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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07oc | Let $N_1 = 12$. The goal in this problem is to obtain the number $N_k = 21$ after performing a finite number of steps according to the following rules. In each step, the numbers $N_i$ and $N_{i+1}$ need to be in *Shamrock Relation*. Two numbers are said to be in *Shamrock Relation* if and only if
(a) one number is obta... | [
"The answer is yes. Here is one way to proceed.\n\n$$\n12 = (12)\n$$\nuse (a) with $N=12, X=12$\n$$\n121212 = (1)(21)(21)2\n$$\nuse (b) with $M = 121212, X = 21, Y = 1$\n$$\n212112 = 212(1)(1)(2)\n$$\nuse (b) with $N = 212112, X = 1, Y = 2$\n$$\n212211 = 2(1)(2)(2)11\n$$\nuse (b) with $M = 212211, X = 2, Y = 1$\n$$... | Ireland | Ireland | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Other"
] | null | proof and answer | Yes | |
0htk | Problem:
The positive integers from $1$ to $100$ are written, in some order, in a $10 \times 10$ square. In each row, the five smallest numbers are crossed out. In each column, the five largest numbers (including those that have already been crossed out) are circled. Prove that at least $25$ numbers will be circled bu... | [
"Solution:\n\nWe generalize to the following:\nLet $m \\geq p$, $n \\geq q$ be integers. Each square of an $m$-row, $n$-column grid is filled with a different positive integer.\nThen at least $p q$ numbers are among the $p$ largest numbers in their columns and the $q$ largest numbers in their rows.\nWe prove this l... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
027g | Problem:
Uma sequência especial - Na sequência $1, 3, 2, \ldots$ cada termo depois dos dois primeiros é igual ao termo precedente, subtraído do termo que o precede, ou seja, se $n > 2$, então $a_{n} = a_{n-1} - a_{n-2}$. Qual é a soma dos cem primeiros termos dessa sequência? | [
"Solution:\n\nInicialmente escrevemos os primeiros termos dessa sequência, como segue.\n$$\n1, 3, 2, -1, -3, -2, 1, 3, 2, \\ldots\n$$\nO sétimo e o oitavo termos são, respectivamente, iguais ao primeiro e ao segundo. Isso significa que a sequência se repete de seis em seis termos. A soma dos seis primeiros termos é... | Brazil | Nível 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 5 | |
03km | Problem:
Let $\triangle ABC$ be an isosceles triangle with $AB = AC$. Suppose that the angle bisector of $\angle B$ meets $AC$ at $D$ and that $BC = BD + AD$. Determine $\angle A$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 100° | |
041x | Suppose that 12 acrobats labeled $1$–$12$ are divided into two circles $A$ and $B$, with six persons in each. Let each acrobat in $B$ stand on the shoulders of two adjacent acrobats of $A$. We call it a *tower* if the label of each acrobat of $B$ is equal to the sum of the labels of the acrobats under his feet. How man... | [
"Denote the sum of labels of $A$ and $B$ by $x$ and $y$, respectively. Then $y = 2x$. Thus, we have\n$$\n3x = x + y = 1 + 2 + \\cdots + 12 = 78, \\quad x = 26.\n$$\nObviously, $1, 2 \\in A$ and $11, 12 \\in B$. Denote $A = \\{1, 2, a, b, c, d\\}$, where $a < b < c < d$. Then $a + b + c + d = 23$, and $a \\ge 3$, $8... | China | China Southeastern Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 6 | |
01pg | Ann and Bob play the following game. They, in turn (Ann starts), replace one of the asterisks in the expression $\overline{*****}$ by one of the digits from $0$ to $9$ (each digit can be used exactly one time). Ann wins if the obtained $5$-digit number is divisible by $11$, otherwise Bob wins.
Who of the players wins i... | [
"Answer: Bob wins.\n\nIt is well-known that a natural number $n$ is divisible by $11$ if and only if $(S_o - S_e) \\div 11$, where $S_o, S_e$ are the sums of the digits on the odd and even positions respectively in the decimal representation of $n$.\n\nLet Ann use the numbers $a$, $c$, and $y$ for her first, second... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | English | proof and answer | Bob wins | |
03vz | For positive integer $n$, let $f_n = [2^n \sqrt{2008}] + [2^n \sqrt{2009}]$. Prove that there are infinitely many odd numbers and even numbers in the sequence $f_1, f_2, \dots$ ($[x]$ represents the biggest integer that does not exceed $x$.) (Posed by Zuming Feng) | [
"We use the dyadic representations of $\\sqrt{2008}$ and $\\sqrt{2009}$:\n$$\n\\sqrt{2008} = \\overline{01}1100.\\overline{a_1}a_2\\cdots\\overline{(2)}, \\quad \\sqrt{2009} = \\overline{01}1100.b_1b_2\\cdots\\overline{(2)}.\n$$\nFirst, we prove that there are infinitely many even numbers by contradiction. Suppose ... | China | China Girls' Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof only | null | |
079l | Consider that we have omitted some of the points of a 2d-lattice (same as $\mathbb{Z}^2$). Look at these points as a graph, two points (vertices) are connected if they are the same in one of their coordinates and differ 1 in another. Each connected component is called a *cluster*. Suppose that for all $n \in \mathbb{N}... | [
"We suppose that the origin is not omitted without loss of generality. Consider the cluster having the origin, if the cluster is finite, its border contains some omitted vertices that are adjacent to at least one of the vertices of the cluster. Among all of the points of the cluster having origin, consider the poin... | Iran | 27th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0dg1 | Prove there exist infinitely many positive integers divisible by $2021$ and each of them containing the same number of digits $0, 1, \ldots, 9$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | English | proof only | null | |
0gzi | Let $P(x)$, $Q(x)$ and $R(x)$ be polynomials, such that $Q(x)$ and $R(x)$ take nonnegative values only. It is known, that the equation
$$
P(x) + \sqrt{Q(x)} + \sqrt{Q(x) + \sqrt{R(x)}} = 0
$$
has infinitely many solutions. Is it true, that every real number is a root of this equation? | [
"Not necessary.\n\nWe construct the following example: $P(x) = x$, $Q(x) = \\frac{1}{4}x^2$, $\\sqrt{Q(x)} = \\frac{1}{2}|x|$, $R(x) = 0$. Then, $\\sqrt{Q(x) + \\sqrt{R(x)}} = \\frac{1}{2}|x|$ and we have: \n$$\nP(x) + \\sqrt{Q(x)} + \\sqrt{Q(x) + \\sqrt{R(x)}} = x + |x| = \\begin{cases} 2x, & x \\ge 0, \\\\ 0, & x... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010) | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | No | |
0l3w | Let $N$ be the product of all the positive integer divisors of $42$. What is the units digit of $N$?
(A) 0 (B) 2 (C) 4 (D) 6 (E) 8 | [
"**Answer (D):** The prime factorization of $42$ is $2 \\cdot 3 \\cdot 7$. A divisor of $42$ is determined by choosing whether or not to include each prime as a factor of the divisor, so there are $2^3 = 8$ divisors. Furthermore, the divisors can be formed into $4$ pairs each of whose product is $42$: $1 \\cdot 42$... | United States | AMC 10 B | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
0jrt | Problem:
Let $x$ and $y$ be complex numbers such that $x+y=\sqrt{20}$ and $x^{2}+y^{2}=15$. Compute $|x-y|$. | [
"Solution:\n\nWe have $(x-y)^{2}+(x+y)^{2}=2\\left(x^{2}+y^{2}\\right)$, so $(x-y)^{2}=10$, hence $|x-y|=\\sqrt{10}$."
] | United States | HMMT February | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | sqrt(10) | |
0jif | Problem:
For an integer $n \geq 0$, let $f(n)$ be the smallest possible value of $|x+y|$, where $x$ and $y$ are integers such that $3x - 2y = n$. Evaluate $f(0) + f(1) + f(2) + \cdots + f(2013)$. | [
"Solution:\n\nAnswer: $2416$\n\nFirst, we can use $3x - 2y = n$ to get $x = \\frac{n + 2y}{3}$. Thus $|x + y| = \\left| \\frac{n + 5y}{3} \\right|$. Given a certain $n$, the only restriction on $y$ is that $3 \\mid n + 2y \\Longleftrightarrow 3 \\mid n + 5y$. Hence the set of possible $x + y$ equals the set of inte... | United States | HMMT November | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2416 | |
022g | Problem:
Um triângulo isósceles tem base medindo $10~\mathrm{cm}$ e dois lados iguais a $13~\mathrm{cm}$. É possível mudar a base do triângulo e obter outro triângulo isósceles com mesma área?
