real numbers. Prove the inequality
$$
\left(\frac{a}{a+b}\right)^{5} + \left(\frac{b}{b+c}\right)^{5} + \left(\frac{c}{c+d}\right)^{5} + \left(\frac{d}{d+a}\right)^{5} \ge \frac{1}{8}
$$ | [] | Bulgaria | Selection test for 51. International Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
046g | Fix a prime number $p \ge 5$, and put $\Omega = \{1, 2, \dots, p\}$. For any $x, y \in \Omega$, define
$$
r(x, y) = \begin{cases} y - x, & \text{if } y \ge x, \\ y - x + p, & \text{if } y < x. \end{cases}
$$
For a nonempty subset $A$ of $\Omega$, define
$$
f(A) = \sum_{x \in A} \sum_{y \in A} (r(x, y))^2.
$$
We say tha... | [
"(1) For an intuitive understanding, place numbers $1, 2, \\dots, p$ equidistantly in a clockwise direction on the circumference of a circle with a perimeter exactly equal to $p$. Then $r(x, y)$ is precisely the distance from $x$ to $y$ in a clockwise direction.\nFor an $m$-element subset $A = \\{x_1, x_2, \\dots, ... | China | Chinese Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2 floor(log2(p+1)) | |
06la | If $57a + 88b + 125c \ge 1148$, where $a, b, c > 0$, what is the minimum value of
$$
a^3 + b^3 + c^3 + 5a^2 + 5b^2 + 5c^2?
$$ | [
"The answer is $466$.\nNote that\n$$\na^3 + 5a^2 - 57a + 99 = (a + 11)(a - 3)^2 \\ge 0,\n$$\n$$\nb^3 + 5b^2 - 88b + 208 = (b + 13)(b - 4)^2 \\ge 0,\n$$\n$$\nc^3 + 5c^2 - 125c + 375 = (c + 15)(c - 5)^2 \\ge 0.\n$$\nAdding these inequalities, we find that\n$$\na^3 + b^3 + c^3 + 5a^2 + 5b^2 + 5c^2 \\ge 57a + 88b + 125... | Hong Kong | HKG TST | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 466 | |
0hrv | Problem:
Find the remainder when $10^{2^{0}} + 10^{2^{1}} + \cdots + 10^{2^{2021}}$ is divided by $\underbrace{44 \cdots 44}_{44}$. | [
"Solution:\nLet $N = 10^{2^{0}} + 10^{2^{1}} + \\cdots + 10^{2^{2021}}$. By the Chinese Remainder Theorem (CRT), it suffices to find $N \\bmod 4$ and $N \\bmod M = \\underbrace{11 \\cdots 11}_{44}$.\n\nModulo $4$, the remainder is $2$.\n\nModulo $M = \\frac{10^{44} - 1}{9}$, $10^{44} \\equiv 1$, so $10^{2^{i}} \\bm... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | final answer only | 2020202020202020202020202020202020202020110 | |
070v | Problem:
Given two points $A$ and $B$, take $C$ on the perpendicular bisector of $AB$. Define the sequence $C_{1}, C_{2}, C_{3}, \ldots$ as follows. $C_{1}=C$. If $C_{n}$ is not on $AB$, then $C_{n+1}$ is the circumcenter of the triangle $ABC_{n}$. If $C_{n}$ lies on $AB$, then $C_{n+1}$ is not defined and the sequenc... | [
"Solution:\n\nAnswer: any $C$ such that $\\angle ACB=180^{\\circ} r / s$, with $r$ and $s$ relatively prime integers and $s$ not a power of $2$.\n\nLet $\\angle AC_{n}B = x_{n}$, where the angle is measured clockwise, so that $x_{n}$ is positive on one side of $AB$ and negative on the other side. Then $x_{n}$ uniqu... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | All points C on the perpendicular bisector such that ∠ACB = 180°·r/s with r and s coprime integers and s not a power of 2. | |
09qi | Problem:
Laat $n \geq 2$ en $k \geq 1$ gehele getallen zijn. In een land zijn $n$ steden en tussen elk paar steden is een busverbinding in twee richtingen. Laat $A$ en $B$ twee verschillende steden zijn. Bewijs dat het aantal manieren waarop je van $A$ naar $B$ kunt reizen met precies $k$ bussen gelijk is aan
$$
\frac{... | [
"Solution:\nZij $\\alpha(k)$ het aantal manieren om van stad $A$ naar stad $B \\neq A$ te reizen met $k$ bussen. Zij $\\beta(k)$ het aantal manieren om van stad $A$ naar stad $A$ te reizen met $k$ bussen. Als we beginnen in stad $A$ en daarna $k$ keer een bus nemen, dan kan dat op $(n-1)^{k}$ manieren. In $\\beta(k... | Netherlands | Dutch TST | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0emh | Let $a$, $b$, $c$, $d > 0$. Find all possible values of the sum
$$
S = \frac{a}{d+a+b} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{c+d+a}.
$$ | [
"Observe that\n$$\n\\begin{aligned}\nS &> \\frac{a}{a+b+c+d} + \\frac{b}{a+b+c+d} + \\frac{c}{a+b+c+d} + \\frac{d}{a+b+c+d} = 1, \\\\\nS &< \\frac{a}{a+b} + \\frac{b}{a+b} + \\frac{c}{c+d} + \\frac{d}{c+d} = 2.\n\\end{aligned}\n$$\nThe function changes smoothly as we vary $a$, $b$, $c$ and $d$. We will prove that i... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (1, 2) | |
0awt | Problem:
How many ways are there to arrange four $3$'s and two $5$'s into a six-digit number divisible by $11$? | [
"Solution:\n\nFrom the divisibility rule for $11$, we know that the difference of the sum of the odd-positioned digits and the even-positioned digits must be equal to a multiple of $11$. The only way this can happen here is that if $3$'s and $5$'s are equally distributed over odd and even positions, i.e., two $3$'s... | Philippines | Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 9 | |
0ku5 | Problem:
It is midnight on April 29th, and Abigail is listening to a song by her favorite artist while staring at her clock, which has an hour, minute, and second hand. These hands move continuously. Between two consecutive midnights, compute the number of times the hour, minute, and second hands form two equal angles... | [
"Solution:\n\nLet $t \\in [0,2]$ represent the position of the hour hand, i.e., how many full revolutions it has made. Then, the position of the minute hand is $12 t$ (it makes 12 full revolutions per 1 revolution of the hour hand), and the position of the second hand is $720 t$ (it makes 60 full revolutions per 1 ... | United States | HMMT November | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 5700 | |
045h | Fix two positive integers $m$ and $n$. Fix a way to color the vertices of a regular $(2m+2n)$-gon so that $2m$ of them are black and the other $2n$ are white. Define the coloring distance $d(B, C)$ between two black points $B$ and $C$ to be the lesser of the numbers of white points on either side of the line $BC$; simi... | [
"**Proof:** Consider $2m+2n$ vertices placed on the unit circle. The Euclidean distances between vertices are irrelevant to the discussion. We define the line segments in the white pairing scheme as white segments and those in the black pairing scheme as black segments.\n\n**Lemma 1:** For any black pairing scheme ... | China | Chinese Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null |
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