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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
068s | Prove that the number $A = 7^{2n} - 48n - 1$ is a multiple of $9$, for all positive integers $n$. | [
"We distinguish three cases mod $3$.\n\n* If $n = 3k$, then $A = 7^{6k} - 48(3k) - 1 = 49^{3k} - 9 \\cdot 16k - 1$. Since $49 \\equiv 4 \\pmod{9}$, it follows that $49^3 \\equiv 4^3 \\pmod{9} \\equiv 1 \\pmod{9}$ and hence $9 \\mid 49^{3k} - 1 \\Rightarrow 9 \\mid A$.\n\n* If $n = 3k+1$, then\n$A = 7^{6k+2} - 48(3k... | Greece | SELECTION EXAMINATION | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
01w6 | Given a quadratic trinomial $p(x)$ with integer coefficients such that $p(x)$ is not divisible by $3$ for all integers $x$.
Prove that there exist polynomials $f(x)$ and $h(x)$ with integer coefficients such that
$$
p(x) \cdot f(x) + 3h(x) = x^6 + x^4 + x^2 + 1.
$$ | [
"**1. It is straightforward to verify the following**\n**Lemma.** Let $x^2 + bx + c$ be a trinomial with integer coefficients which is not divisible by $3$ for any integer $x$. Then this trinomial is congruent modulo $3$ to one of the next three trinomials: $x^2 + 1$, $x^2 + x - 1$ and $x^2 - x - 1$.\nDenote $x^2 +... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
07wb | A function $f : \mathbb{Q}_+ \to \mathbb{Q}_+$, where $\mathbb{Q}_+$ denotes the (strictly) positive rational numbers, satisfies
$$
f(f(x) + f(y)) = \frac{xy}{x+y}
$$
for any $x, y \in \mathbb{Q}_+$. Given $f(1) = 2023$, find $f(2023)$. | [
"We first show that $f$ is injective. Suppose that $f(y) = f(z)$. Then we have:\n$$\n\\frac{xy}{x+y} = f(f(x) + f(y)) = f(f(x) + f(z)) = \\frac{xz}{x+z}\n$$\nTaking reciprocals of the first and last expression implies $y = z$, as required for injectivity.\n\nNote that the right hand side of the functional equation ... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | 1 | |
0j22 | Problem:
For $0 \leq y \leq 2$, let $D_{y}$ be the half-disk of diameter $2$ with one vertex at $(0, y)$, the other vertex on the positive $x$-axis, and the curved boundary further from the origin than the straight boundary. Find the area of the union of $D_{y}$ for all $0 \leq y \leq 2$. | [
"Solution:\nAnswer: $\\pi$\n\nFrom the picture above, we see that the union of the half-disks will be a quarter-circle with radius $2$, and therefore area $\\pi$. To prove that this is the case, we first prove that the boundary of every half-disk intersects the quarter-circle with radius $2... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | π | |
015v | In the very large Baltic-Way-City (in the far future) there are $16$ hospitals. Every night exactly $4$ of them must be on duty for emergencies. Is it possible to arrange the schedule in such a way that after $20$ days every pair of hospitals were on duty exactly once? If not, prove the non-existence. If yes, give a sc... | [
"The answer is yes. Let the hospitals be numbered $1, 2, \\ldots, 16$. The hospitals on duty are the $4$ on the rows.\n\n| 1 | 2 | 3 | 4 |\n|---|---|---|---|\n| 5 | 6 | 7 | 8 |\n| 9 | 10 | 11 | 12 |\n| 13 | 14 | 15 | 16 |\n\n| 1 | 5 | 9 | 13 |\n|---|---|---|---|\n| 2 | 8 | 10 | 15 |\n| 3 | 6 | 11 | 16 |\n| 4 | 7 | ... | Baltic Way | Baltic Way SHL | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes; an explicit schedule exists, for example the 20 quadruples listed in the five 4-by-4 grids in the solution. | |
096c | Problem:
În pocal, secțiunea axială a căruia este graficul funcției $y = x^{4}$, se pune o "vișină" - bilă de rază $R$. Pentru care valori ale lui $R$ bila atinge fundul pocalului (punctul $(0,0)$)? | [
"Solution:\n\nCercetăm secțiunea axială indicată. Deoarece secțiunea axială a bilei este un cerc de rază $R$, atunci acest cerc trebuie să fie tangent la axa $Ox$ în punctul $O = (0, 0)$. Prin urmare, centrul cercului se află pe axa $Oy$ în punctul $O_1 = (0, R)$.\n\nScriem condiția că graficul circumferinței de ra... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 0 <= R <= (3/4) * 2^(1/3) | |
0c86 | Determine the largest integer $N$ satisfying the following condition: for every cell labeling of a $5 \times 5$ array from $1$ through $25$ such that no two cells bear the same number, the numbers in some $2 \times 2$ square add up to at least $N$. | [