 | [
"Solution:\n\nEm um triângulo isósceles, a altura relativa à base coincide com a mediana. Traçando esta altura, obtemos dois triângulos retângulos com catetos medindo $h$ e $5$, e hipotenusa $13$. Pelo Teorema de Pitágoras temos:\n$$\nh^{2} + 5^{2} = 13^{2} \\Rightarrow h^{2} = 13^{2} - 5^{2} = 144 \\Rightarrow h =... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Yes; for example, an isosceles triangle with base 24 cm and equal sides 13 cm has the same area. | |
0104 | Problem:
Let $P$ and $Q$ be polynomials with integer coefficients. Suppose that the integers $a$ and $a+1997$ are roots of $P$, and that $Q(1998)=2000$. Prove that the equation $Q(P(x))=1$ has no integer solutions. | [
"Solution:\n\nSuppose $b$ is an integer such that $Q(P(b))=1$. Since $a$ and $a+1997$ are roots of $P$ we have $P(x) = (x-a)(x-a-1997) R(x)$ where $R$ is a polynomial with integer coefficients. For any integer $b$ the integers $b-a$ and $b-a-1997$ are of different parity and hence $P(b) = (b-a)(b-a-1997) R(b)$ is e... | Baltic Way | Baltic Way 1997 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
01vx | 1. The two lines with slopes $2$ and $1/2$ pass through an arbitrary point $T$ on the axis $Oy$ and intersect the hyperbola $y = 1/x$ at four points.
a) Prove that these four points lie on a circle.
b) The point $T$ runs through the entire $y$-axis. Find the locus of the centers of such circles. | [
"b) the required locus is the line $y = -1.6x$.\n\nLet $a$ be the ordinate of the point $T$. Then the lines from the problem condition have equations $y - a - 2x = 0$ and $y - a - \\frac{1}{2}x = 0$. The union of these lines is defined by $(y - a - 2x)(y - a - \\frac{1}{2}x) = 0$ which is equivalent to $(y - a)^2 +... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | y = -1.6x | |
064t | $ABCD$ is a convex cyclic quadrilateral. The diagonals $AC$ and $BD$ intersect at the point $E$. It is given that $AB = 39$, $AE = 45$, $AD = 60$ and $BC = 56$.
Determine the length of $CD$. | [
"From the similarity of triangles $BEC$ and $AED$ we $BE = 42$.\n\nWe put $DE = x$ and $CE = y$. From the similarity of triangles $ABE$ and $DEC$ we have\n$$\n\\frac{x}{AE} = \\frac{y}{BE} = \\frac{CD}{AB} \\Leftrightarrow \\frac{x}{15} = \\frac{y}{14} = \\frac{CD}{13} = t. \\quad (1)\n$$\nMoreover, from theorem of... | Greece | Mediterranean Mathematical Competition | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 91/5 | |
06ch | Let $ABCD$ be a trapezoid with $AB \parallel CD$, $AB > CD$ and $AD > BC$. Show that
i. $\angle CBA > \angle DAB$,
ii. $AC > BD$. | [
"i. Let $AD$ meet $BC$ at $E$, $AC$ meet $BD$ at $F$, and let $M$ be the midpoint of $AB$. Since $AB > CD$, the points $D$ and $C$ must lie on the segments $AE$ and $BE$ respectively.\nSince $\\triangle EDC \\sim \\triangle EAB$, we have $\\frac{EA}{EB} = \\frac{DA}{CB} > 1$. This implies $EA > EB$, and hence $\\an... | Hong Kong | Test 1 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04kh | Let $ABCD$ be a rectangle such that $|AB| : |AD| = 2 : 3$ and let $E$ be a point on the side $\overline{AD}$ such that $|AE| = |AB|$. Point $F$ lies on the ray $AB$ so that triangle $AFE$ and quadrilateral $CDEF$ have the same areas. Determine $|AB| : |BF|$. | [] | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 1:2 | |
040b | Let the focus and directrix of parabola $y^2 = 2px$ ($p > 0$) be $F$ and $l$, respectively. $A$ and $B$ are moving points on the parabola satisfying $\angle AFB = \frac{\pi}{3}$. Let the projection of $M$, the midpoint of segment $AB$, on $l$ be $N$. Then the maximum value of $\frac{|MN|}{|AB|}$ is ______. | [
"Suppose $\\angle ABF = \\theta$ ($0 < \\theta < \\frac{2\\pi}{3}$). Then by the Law of Sine, we have\n$$\n\\frac{|AF|}{\\sin \\theta} = \\frac{|BF|}{\\sin(\\frac{2\\pi}{3} - \\theta)} = \\frac{|AB|}{\\sin \\frac{\\pi}{3}}\n$$\nAnd then\n$$\n\\frac{|AF| + |BF|}{\\sin \\theta + \\sin(\\frac{2\\pi}{3} - \\theta)} = \... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 1 | |
05so | Problem:
Soit $ABC$ un triangle aux angles aigus. Soit $D$ le pied de la bissectrice de l'angle $\widehat{BAC}$. La perpendiculaire à la droite $(AD)$ passant par le point $B$ recoupe le cercle circonscrit au triangle $ADB$ en un point $E$. Soit $O$ le centre du cercle circonscrit au triangle $ABC$. Montrer que les po... | [
"Solution:\n\n\n\nExaminons les angles que nous pouvons obtenir facilement. Puisque les points $A$, $E$, $D$ et $B$ sont sur un même cercle, on doit pouvoir utiliser le théorème de l'angle inscrit pour avoir une égalité portant sur $\\widehat{EAD}$. Puisque la droite $(BE)$ est perpendicula... | France | Envoi 5: Pot Pourri | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gjp | 三角形 $ABC$ 的 $\angle BAC$ 角平分線、過 $B$ 點的中線以及 $AB$ 中垂線三線共點於 $X$。令 $H$ 為 $ABC$ 的垂心。證明 $\angle BXH = 90^\circ$。
Let $ABC$ be a triangle such that the angular bisector of $\angle BAC$, the $B$-median and the perpendicular bisector of $AB$ intersect at a single point $X$. Let $H$ be the orthocenter of $ABC$. Show that $\angl... | [
"Let $M$ be the midpoint of $AC$, and $X'$ be the reflection of $X$ with respect to $M$. Since $XA = XB$ and $AX$ is the angular bisector of $\\angle BAC$, we know that the circumcircle of $AXB$ is tangent to $AC$ and so $MA^2 = MX \\times MB$ as $M = BX \\cap AC$. Therefore $MA \\times MC = MX \\times MB$ and so $... | Taiwan | IMO 3J, Independent Study 2 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Chinese; English | proof only | null | |
03om | There are real numbers $a$, $b$ and $c$ such that $f(x) = x^3 + a x^2 + b x + c$ has three real roots $x_1$, $x_2$ and $x_3$ satisfying
$$
(1)\quad x_2 - x_1 = \lambda,
$$
$$
(2)\quad x_3 > \frac{1}{2}(x_1 + x_2).
$$
Find the maximum value of $\frac{2a^3 + 27c - 9ab}{\lambda^3}$. | [
"Let $S = \\frac{2a^3 + 27c - 9ab}{\\lambda^3}$, then\n$$\nS = \\frac{27\\left(\\frac{2}{27}a^3 - \\frac{1}{3}ab + c\\right)}{\\lambda^3} = \\frac{27f\\left(-\\frac{1}{3}a\\right)}{\\lambda^3} \\\\\n= \\frac{27\\left(-\\frac{1}{3}a - x_1\\right)\\left(-\\frac{1}{3}a - x_2\\right)\\left(-\\frac{1}{3}a - x_3\\right)}... | China | China Mathematical Competition (Extra Test) | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 3*sqrt(3)/2 | |
0arj | Problem:
$x + \frac{1}{x}$ has a maximum in $x < 0$ and a minimum in $x > 0$. Find the area of the rectangle whose sides are parallel to the axes and two of whose vertices are the maximum and minimum values of $x + \frac{1}{x}$. | [
"Solution:\n\n(ans. area $= 4 \\cdot 2 = 8$.\nVertex at maximum is $(-1, -2)$, vertex at minimum is $(1, 2)$. Thus, width is $2$ and height is $4$. Vertices are obtained from the inequalities $\\frac{(x+1)^2}{x} \\geq 0$ for $x > 0$ and $\\frac{(x-1)^2}{x} \\leq 0$ for $x < 0$.)"
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 8 | |
077b | Problem:
Suppose $P(x)$ is a polynomial with real coefficients satisfying the condition $P(\cos \theta + \sin \theta) = P(\cos \theta - \sin \theta)$, for every real $\theta$. Prove that $P(x)$ can be expressed in the form
$$
P(x) = a_{0} + a_{1} (1 - x^{2})^{2} + a_{2} (1 - x^{2})^{4} + \cdots + a_{n} (1 - x^{2})^{2n}... | [
"Solution:\nChanging $\\theta$ to $\\theta - \\pi / 2$, we see that\n$$\nP(\\sin \\theta + \\cos \\theta) = P(\\sin \\theta - \\cos \\theta)\n$$\nThis shows that $P(x) = P(-x)$ for all $x \\in [-\\sqrt{2}, \\sqrt{2}]$ and as $P$ is a polynomial, in fact,\n$$\nP(x) = P(-x)\n$$\nfor all $x \\in \\mathbb{R}$. Hence $P... | India | INMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
022h | Problem:
Seja $a$ um número inteiro positivo tal que $a$ é múltiplo de $5$, $a+1$ é múltiplo de $7$, $a+2$ é múltiplo de $9$ e $a+3$ é múltiplo de $11$. Determine o menor valor que $a$ pode assumir. | [
"Solution:\n\nAs condições do problema equivalem a dizer que:\n$$\n2a - 5 = 2(a+1) - 7 = 2(a+2) - 9 = 2(a+3) - 11\n$$\né múltiplo de $5$, $7$, $9$ e $11$, donde é múltiplo de $5 \\cdot 7 \\cdot 9 \\cdot 11 = 3465$. Assim, o menor valor de $a$ é tal que $2a - 5 = 3465$, ou seja, $a = 1735$."
] | Brazil | Nível 3 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 1735 | |
0gm0 | Prove that there are no integer pairs $(x, y)$ satisfying
$$
2560x^2 + 5x + 6 = y^5.
$$ | [
"Suppose for the sake of contradiction that an integer pair $(x, y)$ satisfies $2560x^2 + 5x + 6 = y^5$. Thus, $6 \\equiv y^5 \\equiv y \\pmod{5}$, and so $y \\equiv 1 \\pmod{5}$. Let $k$ be an integer such that $y = 5k + 1$. We now have\n$$\n2560x^2 + 5x + 6 = (5k + 1)^5 = 1 + \\sum_{j=1}^{5} \\binom{5}{j} (5k)^j.... | Thailand | The 14th Thailand Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
05u6 | Problem:
Trouver tous les triplets d'entiers naturels non nuls $(a, b, c)$ tels que
$$
2021^{a}+4=3^{b} \times 5^{c}
$$ | [
"Solution:\n\nDans une équation diophantienne, on commence toujours par regarder de petites valeurs des paramètres. Ici, il suffit de factoriser $2021^{a}+4$ en produit de facteurs premiers, donc on regarde les petites valeurs de $a$. Ici, lorsque $a=1$, l'équation devient $2025=3^{b} \\times 5^{c}$, donc $(a, b, c... | France | Préparation Olympique Française de Mathématiques | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | (1,4,2) | |
0fz8 | Problem:
Im Dreieck $ABC$ sei $\angle BAC = 60^{\circ}$. Sei $E$ ein Punkt auf der Geraden $AB$, sodass $B$ zwischen $A$ und $E$ liegt und $BE = BC$ gilt. Analog sei $F$ ein Punkt auf $AC$, sodass $C$ zwischen $A$ und $F$ liegt und $CF = BC$ gilt. Der Umkreis von $ACE$ schneide $EF$ in $K$. Zeige, dass $K$ auf der Win... | [
"Solution:\n\nSei $S$ der Schnittpunkt von $BF$ und $CE$. Es gilt (Aussenwinkelsatz, $BE = BC = CF$):\n$$\n\\angle CSF = \\angle BCS + \\angle CBS = \\frac{1}{2} \\angle ABC + \\frac{1}{2} \\angle ACB = \\frac{1}{2}(180^{\\circ} - \\angle CAB) = 60^{\\circ}\n$$\nHieraus folgt, dass $ABSC$ ein Sehnenviereck ist. Die... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
001i | El cuadrilátero $ABCD$ tiene sus diagonales perpendiculares y está inscrito en una circunferencia $\Gamma$ de centro $O$. Una recta paralela a $BD$ corta los segmentos $AO$ y $AD$ en $P$ y $Q$, respectivamente. Demuestre que las rectas $AP$ y $CQ$ se cortan en un punto de $\Gamma$. | [] | Argentina | XII Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | español | proof only | null | |
0kja | Inside a right circular cone with base radius $5$ and height $12$ are three congruent spheres each with radius $r$. Each sphere is tangent to the other two spheres and also tangent to the base and side of the cone. What is $r$?
(A) $\frac{3}{2}$ (B) $\frac{90 - 40\sqrt{3}}{11}$ (C) $2$ (D) $\frac{144 - 25\sqrt{3}}{... | [] | United States | AMC 10 A | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circl... | null | MCQ | B | |
0aav | A $3 \times 3$ square is divided into $9$ squares. A positive integer is inscribed in every square, such that the sums of the numbers in each row, column and diagonal are equal. Prove that this sum cannot be $2008$. | [
"Let $m$ be the sum in each row, column and diagonal. Then we have the following situation:\n\n| $x$ | $y$ | $m-x-y$ |\n|-------------|-------------|-------------|\n| $a$ | $b$ | $m-a-b$ |\n| $m-x-a$ | $m-b-y$ | $x+a+b+y-m$ |\n\nBecause the sum of the elements in the ... | North Macedonia | Macedonian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Other"
] | null | proof only | null | |
06cp | Suppose there are 6 members in an International Mathematical Olympiad team. Prove that among these 6 members there are three members who either all know each other or all don't know each other. | [
"Suppose the members are $A, B, C, D, E, F$. By the pigeonhole principle, either $A$ knows at least $\\left\\lfloor \\frac{5}{2} \\right\\rfloor = 3$ other members, or $A$ does not know at least 3 other members. WLOG assume it is the former case. Suppose $A$ knows $B, C, D$. If any two of $B, C, D$ know each other,... | Hong Kong | Test 1 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
06pk | The diagonals of a trapezoid $ABCD$ intersect at point $P$. Point $Q$ lies between the parallel lines $BC$ and $AD$ such that $\angle AQD = \angle CQB$, and line $CD$ separates points $P$ and $Q$. Prove that $\angle BQP = \angle DAQ$. | [
"Let $t = \\frac{AD}{BC}$. Consider the homothety $h$ with center $P$ and scale $-t$. Triangles $PDA$ and $PBC$ are similar with ratio $t$, hence $h(B) = D$ and $h(C) = A$.\n\n\n\nLet $Q' = h(Q)$ (see Figure 1). Then points $Q$, $P$ and $Q'$ are obviously collinear. Points $Q$ and $P$ lie o... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07h0 | *Morteza* has $100$ sets. At each step *Mahdi* can choose two distinct sets from them and Morteza tells him the intersection and union of those two sets. Find the least possible number of steps that *Mahdi* can determine all of $100$ sets. | [
"Suppose that the sets are $A_1, A_2, \\dots, A_{100}$. We claim that Mahdi needs $100$ steps. First of all, notice that $100$ would be enough; by calling $(1, 2)$, $(1, 3)$ and $(2, 3)$ he can understand $A_1$, $A_2$ and $A_3$. Because we have\n$$\nA_1 = ((A_1 \\cup A_2) - (A_2 \\cup A_3)) \\cup (A_1 \\cap A_2) \\... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 100 | |
0dx6 | Problem:
Na krožnico s premerom $10~\mathrm{cm}$ sta narisani obe tangenti iz točke $A$ izven krožnice. Točka $A$ je od središča krožnice oddaljena $1{,}3~\mathrm{dm}$. Izračunaj razdaljo med dotikališčema tangent na krožnico in jo izrazi v centimetrih. Rezultat zaokroži na eno decimalno mesto natančno. | [
"Solution:\n\nNarišemo skico: središče krožnice $S$, točka $A$ izven krožnice, dotikališči $B$ in $C$. Označimo kot $\\angle B S A$ z $\\alpha$, razdaljo od $S$ do $A$ z $x$ in razdaljo od $B$ do $C$ z $2y$. Ker je trikotnik $A S B$ pravokoten, uporabimo kotno funkcijo $\\cos \\alpha = \\frac{r}{x}$, iz česar izrač... | Slovenia | 6. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 9.2 cm | |
0eks | Problem:
V kvadratu $ABCD$ so včrtane krožnica $\mathcal{K}$, polkrožnica $\mathcal{P}$ in četrtina krožnice $\mathcal{Q}$. Četrtina krožnice $\mathcal{Q}$ ima središče v oglišču $A$ in vsebuje točki $B$ in $D$. Polkrožnica $\mathcal{P}$ ima središče v razpolovišču stranice $AD$ in vsebuje točki $A$ in $D$. Krožnica $... | [
"Solution:\n\n(a) Označimo dolžino stranice kvadrata $ABCD$ z $a$, polmer krožnice $\\mathcal{K}$ pa z $r$. Naj bo $S$ središče polkrožnice $\\mathcal{P}$, $T$ središče krožnice $\\mathcal{K}$ in $H$ pravokotna projekcija točke $T$ na stranico $AD$. Dolžino daljice $AG$ označimo z $x$. Ker je $|AT| = |AF| - |FT| = ... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | The radius equals one quarter of the side length, and the line through the tangency points is parallel to the side. | |
0714 | Problem:
Call a set of 3 distinct elements which are in arithmetic progression a trio. What is the largest number of trios that can be subsets of a set of $n$ distinct real numbers? | [