"The required maximum is $N = 45$, and is achieved, for instance, by the following extremal cell labeling:\n\n| 25 | 5 | 24 | 6 | 23 |\n|----|---|----|---|----|\n| 11 | 4 | 12 | 3 | 13 |\n| 22 | 7 | 21 | 8 | 20 |\n| 14 | 2 | 15 | 1 | 16 |\n| 19 | 9 | 18 | 10 | 17 |\n\nIn this cell labeling, every $2 \\times 2$ squa... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 45 | |
0jjc | Problem:
A particular coin has a $\frac{1}{3}$ chance of landing on heads (H), $\frac{1}{3}$ chance of landing on tails (T), and $\frac{1}{3}$ chance of landing vertically in the middle (M). When continuously flipping this coin, what is the probability of observing the continuous sequence HMMT before HMT? | [
"Solution:\n$\\frac{1}{4}$\n\nFor a string of coin flips $S$, let $P_{S}$ denote the probability of flipping HMMT before HMT if $S$ is the starting sequence of flips. We know that the desired probability, $p$, is $\\frac{1}{3} P_{H} + \\frac{1}{3} P_{M} + \\frac{1}{3} P_{T}$.\n\nNow, using conditional probability, ... | United States | HMMT November 2014 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 1/4 | |
09dz | All numbers of the form $2^{k}3^{s}$ ($k, s$ are non-negative whole numbers) were written in increasing order $1, 2, 3, 4, 6, 8, 9, 12, \ldots$ and formed the sequence $a_n$. Prove that there are infinitely many numbers $n$ satisfying the condition
$$
\frac{a_{n+1}}{a_n} < 1 + \frac{1}{10^{2014}}
$$ | [
"From the Euler number's definition $(1 + \\frac{1}{10^{2014}})^{10^{2014}} > 2$.\n\nFor any nonnegative whole number $n$, there exists nonnegative $m$ such that $2^m \\le 3^n \\le 2^{m+1}$. Different $n_1, n_2$ correspond to different $m_1, m_2$. Note that $2^m \\le 3^n \\le 2^{m+1} \\Leftrightarrow 1 \\le \\frac{... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof only | null | |
03n7 | One hundred circles of radius one are positioned in the plane so that the area of any triangle formed by the centres of three of these circles is at most $2017$. Prove that there is a line intersecting at least three of these circles. | [
"We will prove that given $n$ circles, there is some line intersecting more than $\\frac{n}{46}$ of them. Let $S$ be the set of centers of the $n$ circles. We will first show that there is a line $\\ell$ such that the projections of the points in $S$ lie in an interval of length at most $\\sqrt{8068} < 90$ on $\\el... | Canada | CMO 2017 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0c4i | Find all the sets $M$, of positive integers, so that every positive integer can be uniquely written as the sum of some distinct elements of $M$ (the sum may contain only one term). | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | M = {1, 2, 4, 8, 16, ...} = {2^k | k ≥ 0} | |
0bp8 | Problem:
Fie $(K,+, \cdot)$ un corp finit cu cel puţin patru elemente. Arătaţi că mulţimea $K^{*}$ poate fi partiţionată în două submulţimi nevide $A$ şi $B$, cu proprietatea că
$$
\sum_{x \in A} x=\prod_{y \in B} y
$$ | [
"Solution:\n\nDeoarece $\\prod_{x \\in K^{*}} x=-1$, dacă $A$ şi $B$ formează o partiţie a lui $K^{*}$, atunci $\\left(\\prod_{x \\in A} x\\right)\\left(\\prod_{x \\in B} x\\right) = -1$, deci $\\sum_{x \\in A} x=\\prod_{y \\in B} y$ dacă şi numai dacă\n$$\n\\left(\\sum_{x \\in A} x\\right) \\cdot\\left(\\prod_{x \... | Romania | Olimpiada Naţională de Matematică, Etapa Naţională | [
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0g9s | 設函數 $f: [0, \infty) \to [0, \infty)$ 滿足：
(1) $\forall x, y \ge 0$, 有 $f(x)f(y) \le y^2 f(\frac{x}{2}) + x^2 f(\frac{y}{2})$;
(2) $\forall 0 \le x \le 1$, $f(x) \le 2016$.
證明: $f(x) \le x^2$, 對於所有 $x \ge 0$ 都成立.
Suppose function $f: [0, \infty) \to [0, \infty)$ satisfies
(1) $\forall x, y \ge 0$, we have $f(x)f(y) \le ... | [
"By (1), and take $x = y = 0$, we have $f(0) = 0$.\n\nNow, suppose there exists $x_0 > 0$ such that $f(x_0) > x_0^2$. By using (1), we have\n$$\nf\\left(\\frac{x_0}{2}\\right) > \\frac{1}{2}x_0^2.\n$$\nIn fact, by mathematical induction, and use (1) repeatedly, it is easy to show that\n$$\nf\\left(\\frac{x_0}{2^k}\... | Taiwan | 二〇一六數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
01mf | Given nonzero real numbers $a, b, c$, with
$$
a + b + c = a^2 + b^2 + c^2 = a^3 + b^3 + c^3. \quad (*)
$$
a) Find $\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) (a + b + c - 2)$.
b) Do there exist pairwise different nonzero $a, b, c$ satisfying $(*)$?
(D. Bazylev) | [
"a) (Solution of M.Mankevich, E.Dovgialo.) Let $\\sigma_1 = a + b + c$, $\\sigma_2 = ab + bc + ca$, $\\sigma_3 = abc$. From the given equalities we have the system\n$$\n\\sigma_1 = \\sigma_1^2 - 2\\sigma_2 \\quad (1), \\quad \\sigma_1^2 = \\sigma_1^3 - 3\\sigma_1\\sigma_2 + 3\\sigma_3. \\quad (2)\n$$\nFrom (1) we h... | Belarus | Selection and Training Session | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | 3; Yes, for example a = 1/2, b = (2 + sqrt(6))/4, c = (2 - sqrt(6))/4. | |
0f7h | Problem:
The positive reals $a$, $b$, $c$, $x$, $y$, $z$ satisfy $a + x = b + y = c + z = k$. Show that $ax + by + cz \leq k^2$. | [] | Soviet Union | 21st ASU | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0guy | Let $n \ge 3$ be an integer and $a_1, a_2, \dots, a_n$ be real numbers. For each $1 \le k \le n$ the real numbers $b_1, b_2, \dots, b_{n+1}$ are defined by
$$
b_k = \frac{a_k + \max\{a_{k+1}, a_{k+2}\}}{2}
$$
and $b_{n+1} = b_1$ ($a_{n+1} = a_1$ and $a_{n+2} = a_2$). Find the smallest $\lambda$ such that the inequalit... | [
"Answer: $\\lambda = 2^{2024}$.\nLet $x_i = a_i - a_{i+1}$. Since for all real numbers\n$$\n\\max\\{x, y\\} = \\frac{x+y}{2} + \\left|\\frac{x-y}{2}\\right|\n$$\nwe have\n$$\nb_i - b_{i+1} = x_i + \\frac{x_{i+1}}{2} + \\frac{x_{i+2}}{2} + \\left| \\frac{x_{i+1}}{2} - \\frac{x_{i+2}}{2} \\right| \\quad (1)\n$$\n\nLe... | Turkey | Team Selection Test for IMO 2024 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 2^{2024} | |
0d7b | Find all positive integer $n$ such that there exists a permutation $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of $\left(1,2,3, \ldots, n\right)$ satisfying the condition:
$$
a_{1}+a_{2}+\cdots+a_{k} \text{ is divisible by } k \text{ for each } k=1,2,3, \ldots, n \text{. }
$$ | [
"It is easy to see that $n=1$, $n=3$ satisfy the given condition, while $n=2$ does not. We will show that all numbers $n>3$ do not satisfy the condition.\n\nFirst, we can see that\n$$\na_{1}+a_{2}+\\cdots+a_{n}=1+2+\\cdots+n=\\frac{n(n+1)}{2}\n$$\nthen $\\frac{n(n+1)}{2}$ is divisible by $n$, which means $\\frac{n+... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 1 and 3 | |
0hyz | Problem:
You are trapped in a room with only one exit, a long hallway with a series of doors and land mines. To get out you must open all the doors and disarm all the mines. In the room is a panel with 3 buttons, which conveniently contains an instruction manual. The red button arms a mine, the yellow button disarms t... | [
"Solution:\n\nClearly we do not want to reset the system at any time. After pressing the red button $r$ times, the yellow button $y$ times, and the green button $g$ times, there will be $3 + r - 2y$ armed mines and $3 + y - 2g$ closed doors, so we want the values of $r$, $y$, and $g$ that make both of these quantit... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 9 | |
0cyk | On a semicircle of diameter $A B$ and center $C$ consider variable points $M$ and $N$ such that $M C \perp N C$. The circumcircle of triangle $M N C$ intersects $A B$ for the second time at $P$. Prove that $\frac{|P M-P N|}{P C}$ is a constant and find its value. | [
"Consider the case when point $P$ is between $C$ and $B$. Quadrilateral $M C P N$ is cyclic and from Ptolemy's relation it follows\n$$\nP M \\cdot C N = P C \\cdot M N + C M \\cdot P N \\tag{1}\n$$\n\n\n\nWe have $C M = C N = R$, $M N = R \\sqrt{2}$, and replacing in (1) we obtain $P M = P ... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | sqrt(2) | |
07hg | Cyclic quadrilateral $ABCD$ with circumcenter $O$ is given. Point $P$ is the intersection of diagonals $AC$ and $BD$. Let $M$ and $N$ be midpoints of the sides $AD$ and $BC$, respectively. Suppose that $\omega_1$, $\omega_2$ and $\omega_3$ are circumcircles of triangles $ADP$, $BCP$ and $OMN$, respectively. Let $E$ and... | [
"Let $E'$ be the second intersection of $NP$ and $\\omega_1$. Since $\\triangle BPC \\sim \\triangle APD$ we have $\\angle MPD = \\angle NPC = \\angle APE'$. So $APDE'$ is a harmonic quadrilateral. Then\n$$\n\\begin{aligned}\n\\angle E'MA &= \\angle PMA = \\angle PNB \\implies \\angle E'MO = \\angle E'MA + 90^\\cir... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0eu2 | For a convex hexagon $ABCDEFG$, three triangles $ABC$, $CDE$, $EFA$ are similar. That is,
$$
\angle BAC = \angle DCE = \angle FEA
$$
$$
\angle BCA = \angle DEC = \angle FAE.