"Solution:\n\nLet $X$ be one of the elements. What is the largest number of trios that can have $X$ as middle element? Obviously, at most $\\max\\{b, a\\}$, where $b$ is the number of elements smaller than $X$ and $a$ is the number larger. Thus if $n=2m$, the number of trios is at most $0+1+2+\\ldots+m-1+m-1+m-2+\\... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | If the set size is 2m, the maximum is m(m−1); if the set size is 2m+1, the maximum is m^2. | |
0jqu | Problem:
Let $f:[0,1] \rightarrow \mathbb{C}$ be a nonconstant complex-valued function on the real interval $[0,1]$. Prove that there exists $\epsilon>0$ (possibly depending on $f$) such that for any polynomial $P$ with complex coefficients, there exists a complex number $z$ with $|z| \leq 1$ such that $|f(|z|)-P(z)| \... | [
"Solution:\nWe claim we can choose $\\epsilon(f)=\\max \\left(|f(x_{1})-f(x_{2})|\\right) / 2$. Fix $P$ and suppose for the sake of contradiction that for all $z$ with $|z| \\leq 1$ it is the case that $|f(z)-P(z)|<\\epsilon$. We can write $z=r e^{i \\theta}$ so that\n$$\n|f(r)-P(r e^{i \\theta})|<\\epsilon\n$$\nLe... | United States | HMMT February 2015 | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
006m | Axel y Franco juegan al siguiente juego. Inicialmente Axel piensa un número natural $N$. A partir de ahí, en cada jugada, Franco elige 4 números distintos $a, b, c, d$ del conjunto $\{1, 2, 3, 4, 5, 6, 7, 8\}$ y se los dice a Axel. A continuación Axel anuncia una de las sumas $N+a, N+b, N+c, N+d$, a su elección. (Por e... | [] | Argentina | Argentina 2008 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Spanish | proof and answer | 3 | |
083a | Problem:
In un'isola ci sono due categorie di persone: i cavalieri, che dicono sempre la verità, e i furfanti, che mentono sempre. Un certo giorno si riunisce il gran consiglio, costituito da 2003 membri. Essi si siedono a caso intorno ad una tavola rotonda, e durante la riunione ciascuno di essi dichiara: "Entrambi $i... | [
"Solution:\nIl malato era un cavaliere.\nDalla dichiarazione del primo giorno si deduce che accanto ad ogni cavaliere ci sono due furfanti, mentre non è possibile che ci siano due furfanti accanto ad un furfante. In particolare, non ci possono essere tre furfanti consecutivi, e pertanto, su 3 persone sedute consecu... | Italy | XIX Gara Nazionale di Matematica | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | The sick person was a knight. | |
06a1 | We consider the segment $AB$ and the point $\Gamma$ on it such that $AB = 3 \cdot AG$. We construct the parallelogram $A\Gamma\Delta E$ with $A\Gamma = \Delta E = \Gamma E > AE$. We consider point $Z$ on the side $A\Gamma$ such that $A\hat{E}Z = A\Gamma E = \omega$. Prove that the line perpendicular from $B$ to the lin... | [
"If $A\\Gamma E = \\omega$, from the right angled triangle $\\Gamma\\Theta B$ we have:\n$$\n\\Gamma \\hat{B} \\Theta = 90^{\\circ} - B \\hat{\\Gamma} \\Theta = 90^{\\circ} - \\omega, \\qquad (1)\n$$\nsince $B\\Gamma\\Theta = A\\Gamma E = \\omega$, and moreover $B\\hat{E}K = 90^{\\circ}$, and so\n$$\n\\Gamma \\hat{B... | Greece | 37th Hellenic Mathematical Olympiad 2020 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
05qe | Problem:
Trouver la somme des $n$ tels que $n^{2}+8 n+44$ soit un carré parfait. | [
"Solution:\n\nSoit $f$ la fonction étudiée, une idée dans ce genre d'exercice est de coincer $f(n)$ entre 2 carrés d'entiers consécutifs $k^{2}<f(n)<(k+1)^{2}$ et de dire par l'absurde que si $f(n)=i^{2}$ alors $k<i<k+1$ ce qui est impossible car on a des entiers.\n\nIci les calculs peuvent s'abréger en remarquant ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2 | |
0aga | Let every one of the numbers $x_1$, $x_2$, ..., $x_n$ be equal to $1$ or $-1$ and also:
$$
x_1x_2x_3x_4 + x_2x_3x_4x_5 + x_3x_4x_5x_6 + ... + x_{n-2}x_{n-1}x_nx_1 + x_{n-1}x_nx_1x_2 + x_nx_1x_2x_3 = 0
$$
Prove that $n$ is divisible by $4$. | [
"Let $y_k = x_k x_{k+1} x_{k+2} x_{k+3}$ (clearly $y_{n-2} = x_{n-2}x_{n-1}x_nx_1$, $y_{n-1} = x_{n-1}x_nx_1x_2$, $y_n = x_nx_1x_2x_3$). All $y_k$ are $1$ or $-1$. Then, by the conditions of the problem, we get $y_1 + ... + y_n = 0$. Therefore $n = 2k$ and moreover exactly $k$ of the numbers $y_1, ..., y_n$ are equ... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
03co | A convex $2019$-gon is given, for which no three diagonals intersect in a point. An intersection point of two diagonals, internal for the given polygon will be called “knot”. Two knots are called “neighbors” if they share a common diagonal. A closed path between neighboring knots, such that no three consecutive knots s... | [
"Answer: $\\frac{2019(2019-3)}{2} - 1 = 2035151$.\nLet us consider the more general problem, where $2019$ is replaced with $n \\ge 4$. We will prove that in a convex $n$-gon, neither three of which diagonals have a common point, we can color at most $\\frac{n(n-3)}{2} - 1$ of the knots without generating a colored ... | Bulgaria | 68. National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2035151 | |
0ibt | Problem:
I have written a strictly increasing sequence of six positive integers, such that each number (besides the first) is a multiple of the one before it, and the sum of all six numbers is $79$. What is the largest number in my sequence? | [
"Solution:\n\nIf the fourth number is $\\geq 12$, then the last three numbers must sum to at least $12 + 2 \\cdot 12 + 2^{2} \\cdot 12 = 84 > 79$. This is impossible, so the fourth number must be less than $12$. Then the only way we can have the required divisibilities among the first four numbers is if they are $1... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 48 | |
0jdo | Problem:
How many functions $f:\{1,2, \ldots, 2013\} \rightarrow \{1,2, \ldots, 2013\}$ satisfy $f(j) < f(i) + j - i$ for all integers $i, j$ such that $1 \leq i < j \leq 2013$? | [
"Solution:\nAnswer: $\\dbinom{4025}{2013}$\n\nNote that the given condition is equivalent to $f(j) - j < f(i) - i$ for all $1 \\leq i < j \\leq 2013$. Let $g(i) = f(i) - i$, so that the condition becomes $g(j) < g(i)$ for $i < j$ and $1 - i \\leq g(i) \\leq 2013 - i$. However, since $g$ is decreasing, we see by ind... | United States | HMMT November 2013 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | C(4025, 2013) | |
0dmk | Problem:
Одредити најмањи природан број који је дељив са $2009$ и коме је збир цифара једнак $2009$. | [
"Solution:\n\nПошто је $2009 = 223 \\cdot 9 + 2$, тражени број има бар $224$ цифре. Посматраћемо $224$-цифрене бројеве $x = \\overline{c_{223} c_{222} \\ldots c_{1} c_{0}}$. Јасно је да је $c_{223} \\geq 2$. Притом, ако је $c_{223} = 2$ онда је $c_{222} = \\cdots = c_{0} = 9$ и $x = 3 \\cdot 10^{223} - 1 \\equiv 3 ... | Serbia | Српска математичка олимпијада | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | The number with decimal form 4998 followed by 170 nines, then 8, then 49 nines (equivalently 5*10^223 − 10^220 − 10^49 − 1). | |
08nf | Problem:
If $a$, $b$, $c$, $d$ are integers and $A = 2(a-2b+c)^{4} + 2(b-2c+a)^{4} + 2(c-2a+b)^{4}$, $B = d(d+1)(d+2)(d+3) + 1$, prove that $(\sqrt{A} + 1)^{2} + B$ cannot be a perfect square. | [
"Solution:\nFirst we prove the following Lemma\nLemma: If $x$, $y$, $z$ are real numbers such that $x + y + z = 0$, then $2\\left(x^{4} + y^{4} + z^{4}\\right) = \\left(x^{2} + y^{2} + z^{2}\\right)^{2}$.\nProof of the Lemma:\n$$\n\\begin{aligned}\nx^{4} + y^{4} + z^{4} & = x^{2} x^{2} + y^{2} y^{2} + z^{2} z^{2} =... | JBMO | JBMO Shortlist | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0f36 | Problem:
In a group of people every person has less than $4$ enemies. Assume that $A$ is $B$'s enemy iff $B$ is $A$'s enemy. Show that we can divide the group into two parts, so that each person has at most one enemy in his part. | [
"Solution:\n\nWe model the group as a graph where each person is a vertex and an edge connects two people if they are enemies. Each vertex has degree less than $4$.\n\nWe need to partition the vertices into two sets so that each vertex has at most one neighbor (enemy) in its own set.\n\nWe proceed by induction on t... | Soviet Union | ASU | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
053m | Some cities of a country are connected with roads. We say that a city $A$ belongs to a cycle of length $n$ if one can travel from $A$ through exactly $n-1$ other cities and return back to $A$. It is known that each city of the country belongs to a cycle of length 4 and also to a cycle of length 5.