$$
Find conditions on these three triangles under which $\triangle ACE$ is an equilateral triangle if and only if $\triangle BDF$ is an equilatera... | [
"Suppose $\\triangle ACE$ is equilateral, then $\\triangle BDF$ is always equilateral, because three triangles $ACE$, $CDE$, $EFA$ are identical and shape of the hexagon $ABCDEF$ is invariable under the $120^\\circ$ rotation around the center of $\\triangle ACE$. That is, since $B \\to D \\to F$ is obtained from th... | South Korea | Korean Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | The equivalence holds if and only if the three similar triangles are not the isosceles type with two equal angles of thirty degrees (i.e., not 30–30–120). | |
0kwr | Problem:
Let $a_{1}, a_{2}, \ldots$ be an infinite sequence of positive integers such that, for all positive integers $m$ and $n$, we have that $a_{m+n}$ divides $a_{m} a_{n}-1$. Prove that there exists an integer $C$ such that, for all positive integers $k>C$, we have $a_{k}=1$. | [
"Solution:\n\nFor convenience, define $g(x): \\mathbb{N} \\rightarrow \\mathbb{N}$ to be $g(n)=a_{n}$.\nWe first prove that $1 \\in \\operatorname{Im}(g)$. Assume otherwise. First, note that $\\operatorname{gcd}(g(m+n), g(m))=1$ for all positive integers $m, n$, so thus if $g$ never takes the value $1$ then $g$ is ... | United States | HMIC 2023 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof only | null | |
00om | Let $a$, $b$ and $c$ be real numbers larger than $1$. Prove the inequality
$$
\frac{ab}{c-1} + \frac{bc}{a-1} + \frac{ca}{b-1} \geq 12.
$$
When does equality hold? | [
"By the AM-GM inequality, we know that\n$$\n\\sqrt{(c-1) \\cdot 1} \\le \\frac{c-1+1}{2},\n$$\ntherefore\n$$\nc - 1 \\le \\frac{c^2}{4}\n$$\nwith equality for $c=2$. With the two analogous inequalities for $a$ and $b$ we obtain\n$$\n\\frac{ab}{c-1} + \\frac{bc}{a-1} + \\frac{ca}{b-1} \\ge \\frac{4ab}{c^2} + \\frac{... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds when a = b = c = 2. | |
0bwg | Let $x, y, z$ be positive real numbers such that $\frac{(x+y)(y+z)(z+x)}{xyz} = 9$. Show that $31.25 \le \frac{(x+y+z)^3}{xyz} \le 32$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0l2z | Problem:
Let $ABCD$ be a trapezoid with $AB \parallel CD$, $AB = 20$, $CD = 24$, and area $880$. Compute the area of the triangle formed by the midpoints of $AB$, $AC$, and $BD$. | [
"Solution:\n\n\n\nWe first compute the height of the trapezoid. If $h$ is the height, then the area is\n$$\n880 = \\frac{1}{2} h (20 + 24)\n$$\nso $h = 40$. Now, let $M$, $N$, $P$ be the midpoints of $AB$, $AC$, and $BD$. Notice that $PN$ is parallel to $AB$. Thus, the altitude from $M$ to ... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 20 | |
0fgp | Problem:
Si $a$ y $b$ son dos números reales diferentes, resolver el sistema
$$
\begin{aligned}
x+y & =1 \\
(a x+b y)^2 & \leq a^2 x+b^2 y
\end{aligned}
$$
Resolver también el sistema
$$
\begin{aligned}
x+y & =1 \\
(a x+b y)^4 & \leq a^4 x+b^4 y
\end{aligned}
$$ | [
"Solution:\n\nEn el primer caso, de la primera ecuación despejamos $y=1-x$, sustituimos en la inecuación y hacemos operaciones, lo que nos proporciona $(a-b)^2 x^2 \\leq (a-b)^2 x$, cuya solución es el intervalo $[0,1]$ y por lo tanto $y$ recorre el mismo intervalo. Geométricamente es el segmento cuyos extremos tie... | Spain | OME 23 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | For both systems, all solutions are the pairs with x between zero and one and y equal to one minus x; equivalently, the segment from (zero, one) to (one, zero). | |
06zo | Problem:
$ABC$ is an equilateral triangle with side $2$. Show that any point $P$ on the incircle satisfies $PA^{2} + PB^{2} + PC^{2} = 5$. Show also that the triangle with side lengths $PA$, $PB$, $PC$ has area $\sqrt{3}/4$. | [
"Solution:\n\nTake vectors centered at the center $O$ of the triangle. Write the vector $OA$ as $\\mathbf{A}$ etc. Then\n$$\nPA^{2} + PB^{2} + PC^{2} = (\\mathbf{P} - \\mathbf{A})^{2} + (\\mathbf{P} - \\mathbf{B})^{2} + (\\mathbf{P} - \\mathbf{C})^{2} = 3 P^{2} + (A^{2} + B^{2} + C^{2}) - 2 \\mathbf{P} \\cdot (\\ma... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof and answer | PA^2 + PB^2 + PC^2 = 5; area = sqrt(3)/4 | |
0fga | Problem:
Indicaremos por $[x],\{x\}$ las partes entera y decimal del número real $x$. Definimos una distancia entre los números reales $x$ e $y$
$$
d(x, y)=\sqrt{([x]-[y])^{2}+(\{x\}-\{y\})^{2}}
$$
Determinar (como unión de intervalos) el conjunto de los números reales que distan del número $3 / 2$ menos de $202/100$. | [
"Solution:\n\nConsideramos la función $f: \\mathbb{R} \\rightarrow \\mathbb{R}^{2}$ dada por $f(x)=([x],\\{x\\})$ y observamos que la distancia definida en el enunciado entre dos puntos de $\\mathbb{R}$, es la distancia euclídea entre sus imágenes por $f$ en el plano. De hecho, el conjunto imagen $f(\\mathbb{R})$ q... | Spain | OME 22 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | (-1/2 - sqrt(201)/50, -1/2 + sqrt(201)/50) ∪ [0,3) ∪ (7/2 - sqrt(201)/50, 7/2 + sqrt(201)/50) | |
02d4 | The remainder on dividing the polynomial $p(x)$ by $x^2 - (a+b)x + ab$ (where $a$ and $b$ are unequal) is $mx + n$. Find the coefficients $m, n$ in terms of $a, b$. Find $m, n$ for the case $p(x) = x^{200}$ divided by $x^2 - x - 2$ and show that they are integers. | [
"Let $p(x) = q(x)(x-a)(x-b) + mx + n$. So putting $x = a, b$ we get $p(a) = ma + n$, $p(b) = mb + n$. Solving, $m = \\frac{p(a)-p(b)}{a-b}$, $n = \\frac{p(b)a-p(a)b}{a-b}$.\n\nIn the case given $a = 2$, $b = -1$, so $m = \\frac{2^{200}-1}{3}$, $n = \\frac{2^{200}+2}{3}$. Note that $2 \\equiv -1 \\pmod{3}$, so $2^{2... | Brazil | I OBM | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | General: m = (p(a) − p(b)) / (a − b), n = (a p(b) − b p(a)) / (a − b). Specific (p(x) = x^200, divisor x^2 − x − 2): m = (2^200 − 1) / 3, n = (2^200 + 2) / 3, which are integers. | |
0aex | Учениците од IVª одделение членуваат во еколошката, литературната или математичката секција. Пет ученици членуваат во сите три секции, а девет ученици членуваат во по две секции. Во еколошката и литературната членуваат 8 ученици, и исто толку во литературната и математичката секција. Исто така, 20 ученици членуваат сам... | [
"**I начин.** Со помош на Ојлер – Венов дијаграм\n\n$6 (IV^{a}) = 34$\n\n\n**II начин.**\nВо сите три секции членуваат $5$ ученици. Бидејќи во секои две членуваат по $8$ ученици, тогаш само во по две секции (без учениците кои членуваат во сите три секции истовремено) членуваат по $3$ ученик... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | Macedonian, English | proof and answer | 34 | |
083g | Problem:
Una lampada, che si suppone puntiforme, è collocata in un punto $V$. Essa proietta su un piano $a$ situato a 2 metri da $V$ un fascio di luce avente la forma di cono circolare con asse perpendicolare ad $a$, le cui generatrici formano un angolo di $30^{\circ}$ con l'asse del cono. Sul piano $a$, al centro del... | [] | Italy | UNIONE MATEMATICA ITALIANA Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO BIENNIO | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | MCQ | D | |
0ish | Problem:
Let $\omega_1$, $\omega_2$, and $\omega_3$ be three circles passing through the origin $O$ of the coordinate plane but not tangent to each other or to either axis. Denote by $(x_i, 0)$ and $(0, y_i)$, $1 \leq i \leq 3$, the respective intersections (besides $O$) of circle $\omega_i$ with the $x$ and $y$ axes.... | [
"Solution:\n\nFirst of all, note that if a circle passes through $(0,0)$, $(x_i, 0)$, and $(0, y_i)$, its center must be $\\left(\\frac{x_i}{2}, \\frac{y_i}{2}\\right)$, the midpoint of the side opposite the right angle at $O$. Also note that the three points $\\left(\\frac{x_i}{2}, \\frac{y_i}{2}\\right)$ are rela... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocent... | null | proof only | null | |
0dmn | Problem:
Нека су $x, y, z$ позитивни реални бројеви такви да је $x y + y z + z x = x + y + z$. Доказати неједнакост
$$
\frac{1}{x^{2} + y + 1} + \frac{1}{y^{2} + z + 1} + \frac{1}{z^{2} + x + 1} \leqslant 1
$$
Када се у претходној неједнакости достиже знак једнакости? | [
"Solution:\n\nКоши-Шварцова неједнакост за тројке $(x, \\sqrt{y}, 1)$ и $(1, \\sqrt{y}, z)$ даје $\\frac{1}{x^{2} + y + 1} \\leqslant \\frac{1 + y + z^{2}}{(x + y + z)^{2}}$. Аналогно важи $\\frac{1}{y^{2} + z + 1} \\leqslant \\frac{1 + z + x^{2}}{(x + y + z)^{2}}$ и $\\frac{1}{z^{2} + x + 1} \\leqslant \\frac{1 + ... | Serbia | Српска математичка олимпијада | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds only when x = y = z = 1 | |
06j2 | Let $ABC$ be an equilateral triangle, and let $D$ be a point on $AB$ between $A$ and $B$. Next, let $E$ be a point on $AC$ with $DE$ parallel to $BC$. Further, let $F$ be the midpoint of $CD$ and $G$ the circumcentre of triangle $ADE$. Determine the interior angles of triangle $BFG$. | [
"$\\angle GFB = 90^\\circ$, $\\angle FBG = 30^\\circ$ and $\\angle BGF = 60^\\circ$.\nLet $H$ be the point such that $BCHD$ is a parallelogram. Then $E$ is contained in the segment $DH$, and $\\triangle ECH$ is equilateral. Thus, we have $BD = CH = HE$. Moreover, it is clear that $DG = EG$ and $\\angle BDG = 150^\\... | Hong Kong | 1997-2023 IMO HK TST | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | angle GFB = 90 degrees, angle FBG = 30 degrees, angle BGF = 60 degrees | |