Is it sure that
a) a... | [
"Let there be 10 cities and let cities be connected as in the Fig. 19. Then every city belongs to a cycle of length 5 and also to a cycle of length 4. On the other hand, no cycles of length 3 exist.\n\n\nFig. 19"
] | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | a) No. b) No. | |
0jvu | Problem:
How many 3-element subsets of the set $\{1,2,3, \ldots, 19\}$ have sum of elements divisible by 4? | [
"Solution:\nConsider the elements of the sets mod 4. Then we would need to have sets of the form $\\{0,0,0\\}$, $\\{0,2,2\\}$, $\\{0,1,3\\}$, $\\{1,1,2\\}$, or $\\{2,3,3\\}$. In the set $\\{1,2, \\ldots, 19\\}$ there are four elements divisible by 4 and 5 elements congruent to each of $1,2,3 \\bmod 4$. Hence the de... | United States | HMMT November | [
"Number Theory > Modular Arithmetic"
] | null | final answer only | 244 | |
006a | Diremos que un entero positivo es *afortunado* si la suma de sus dígitos es divisible por $31$. ¿Cuál es la máxima diferencia posible entre dos números afortunados consecutivos? | [] | Argentina | Argentina 2008 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | Spanish | proof and answer | 9999999 | |
03k5 | Problem:
For $x, y, z \geq 0$, establish the inequality
$$
x(x-z)^2 + y(y-z)^2 \geq (x-z)(y-z)(x+y-z)
$$
and determine when equality holds. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Equations and Inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | Equality holds if and only if either all three variables are equal, or exactly one of the variables is zero and the other two are equal. | |
0lb8 | Given two arithmetic progressions $(a_n)$, $(b_n)$ and an integer $m > 2$. Consider $m$ quadratics $P_k(x) = x^2 + a_k x + b_k$, $k = 1, 2, \dots, m$. Suppose that neither $P_1(x)$ nor $P_m(x)$ has a real root. Prove that none of the given quadratics contains a real root. | [
"Denote by $\\alpha$ and $\\beta$ the common differences of the arithmetic progressions $a_n$ and $b_n$, respectively. The quadratics $P_k$ ($k = 1, 2, \\dots, m$) can be written as\n$$\nP_k(x) = x^2 + (a_1 + (k-1)\\alpha)x + b_1 + (k-1)\\beta, \\quad k = 1, 2, \\dots, m.\n$$\nSince $P_1$ and $P_m$ have no real roo... | Vietnam | Vietnam Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Vietnamese | proof only | null | |
04gw | Let $\triangle ABC$ be a right triangle. Point $M$ is chosen on the bisector of the right angle $\angle ACB$. If we have $\sin \angle MAC = \frac{8}{17}$ and $\cos \angle MBC = \frac{3}{5}$, determine the ratio $|AB| : |CM|$. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 5√58 : 16 | |
095m | Problem:
Găsiți toate valorile parametrului $a \in \mathbb{R}$, pentru care sistemul
$$
\left\{\begin{array}{c}
y=\sqrt{5+4x-x^{2}}+2 \\
y=\sqrt{9-a^{2}+2a x-x^{2}}+a
\end{array}\right.
$$
are exact o soluție în $\mathbb{R} \times \mathbb{R}$. | [
"Solution:\nSistemul inițial este echivalent cu următorul sistem:\n$$\n\\left\\{\\begin{array}{c}\n(y-2)^{2}=5+4x-x^{2} \\\\\ny-2 \\geq 0, \\quad 5+4x-x^{2} \\geq 0, \\\\\n(y-a)^{2}=9-a^{2}+2a x-x^{2}, \\\\\ny-a \\geq 0, \\quad 9-a^{2}+2a x-x^{2} \\geq 0\n\\end{array}\\right. \\Leftrightarrow \\left\\{\\begin{array... | Moldova | A 62 - A OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | [-1, 5] \ {2} | |
0cs0 | Let $ABCD$ be an inscribed quadrilateral for which the rays $AB$ and $DC$ meet at $K$. It happened that the points $B$, $D$, and the midpoints of the segments $AC$ and $KC$ are concyclic. Find all possible values the angle $ADC$ may get. | [
"90°."
] | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 90° | |
09xw | We order the positive odd integers as follows:
| | column 1 | column 2 | column 3 | column 4 | column 5 | column 6 | ... |
|--------|----------|----------|----------|----------|----------|----------|-----|
| row 1 | 1 | 3 | 11 | 13 | 29 | 31 | ... |
| row 2 | 5 | 9... | [
"D) $2027$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | MCQ | D | |
0i3f | Problem:
Find the coefficient of $x^{12}$ in the Maclaurin series (i.e. Taylor series around $x=0$) for $\frac{1}{1-3x+2x^{2}}$. | [
"Solution:\nIf you know formal power series, then this is not such a hard question, but since this is a calculus test... Use partial fractions to get\n$$\n\\frac{1}{1-3x+2x^{2}} = \\frac{1/2}{1-2x} - \\frac{1}{1-x}.\n$$\nNow each of these can be expanded as a geometric series (or take derivatives and get the same r... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2047 | |
0bsf | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
$$
f(a^2) - f(b^2) \leq (f(a) + b)(a - f(b)), \quad \text{for all } a, b \in \mathbb{R}.
$$ | [
"Plugging $a = b = 0$ yields $f^2(0) \\leq 0$, hence $f(0) = 0$. Again, plugging $b = 0$, and then $a = 0$, we obtain $f(a^2) \\leq af(a)$, $\\forall a \\in \\mathbb{R}$ and $f(b^2) \\geq bf(b)$, $\\forall b \\in \\mathbb{R}$, therefore $f(x^2) = xf(x)$, $\\forall x \\in \\mathbb{R}$. Replacing the latter in the gi... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = x for all real x, and f(x) = -x for all real x | |
097l | Problem:
Bisectoarele $B B_{1}$ şi $C C_{1}$ ale triunghiului $A B C$ se intersectează în punctul $O$. Măsura unghiului $\angle B O C$ este egală cu $120^{\circ}$. Cercul circumscris triunghiului $B C_{1} O$ intersectează latura $B C$ în punctul $D$. Să se demonstreze că $A D \perp B_{1} C_{1}$. | [
"Solution:\n$\\angle B O C=120^{\\circ} \\Rightarrow \\angle O B C+\\angle O C B=(\\angle A B C+\\angle A C B) / 2=60^{\\circ} \\Rightarrow \\angle B A C=\\angle B_{1} A C_{1}=60^{\\circ} ;$\n\n\n\n$\\angle B_{1} O C_{1}=\\angle B O C=120^{\\circ}$. Astfel, patrulaterul $A B_{1} O C_{1}$ es... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0epd | Anelize put $120$ identical cubes together to form a rectangular prism. She then painted all $6$ faces of the prism. Once the paint had dried she disassembled the cubes and found that $24$ of the cubes had not been painted on any face. The surface area of the prism she formed was
(A) $148$ (B) $150$ (C) $152$ (D) $154$... | [
"Let $l$, $b$, $h$ denote the numbers of cubes in the length, breadth and height respectively. Then $lbh = 120$. To find the unpainted cubes we must remove a layer at both ends of the length, breadth and height, giving $(l-2)(b-2)(h-2) = 24$. By trying various factorizations of $120$, we soon find that $l$, $b$, $h... | South Africa | South African Mathematics Olympiad | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > 3D Shapes",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | A | |
0jma | A country has $n$ cities, labeled $1$, $2$, $3$, $\dots$, $n$. It wants to build exactly $n-1$ roads between certain pairs of cities so that every city is reachable from every other city via some sequence of roads. However, it is not permitted to put roads between pairs of cities that have labels differing by exactly $... | [
"**Solution 1** (by Jon Schneider). Let $A$ be the set of good trees. Consider the operation $S$ which takes a tree, and replaces each edge connecting vertices $x$ and $y$ with an edge connecting vertices $x+1$ and $y+1$ (labels taken modulo $n$). Note that $S$ is invertible and sends good trees to good trees, so i... | United States | IMO Team Selection Team Selection Test | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0akd | Let $n$ three-digit numbers satisfy the following properties:
(1) No number contains the digit $0$.