040z | Find the number of integers $k$ in the set $\{0, 1, 2, \ldots, 2012\}$ such that the combination number $\binom{2012}{k} = \frac{2012!}{k!(2012-k)!}$ is a multiple of $2012$. | [] | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | 1370 | |
0h1u | The number $n = 11^{2011} \cdot 2011^{11}$ is given. How many divisors that are less than $n$ and are not divisors of $n$, does $n^2$ have? | [
"The number of divisors of $n^2 = 11^{4022} \\cdot 2011^{22}$ is:\n$$\nN = (4022 + 1)(22 + 1) = 92529.\n$$\nWe can pair the divisors of $n^2$ in a way $d$ and $\\frac{n^2}{d}$ where one divisor is greater than $n$, another one is less. The number $n$ does have a pair, therefore the number of divisors that are less ... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 22121 | |
0e5p | Let $ABCD$ be a tangential quadrilateral whose inscribed circle touches the sides $AB$, $BC$, $CD$ and $DA$ in points $P$, $Q$, $R$ and $S$, respectively. Let the lines $AB$ and $CD$ intersect at a point $E$, and let the lines $AD$ and $BC$ intersect at a point $F$. Denote $I_1$ the intersection point of the bisectors ... | [
"First suppose that the lines $PQ$ and $RS$ intersect. Denote $T$ their point of intersection. Due to symmetry we may assume that $Q$ lies between $P$ and $T$ and $R$ lies between $S$ and $T$. We shall prove that the lines $EF$, $I_1I_2$ and $AC$ pass through the point $T$.\n\nIn the construction, $I_1$ is the inte... | Slovenia | Selection Examinations for the IMO 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, ... | null | proof only | null | |
0izu | Problem:
Jessica has three marbles colored red, green, and blue. She randomly selects a non-empty subset of them (such that each subset is equally likely) and puts them in a bag. You then draw three marbles from the bag with replacement. The colors you see are red, blue, red. What is the probability that the only marb... | [
"Solution:\n\n$\\boxed{\\dfrac{27}{35}}$\n\nThere are two possible sets of marbles in the bag, $\\{\\text{red}, \\text{blue}\\}$ and $\\{\\text{red}, \\text{blue}, \\text{green}\\}$. Initially, both these sets are equally likely to be in the bag. However, the probability of red, blue, red being drawn from a set $S$... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 27/35 | |
0d58 | How many polynomials $P$ of integer coefficients and degree at most $4$ satisfy $0 \leq P(x) < 72$ for all $x \in \{0,1,2,3,4\}$? | [
"Remember that if $P_{0}(X), P_{1}(X), P_{2}(X), \\ldots$ are monic polynomials with integer coefficients such that $\\deg P_{k}(X) = k$ for all $k = 0, 1, 2, \\ldots$, then any polynomial $P(X)$ with integer coefficients can be written in a unique way as\n$$\nP(X) = a_{0} P_{0}(X) + a_{1} P_{1}(X) + \\cdots + a_{n... | Saudi Arabia | SAMC 2015 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | English, Arabic | proof and answer | 6718464 | |
0402 | Let $A = \{2, 2^2, \dots, 2^n, \dots\}$. Prove:
(1) For any $a \in A$, $b \in \mathbb{N}^*$, if $b < 2a - 1$, then $b(b+1)$ will not be a multiple of $2a$.
(2) For any $a \in \bar{A} (= \mathbb{N}^* - A)$ satisfying $a \neq 1$, there exists $b \in \mathbb{N}^*$ satisfying $b < 2a - 1$, such that $b(b+1)$ is a multipl... | [
"(1) For any $a \\in A$, $a = 2^k$ ($k \\in \\mathbb{N}^*$). Then $2a = 2^{k+1}$. Let $b$ be any positive integer strictly less than $2a - 1$. Then $(b+1) \\le 2a - 1$.\n\nBetween $b$ and $b+1$, one is an odd number that contains no prime factor $2$, and the other is an even number that contains at most the $k$th p... | China | China Mathematical Competition (Complementary Test) | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory ... | English | proof only | null | |
0b3y | Problem:
Find all positive integers $n$ for which there exists a set of exactly $n$ distinct positive integers, none of which exceed $n^{2}$, whose reciprocals add up to $1$. | [
"Solution:\nThe answer is all $n \\neq 2$. For $n=1$, the set $\\{1\\}$ works. For $n=2$, no set exists, simply because the sum of reciprocals of two distinct integers cannot be equal to $1$. For $n=3$, take $\\{2,3,6\\}$.\n\nFor $n>3$, the identity\n$$\n\\frac{1}{k} = \\frac{1}{k+r} + \\frac{1}{k(k+1)} + \\frac{1}... | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | all positive integers n except 2 | |
0i3k | Problem:
A polynomial $P$ has four roots, $\frac{1}{4}, \frac{1}{2}, 2, 4$. The product of the roots is $1$, and $P(1) = 1$. Find $P(0)$. | [
"Solution:\nA polynomial $Q$ with $n$ roots, $x_{1}, \\ldots, x_{n}$, and $Q\\left(x_{0}\\right) = 1$ is given by\n$$\nQ(x) = \\frac{(x - x_{1})(x - x_{2}) \\cdots (x - x_{n})}{(x_{0} - x_{1})(x_{0} - x_{2}) \\cdots (x_{0} - x_{n})}\n$$\nso\n$$\nP(0) = \\frac{1}{\\frac{3}{4} \\cdot \\frac{1}{2} \\cdot (-1) \\cdot (... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | 8/9 | |
05nd | Problem:
Soit $n \geqslant 2$ un nombre entier et soient $x_{1}, \ldots, x_{n}$ des nombres réels positifs tels que $x_{1} x_{2} \cdots x_{n}=1$. Montrer que
$$
\frac{1}{n-1+x_{1}}+\frac{1}{n-1+x_{2}}+\cdots+\frac{1}{n-1+x_{n}} \leqslant 1
$$ | [
"Solution:\n\nEn réduisant au même dénominateur, il s'agit de montrer que\n$$\n\\sum_{k=1}^{n} \\prod_{j \\neq k}\\left(x_{j}+n-1\\right) \\leqslant \\prod_{k=1}^{n}\\left(x_{k}+n-1\\right)\n$$\nOn note\n$$\n\\sigma_{k}=\\sum_{1 \\leqslant i_{1}<\\cdots<i_{k} \\leqslant n} x_{i_{1}} x_{i_{2}} \\cdots x_{i_{k}}\n$$\... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES - ENVOI No. 3 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
023r | Problem:
Julian treina em uma pista de $3~\mathrm{km}$. Ele percorre o primeiro quilômetro caminhando, o segundo correndo, e o terceiro em bicicleta. Se ele tivesse percorrido toda a pista em bicicleta, haveria demorado 10 minutos a menos. Julian corre ao dobro da velocidade com que caminha, e vai em bicicleta ao tripl... | [
"Solution:\nSeja $t$ o tempo, em minutos, que Julian demora para percorrer de bicicleta um quilômetro. Como ele vai de bicicleta ao triplo da velocidade com que caminha, então ele caminha um quilômetro em $3t$ minutos, e como ele corre ao dobro da velocidade com que caminha, então ele corre um quilômetro em $\\dfra... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 6 minutes | |
02d3 | Show that if $a < b$ are in the interval $[0, \pi/2]$ then $a - \sin a < b - \sin b$. Is this true for $a < b$ in the interval $[\pi, 3\pi/2]$? | [
"We have $\\sin b - \\sin a = 2 \\sin \\frac{b-a}{2} \\cos \\frac{b+a}{2} = 2 \\sin \\frac{b-a}{2} < 2(b-a)/2 = b-a$.\n\nThe second case is trivial because both $x$ and $-\\sin x$ are increasing in the interval $[\\pi, 3\\pi/2]$."
] | Brazil | I OBM | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | English | proof and answer | Yes | |
06lv | On the table there are 20 coins of weights $1$, $2$, $3$, $\ldots$, $15$, $37$, $38$, $39$, $40$ and $41$ grams. They all look alike but their colours are all distinct. Now Miss Adams knows the weight and colour of each coin, but Mr. Bean knows only the weights of the coins. There is also a balance on the table, and ea... | [
"Miss Adams needs to perform at least two operations.\nA strategy is as follows. In the first operation, she puts the coins of weights $1$ to $15$ on one side, and the three coins of weights $39$, $40$ and $41$ on the other side.\n\nThen the two groups will be of the same weight ($120$). Since the total weight of $... | Hong Kong | Year 2021 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | 2 | |
07ip | Two triangles $ABC$ and $C'B'A$ are congruent. $M$ is the midpoint of $AB$ and the extension of $CB'$ beyond $B'$ passes through $M$. Point $F$ lies on the arc $MC$ (which does not contain $B$) of the circumcircle of $BMC$ in such a way that $\angle FB'A = 90^\circ$, $\angle C'CB' \neq 90^\circ$. Prove that $\angle B'C... | [
"Let $F'$ be the intersection of the perpendicular bisector of $AB$ and the circumcircle of $BMC$. Let the circumcircle of $AF'M$ intersect $CM$ at $K$. It follows that\n$$\n\\begin{align*}\n\\angle KAF' &= \\angle KMF' = \\angle CBF' \\\\\n\\angle F'KA &= \\angle F'MA = \\angle F'CB = 90^\\circ \\\\\nF'A &= F'B\n\... | Iran | 41th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
07zc | Problem:
Fissiamo un punto $O$ nell'intersezione di due linee di un foglio a quadretti e indichiamo le quattro direzioni parallele alle linee come Nord, Sud, Est, Ovest (il Nord in alto). Muoviamoci, partendo da $O$ di un quadretto verso Est, poi due verso Nord, tre verso Ovest, quattro verso Sud, cinque verso Est e c... | [] | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | B | |
02it | Problem:
André, Bruno, Celina e Dalva ganharam juntos 21 medalhas num concurso. André foi o que mais ganhou medalhas, Bruno ganhou o dobro de Celina e Dalva 3 a mais que Bruno. Quantas medalhas cada um pode ter ganhado? | [
"Solution:\n\nDenotemos por $A$, $B$, $C$ e $D$ o número de medalhas ganhas por André, Bruno, Celina e Dalva respectivamente, então $A+B+C+D=21$. Agora, temos:\n- \"Bruno ganhou o dobro de Celina\" $\\Rightarrow B=2C$\n- \"Dalva 3 a mais que Bruno\": $\\Rightarrow D=B+3$\n\nDaí obtemos:\n$$\nA + B + C + D = 21\n$$\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Possible distributions: André 8, Bruno 4, Celina 2, Dalva 7; André 13, Bruno 2, Celina 1, Dalva 5; André 18, Bruno 0, Celina 0, Dalva 3. | |
0blb | Prove that if $x \in (0, \frac{\pi}{3})$, then $\sin 2x < \frac{2}{3x^2 - x^6}$.