(2) The sum of the digits of each number is $9$.
(3) The units digits of any two numbers are different.
(4) The tens digits of any two numbers are different.
(5) The hundreds digits of any two numbers are different.
Find... | [
"Let $S$ denote the set of three-digit numbers that have digit sum equal to $9$ and no digit equal to $0$. We will first find the cardinality of $S$. We start from the number $111$ and each element of $S$ can be obtained from $111$ by a string of $6A'$s (which means that we add $1$ to the current digit). Then for e... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 5 | |
0c1i | Consider real numbers $a, b, c$ satisfying $1 < b \le c^2 \le a^{10}$, and
$$
\log_a b + 2 \log_b c + 5 \log_c a = 12.
$$
Show that
$$
2 \log_a c + 5 \log_b c + 10 \log_b a \ge 21.
$$ | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0j6h | Problem:
Let $XYZ$ be an equilateral triangle, and let $K, L, M$ be points on sides $XY, YZ, ZX$, respectively, such that $XK / KY = B$, $YL / LZ = 1 / C$, and $ZM / MX = 1$. Determine the ratio of the area of triangle $KLM$ to the area of triangle $XYZ$. | [
"Solution:\n\nAnswer: $\\frac{1}{5}$\n\nFirst, we note that\n$$\n[KLM] = [XYZ] - [XKM] - [YLK] - [ZML]\n$$\nThen, note that\n$$\n\\begin{gathered}\n[XKM] = \\frac{XK}{XY} \\cdot \\frac{XM}{XZ} \\cdot [XYZ] = \\frac{B}{B+1} \\cdot \\frac{1}{2} \\cdot [XYZ] \\\\\n[YLK] = \\frac{YL}{YZ} \\cdot \\frac{YK}{YX} \\cdot [X... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 1/5 | |
0dzr | Problem:
Naj bo $K$ podmnožica naravnih števil. Za vsaki dve števili $a$ in $b$ iz množice $K$ velja, da $a$ deli $b$ ali $b$ deli $a$. Dokaži, da je tedaj vsako število $c$ iz množice $K$ večje od vsote vseh tistih števil iz množice $K$, ki so manjša od $c$. | [
"Solution:\n\nElemente množice $K$ označimo z $a_{1}, a_{2}, a_{3}, \\ldots$ tako, da velja $a_{1}<a_{2}<a_{3}<\\ldots$. Dokažimo trditev z indukcijo. Očitno je vsota vseh števil, ki so manjša od $a_{2}$, enaka $a_{1}$ in torej manjša od $a_{2}$. Denimo sedaj, da za naravno število $n$ velja\n$$\na_{n}>a_{n-1}+a_{n... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
02sn | Problem:
A figura mostra um retângulo $ABCD$ de lado $\overline{AD}=4$ e onde o ponto $M$ é o ponto médio do segmento $AB$.
Sabendo que os segmentos $AC$ e $DM$ são ortogonais, calcule a área do retângulo $ABCD$. | [
"Solution:\nChamemos de $P$ o ponto de interseção dos segmentos $AC$ e $DM$. Como a soma dos ângulos internos do triângulo $APM$ é $180^\\circ$ então\n$$\n\\measuredangle MAP = 90^\\circ - \\measuredangle AMP\n$$\nMas $\\measuredangle MAP + \\measuredangle PAD = 90^\\circ$. Então, concluímos que $\\measuredangle PA... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 16√2 | |
09jm | John walked home with his dog after visiting the store. The dog, being four times faster than John, arrived home first and then came back to join him. How many meters did the dog travel on the 1-kilometer journey from the store to home? | [
"Let the distance from the store to home be $1000$ meters.\n\nLet $v$ be John's speed, so the dog's speed is $4v$.\n\nLet $t$ be the time it takes John to walk home. Then $vt = 1000$, so $t = \\frac{1000}{v}$.\n\nIn time $t$, the dog can travel $4v \\cdot t = 4v \\cdot \\frac{1000}{v} = 4000$ meters.\n\nBut the dog... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 1600 | |
07fr | $$
P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0
$$
with integer coefficients *primitive* if and only if $\text{gcd}(a_n, a_{n-1}, \dots, a_1, a_0) = 1$.
a) Let $P(x)$ be a primitive polynomial with degree less than $1398$ and $S$ be a subset of primes greater than $1398$. Prove that there is a positive intege... | [
"a) We know that for every polynomial $P(x)$ with degree $n$, the equation $P(x) \\equiv 0 \\pmod{p}$ has at most $n$ distinct roots modulo $p$, for every prime number $p$. Then for every $p_i \\in S$ we have some $t_i \\in \\{1, 2, \\dots, p\\}$ where $p_i \\nmid P(t_i)$. Now choose $t$ by Chinese Remainder Theore... | Iran | 37th Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Poly... | English | proof only | null | |
0423 | Let $f(x) = \lfloor \frac{x}{1!} \rfloor + \lfloor \frac{x}{2!} \rfloor + \dots + \lfloor \frac{x}{2013!} \rfloor$, where $\lfloor x \rfloor$ is the greatest integer no greater than $x$. Call an integer $n$ a good number if the equation $f(x) = n$ has a real solution $x$. Find the number of good numbers in the set $\{1... | [
"First, we point out two obvious facts:\n(a) If $m$ is a positive integer and $x$ is real, then\n$$\n\\lfloor \\frac{x}{m} \\rfloor = \\lfloor \\frac{\\lfloor x \\rfloor}{m} \\rfloor.\n$$\n(b) For any integer $l$ and positive even number $m$, we have\n$$\n\\lfloor \\frac{2l + 1}{m} \\rfloor = \\lfloor \\frac{2l}{m}... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 587 | |
0iq7 | Problem:
Let $\ell$ be the line through $(0,0)$ and tangent to the curve $y = x^{3} + x + 16$. Find the slope of $\ell$. | [
"Solution:\n\nLet the point of tangency be $(t, t^{3} + t + 16)$, then the slope of $\\ell$ is $\\dfrac{t^{3} + t + 16}{t}$. On the other hand, since $\\dfrac{dy}{dx} = 3t^{2} + 1$, the slope of $\\ell$ is $3t^{2} + 1$. Therefore,\n$$\n\\frac{t^{3} + t + 16}{t} = 3t^{2} + 1\n$$\nSimplifying, we get $t^{3} = 8$, so ... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof and answer | 13 | |
0dpi | Let $Q$ be a square of size $12 \times 12$ given on an infinite cell grid. Student wants to paint some (not necessary all!) cells with seven colors (each cell with one color), such that no two three-cell rectangles (out of 288) with centers in $Q$ are colored identically. Is this possible?
(Two three-cell rectangles ar... | [
"Answer: impossible.\nSuppose, that there is a coloring, that satisfy above conditions. Let's count the number of different colorings of $1 \\times 3$ rectangle using 8 colors.\nLet's fix the center cell of rectangle, it can be colored in 8 ways. Other 2 cells can be colored in 8 ways, if they have the same color. ... | Silk Road Mathematics Competition | XVI Silk Road Math Competition | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | impossible | |
0j4f | Problem:
The integers from $1$ to $n$ are written in increasing order from left to right on a blackboard. David and Goliath play the following game: starting with David, the two players alternate erasing any two consecutive numbers and replacing them with their sum or product. Play continues until only one number on t... | [
"Solution:\n\nAnswer: $4022$\n\nIf $n$ is odd and greater than $1$, then Goliath makes the last move. No matter what two numbers are on the board, Goliath can combine them to make an even number. Hence Goliath has a winning strategy in this case.\n\nNow suppose $n$ is even. We can replace all numbers on the board b... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 4022 | |
07wk | Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that $f(a+b) = ab + f(a) + f(b)$ for $a, b \in \mathbb{Z}$. | [
"The general solution is $f(a) = \\frac{a(a+k)}{2}$ for odd integers $k$. It is easy to show by substitution that the stated formula is a solution. We now show it is the only solution.\n\nPutting $a = b = 0$ gives $f(0) = 0$. Putting $b = 1$ gives\n$$\nf(a + 1) = f(a) + f(1) + a.\n$$\nBy induction on $a$ it now fol... | Ireland | IRL_ABooklet_2024 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All functions are f(n) = n(n + k)/2 for an odd integer k. | |
0689 | Square $ABCD$ is divided into $n^2$ equal elementary squares, by drawing parallel lines to their sides. We call the vertices of the elementary squares points of the lattice. One rhombus will be called “good”, if it is not a square, their vertices are points of the lattice and their diagonals are parallel to the sides o... | [
"Drawing from the vertices of the rhombus parallel lines to the sides of the lattice, we find the least rectangle in which the rhombus is inscribed. Since the rhombus is symmetric with respect to its center, this rectangle will have sides with even length. Therefore it is enough to count the rectangles with even si... | Greece | 33rd Hellenic Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | If n is even: n(n−2)(3n^2 − 2n − 4)/48. If n is odd: (n^2 − 1)(3n^2 − 8n − 3)/48. | |
0ec9 | Find all triples of positive integers $(a, b, p)$ such that $p$ is a prime and
$$
\frac{5^a + 2^b p}{5^a - 2^b p}
$$
is a perfect square of a positive integer. | [
"Let $\\frac{5^a + 2^b p}{5^a - 2^b p} = n^2$. Since $5^a - 2^b p$ is odd, we also have $5^a + 2^b p$ is even. Thus $5^a - 2^b p$ and $5^a + 2^b p$ are both positive integers. However, the number $5^a - 2^b p$ is odd, hence $5^a - 2^b p$ is odd. Thus we can write $5^a - 2^b p = 2 \\cdot 5^a$. However, the number $5... | Slovenia | Selection Examinations for the IMO 2015 | [
"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 | (1, 1, 2), (2, 3, 3), (2, 2, 5) | |
07hm | Consider $n \ge 6$ points $x_1, x_2, \dots, x_n$ on the plane such that no three of them are colinear. We call a graph with vertices $x_1, x_2, \dots, x_n$ a "road network" if it is connected, each of its edges is a line segment, and no two edges intersect at a point other than its vertices. Prove that there are three ... | [
"For $n = 6$ there are 16 combinatorially different point sets (order types). It is easy to check that all these 16 cases allow for 3 edge-disjoint plane spanning trees.\n\nLet $S$ be a set of $n$ points in the plane and let $e = (r, b)$ be an edge spanned by $S$ having exactly 4 points of $S$ on one side (i.e., on... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0chd | In the exterior of the acute-angled triangle $ABC$, we construct the isosceles triangles $DAB$ and $EAC$ with bases $AB$ and $AC$, respectively, such that $\angle DBC = \angle ECB = 90^\circ$. Let $M$ and $N$ be the reflections of $A$ with respect to $D$ and $E$, respectively. Prove that the line $MN$ passes through th... | [
"\nSince $AH \\perp BC$ and $CP \\perp BC$, we deduce that $AH \\parallel CP$. Therefore $AHCP$ is a parallelogram, and $CP = AH$.\n\nSimilarly, we deduce that $AHFC'$ is a parallelogram, hence $C'F = AH = CP$. Since $C'E = CE$, we obtain that $EP = EF$.\n\nConsequently, $D$ and $E$ are the... | Romania | 74th NMO Selection Tests for JBMO | [
"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 > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Co... | English | proof only | null | |
0g7h | 設 $x, y, z$ 為正整數, 且 $z(xz+1)^2 = (5z+2y)(2z+y)$, 試證: $z$ 必為奇數且 $z$ 為完全平方數。 | [
"(1) 利用反證法, 假設 $z$ 為偶數, 即 $z = 2z_1$. 原式改寫為: $z_1(2xz_1+1)^2 = (5z_1+y)(4z_1+y)$.\n令 $w = (y, z_1)$, 所以令 $y = wy_0, z_1 = wz_0$, 其中 $(y_0, z_0) = 1$.\n$$ \\Rightarrow z_0(2xwz_0+1)^2 = w(5z_0+y_0)(4z_0+y_0) $$\n因為 $(z_0, 5z_0+y_0) = (z_0, 4z_0+y_0) = 1$ 且 $(w, (2xwz_0+1)^2) = 1 \\Rightarrow w|z_0, z_0|w$\n所以 $w = z... | Taiwan | 二0一三數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0904 | Let $0 \le n \le 5^5$ be an integer. Arrange $n$ black stones and $5^5-n$ white stones in a row, and repeat the following operation 5 times:
Divide the row of stones into groups of 5 stones from left to right. For each group, replace the 5 stones in the group with one stone of the color that appears more frequently amo... | [
"2883\n\nLet $k$ be a positive integer. Consider arranging black and white stones in a single row, with a total of $5^k$ stones. Define a **bad arrangement** as one where, after $k$ repetitions of an operation, the last remaining stone is white.\n\nDefine $f(k)$ as the minimum non-negative integer $m$ such that a b... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2883 | |
0fd6 | Problem:
Dos esferas de radio $r$ son tangentes exteriores. Otras tres esferas de radio $R$ son tangentes exteriores entre sí, dos a dos. Cada una de estas tres esferas es, además, tangente exterior a las dos primeras. Encuentra la relación entre $R$ y $r$. | [
"Solution:\n\nLos centros de las tres esferas de radio $R$, $O_{1}$, $O_{2}$ y $O_{3}$, son los vértices de un triángulo\n\nequilátero de lado $2R$. El punto de tangencia, $T$, de las dos esferas de radio $r$ es el centro de ese triángulo y, por tanto, dista de los vértices dos tercios de l... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | R = 6r | |
07z1 | Problem:
Sia $f$ una funzione reale di variabile reale che verifica le condizioni
(i) $f(10+x)=f(10-x)$
(ii) $f(20+x)=-f(20-x)$
per ogni valore reale di $x$. Si dimostri che $f$ è dispari e periodica.
Si ricorda che $f$ si dice dispari se $f(-x)=-f(x)$ per ogni $x$;
$f$ si dice periodica se esiste $T>0$ tale che $f(x+... | [
"Solution:\n\nSi ha: $f(20+x)=f(10+(10+x))=f(10-(10+x))=f(-x)$ dalla (i)\n\n$f(20+x)=-f(20-x)=-f(x)$ dalla (ii) e dalla precedente\n\nda cui $f(-x)=-f(x)$ per ogni $x$ e quindi $f$ è dispari.\n\nInoltre $f(40+x)=f(20+(20+x))=-f(20-(20+x))=-f(-x)=f(x)$\ne quindi essendo $f(40+x)=f(x)$ se ne deduce che $f$ è periodic... | Italy | XIII GARA NAZIONALE di MATEMATICA | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
0afm | Дванаесетте витези на кружната маса треба да изберат двочлена делегација за посета на кралот Артур. На колку начини може тоа да се направи, така што во двочлената делгација да не бидат витези кои седат еден до друг на кружната маса. | [
"Нека претпоставиме дека кружната маса е кружница, а витезите кои седат на кружната маса се темиња на 12-аголник. Секој пар витези што седат на масата кои прават двочлена делагација според условот на задачата, определуваат една дијагонала на 12-аголникот. Според тоа, кралот Артур може да формира онолку делегации ко... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | Macedonian, English | proof and answer | 54 | |
06l5 | Find an integral solution of the equation
$$
\left\lfloor \frac{x}{1!} \right\rfloor + \left\lfloor \frac{x}{2!} \right\rfloor + \left\lfloor \frac{x}{3!} \right\rfloor + \cdots + \left\lfloor \frac{x}{10!} \right\rfloor = 2019.
$$
(Note $\lfloor u \rfloor$ stands for the greatest integer less than or equal to $u$.) | [
"The answer is $2019$.\nClearly, the answer should not be an integer greater than $2018$. Hence we may remove the last $4$ terms on the left-hand side of the equation.\n\nNext, we may assume that $x = a(6!) + b(5!) + c(4!) + d(3!) + e(2!) + f(1!)$, where $a, b, c, d, e, f$ are nonnegative integers and $a \\le 6, b ... | Hong Kong | HKG TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1176 | |
0geu | 坐標平面上有 2020 個點 $\{A_i = (x_i, y_i) : i = 1, \dots, 2020\}$，滿足
$$
0 = x_1 < x_2 < \cdots < x_{2020},
$$
$$
0 = y_{2020} < y_{2019} < \cdots < y_1.
$$
又令 $O = (0,0)$ 為座標原點。我們依序連邊 $OA_1, A_1A_2, \dots, A_{2019}A_{2020}, A_{2020}O$，構成一個 2021 邊形 $C$。
濤哥想要將 $C$ 整個塗黑。他每一次可以指定一個點 $(x, y)$，支付 $xy$ 元，並將 $\{(x', y') : 0 \le x' \l... | [
"題目可以擴張為：在第一象限中，有一個連續有界遞減函數 $f$，並令 $C$ 為此曲線與 $x$、$y$ 兩軸所夾區域，則我們可以用題設之方式，在 $4|C|$ 元內將此區域塗黑。我們可以直接構造此塗法：\n\n\n\n以上構造中，$x$ 為讓 $f(x) \\ge f(0)/2$ 之正實數（由連續性此 $x$ 必存在）。則我們有：\n- 1 號彩色區域面積 $\\times 4 = 4x f(x) \\ge x f(0) + 2x f(x) = \\text{選 } (x, f(0)) \\text{ 加上選 } (2x, f(x)) \\text{ 時需付金額。}$\n-... | Taiwan | 2021 數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0awf | Problem:
Find all real solutions of the following nonlinear system:
$$
\begin{aligned}
x + 4y + 6z &= 16 \\
x + 6y + 12z &= 24 \\
x^{2} + 4y^{2} + 36z^{2} &= 76
\end{aligned}
$$ | [
"Solution:\nUsing the first two equations, we can obtain $y = 4 - 3z$ (by eliminating $x$) and $x = 6z$ (by eliminating $y$).\n\nSubstituting to the third equation gives $(6z)^2 + 4(4 - 3z)^2 + 36z^2 = 76$, which simplifies to $9z^2 - 8z - 1 = 0$.\n\nThis gives $z = -1/9$ or $z = 1$.\n\nSubstituting back to the exp... | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | (6, 1, 1) and (-2/3, 13/3, -1/9) | |
0ia2 | Problem:
Let $a_{1}=1$, and let $a_{n}=\left\lfloor n^{3} / a_{n-1}\right\rfloor$ for $n>1$. Determine the value of $a_{999}$. | [
"Solution:\nWe claim that for any odd $n$, $a_{n}=n$. The proof is by induction. To get the base cases $n=1$, $3$, we compute $a_{1}=1$, $a_{2}=\\left\\lfloor 2^{3} / 1\\right\\rfloor=8$, $a_{3}=\\left\\lfloor 3^{3} / 8\\right\\rfloor=3$.\n\nAnd if the claim holds for odd $n \\geq 3$, then $a_{n+1}=\\left\\lfloor(n... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 999 | |
007p | There are given $1000$ distinct points on a circle. We have to select $k$ of them so that no two chosen points are adjacent. In how many ways can this be done? | [
"Label the points clockwisely $A_1, A_2, \\dots, A_{1000}$ starting at any desired position. To every selection of several points $A_j$ there corresponds bijectively a sequence $a_1a_2 \\dots a_{1000}$ of zeros and ones in which $a_j = 1$ or $a_j = 0$ according as $A_j$ is selected or not. So we may argue about $0$... | Argentina | Mathematical Olympiad Rioplatense | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | binom(1000 - k, k) + binom(999 - k, k - 1) | |
0gx7 | We know that at some natural $n$ the number $n^3 + 2009n^2 + 27n$ written in decimal notation ends with digit 3. Find what digits are in the hundred's and ten's place of the number. | [
"It's clear that the number $2000n^2$ does not influence the answer therefore the sought digits are the same for numbers $A = n^3 + 2009n^2 + 27n$ and $B = n^3 + 9n^2 + 27n$.\n\nAs the number $(B + 27)$ equals $(n + 3)^3$ (being the cube of the natural number) and ends in 0, this number should end in 000. Thus $B =... | Ukraine | Ukrajina 2008 | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof and answer | hundreds digit = 9; tens digit = 7 | |
0dtx | Find all functions $f: \mathbb{Z}^+ \to \mathbb{Z}^+$, where $\mathbb{Z}^+$ is the set of all positive integers, such that for any positive integers $m, n$, $f^{f(m)}(n) + n = f^m(n) + f(n)$. | [
"We claim that the only solutions are:\n1. $f(n) \\equiv n + 1$. This works because\n$$\nf^{f(m)}(n) + n = f^{m+1}(n) + n = m + 1 + 2n\n$$\nand\n$$\nf^m(n) + f(n) = m + n + n + 1 = m + 1 + 2n.\n$$\n2. $f(n) \\equiv n$. This works because\n$$\nf^{f(m)}(n) + n = 2n = f^m(n) + f(n).\n$$\nWe now show these are the only... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | The only functions are f(n) = n for all positive integers n and f(n) = n + 1 for all positive integers n. | |
0bp0 | Consider the integral lattice $\mathbb{Z}^n$, $n \ge 2$, in the Euclidean $n$-space. Define a line in $\mathbb{Z}^n$ to be a set of the form $a_1 \times \dots \times a_{k-1} \times \mathbb{Z} \times a_{k+1} \times \dots \times a_n$, where $k$ is an integer in the range $1, 2, \dots, n$, and the $a_i$ are arbitrary inte... | [
"(a) Let $A$ be an admissible set in $\\mathbb{Z}^2$, choose a point $a_0$ of $A$, and, for each positive integer $k$, choose a point $a_k$ of $A$ different from $a_{k-1}$, having the same first coordinate as the latter if $k$ is odd, and the same second coordinate if $k$ is even. Eventually we must choose an $a_n ... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0l4g | Problem:
Compute the number of ways to color each cell of an $18 \times 18$ square grid either ruby or sapphire such that each contiguous $3 \times 3$ subgrid has exactly 1 ruby cell. | [
"Solution:\n\nSubdivide the grid into 36 subgrids of size $3 \\times 3$. Each contains exactly one ruby cell.\nConsider four adjacent subgrids\n$$\n\\begin{array}{c|c}\nA & B \\\\\n\\hline\nC & D\n\\end{array}\n$$\nand the relative positions of the ruby cells within their respective $3 \\times 3$ subgrids. Between ... | United States | HMMT November 2024 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 4365 | |
05bf | Find the least number of buttons that can be placed on the squares of a $4 \times 4$ grid so that no two buttons are on the same square or on squares with a common side (buttons may be on squares with a common vertex) and no buttons can be added to the grid under the same conditions. | [
"If there are 3 or fewer buttons placed, then it is always possible to add more, as each button blocks up to 5 squares: the square it is in and its 4 neighbors. Thus 3 buttons cover up to $3 \\cdot 5 = 15$ of the $4 \\cdot 4 = 16$ squares of the grid, meaning that a button can be placed in at least one of the squar... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 4 | |
0k0o | Problem:
Consider a convex $n$-gon $A_{1} A_{2} \cdots A_{n}$. (Note: In a convex polygon, all interior angles are less than $180^{\circ}$.) Let $h$ be a positive number. Using the sides of the polygon as bases, we draw $n$ rectangles, each of height $h$, so that each rectangle is either entirely inside the $n$-gon or... | [
"Solution:\n\nFor every point $P$ in the interior of $\\Pi$, let $f(P)$ be the minimum distance from $P$ to a point on the perimeter of $\\Pi$. Let $r$ denote the maximum value of $f(P)$, which necessarily exists, and let this maximum be achieved at some (not necessarily unique) point $P=I$. (We remark that in the ... | United States | 19th Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0dv8 | Problem:
Kateta $a$ in hipotenuza pravokotnega trikotnika sta v razmerju $12:13$. Če skrajšamo hipotenuzo za $23~\mathrm{cm}$ in kateto $a$ za $27~\mathrm{cm}$, dobimo nov pravokotni trikotnik, ki se mu dolžina druge katete ni spremenila. Izračunaj stranice prvotnega trikotnika. | [
"Solution:\n\nZapisane stranice prvega trikotnika: $a = 12x$, $b = b$, $c = 13x$\n\nZapisane stranice drugega trikotnika: $a' = 12x - 27$\n\n$$\n\\begin{aligned}\nb' &= b \\\\\nc' &= 13x - 23\n\\end{aligned}\n$$\n\nZapisan Pitagorov izrek $b^2 = c^2 - a^2 = c'^2 - a'^2$\n\nVstavljeni podatki $(13x)^2 - (12x)^2 = (1... | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a = 48 cm, b = 20 cm, c = 52 cm | |
004h | Hallar todas las parejas de enteros $ (x, y) $ que cumplan
$$
x^3 y + x + y = xy + 2xy^2
$$ | [] | Argentina | XVIII Olimpiada Matemática del Cono Sur | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Español | proof and answer | (-1, -1), (0, 0), (1, 1), (2, 2) | |
0h5j | Prove the inequality for positive $a$, $b$, $c$, $d$
$$
\frac{(a+b)^2}{cd} + \frac{(c+d)^2}{ab} \ge 8.
$$ | [
"Use two inequalities of means:\n$$\n\\frac{(a+b)^2}{cd} + \\frac{(c+d)^2}{ab} \\ge \\frac{4ab}{cd} + \\frac{4cd}{ab} \\ge 2 \\cdot 4 \\sqrt{\\frac{ab}{cd} \\cdot \\frac{cd}{ab}} = 8.\n$$"
] | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0ktn | Problem:
Let $\Gamma_{1}$ and $\Gamma_{2}$ be two circles externally tangent to each other at $N$ that are both internally tangent to $\Gamma$ at points $U$ and $V$, respectively. A common external tangent of $\Gamma_{1}$ and $\Gamma_{2}$ is tangent to $\Gamma_{1}$ and $\Gamma_{2}$ at $P$ and $Q$, respectively, and int... | [
"Solution:\nBy Archimedes lemma, we have $M, Q, V$ are collinear and $M, P, U$ are collinear as well. Note that inversion at $M$ with radius $M X$ shows that $P Q U V$ is cyclic. Thus, we have $M P \\cdot M U = M Q \\cdot M V$, so $M$ lies on the radical axis of $(P U Z)$ and $(Q V Z)$, thus $T$ must lie on the lin... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem"... | null | proof and answer | 2 (R r1 + R r2 - 2 r1 r2) sqrt(r1 r2) / (|r1 - r2| sqrt((R - r1)(R - r2))) |
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