Dumitru Săvulescu, Lucian Tuțescu | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
082b | Problem:
Sia $P(X) = a X^{2} + b X + c$ un polinomio di secondo grado con coefficienti reali (cioè $a, b, c$ sono numeri reali e $a \neq 0$). Se $P(2000) = 2000$ e $P(2001) = 2001$, allora $P(2002)$ non può essere uguale a:
(A) 2000
(B) 2001
(C) 2002
(D) 2003
(E) 2004. | [
"Solution:\n\nLa risposta è $(\\mathbf{C})$. Per semplicità, poniamo $Q(X) = P(X + 2000) - 2000$ : notiamo che $Q(X)$ è sempre un polinomio dello stesso grado di $P(X)$. Sia $Q(X) = \\alpha X^{2} + \\beta X + \\gamma$. Le condizioni date diventano $Q(0) = 0$ e $Q(1) = 1$, cioè $\\gamma = 0$ e $\\alpha + \\beta = 1$... | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | C | |
04ia | Let $a$, $b$, $c$ and $d$ be four pairwise distinct real numbers. If $a$ and $b$ are solutions of the equation $x^2 - 10c x - 11d = 0$, and $c$ and $d$ are solutions of the equation $x^2 - 10a x - 11b = 0$, determine the sum $a + b + c + d$. | [
"From Vieta's formulae we have $a + b = 10c$ and $c + d = 10a$. By adding these equations we get\n$$\na + b + c + d = 10(a + c).\n$$\nSince $a$ is a solution of the equation $x^2 - 10c x - 11d = 0$, and $d = 10a - c$, it follows that\n$$\n0 = a^2 - 10a c - 11d = a^2 - 10a c - 11(10a - c) = a^2 - 110a + 11c - 10a c.... | Croatia | Croatia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 1210 | |
0jy0 | Problem:
Yang has the sequence of integers $1,2, \ldots, 2017$. He makes 2016 swaps in order, where a swap changes the positions of two integers in the sequence. His goal is to end with $2,3, \ldots, 2017,1$. How many different sequences of swaps can Yang do to achieve his goal? | [
"Solution:\n\nLet $n=2017$. The problem is asking to write a cycle permutation of $n$ integers as the product of $n-1$ transpositions. Say that the transpositions Yang uses are $(a_{i}, b_{i})$ (i.e. swapping the $a_{i}$-th integer in the sequence with the $b_{i}$-th integer in the sequence). Draw the graph with ed... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | final answer only | 2017^2015 | |
0izi | Problem:
Let $p(x)$ and $q(x)$ be two cubic polynomials such that $p(0) = -24$, $q(0) = 30$, and
$$
p(q(x)) = q(p(x))
$$
for all real numbers $x$. Find the ordered pair $(p(3), q(6))$. | [
"Solution:\nAnswer: $(3, -24)$\n\nNote that the polynomials $f(x) = a x^{3}$ and $g(x) = -a x^{3}$ commute under composition. Let $h(x) = x + b$ be a linear polynomial, and note that its inverse $h^{-1}(x) = x - b$ is also a linear polynomial. The composite polynomials $h^{-1} f h$ and $h^{-1} g h$ commute, since f... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | (3, -24) | |
0ant | Problem:
Each of 12 students has a ticket to one of 12 chairs in a row at a theater. How many ways are there to seat the students so that each student sits either in the chair specified on his/her ticket or in one to the left or to the right of the specified chair?
(a) 233
(b) 225
(c) 187
(d) 252 | [] | Philippines | Qualifying Round | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | MCQ | a | |
0911 | Problem:
Determine the smallest positive integer $n$ for which the following statement holds true: From any $n$ consecutive integers one can select a non-empty set of consecutive integers such that their sum is divisible by $2019$. | [
"Solution:\n\nAnswer. $n=340$.\n\nThe prime factorization of $2019$ is $3 \\cdot 673$. Let $p=673$.\n\nFor each integer $k$, color the three numbers $k p-1$, $k p$, $k p+1$ red, and the six numbers $k p+\\frac{p-5}{2}$, $k p+\\frac{p-3}{2}$, $k p+\\frac{p-1}{2}$, $k p+\\frac{p+1}{2}$, $k p+\\frac{p+3}{2}$, $k p+\\f... | Middle European Mathematical Olympiad (MEMO) | MEMO Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 340 | |
01dm | Let $a_0, a_1, \dots$ be a sequence of positive integers such that $a_n = a_{n-1}^{2^n}$ for all $n = 1, 2, \dots$. Prove that for each prime $p$, $p > 3$, with residue $3$ modulo $4$ there exists a positive integer $a_0$ such that the sequence $a_N, a_{N+1}, a_{N+2}, \dots$ is not constant modulo $p$ for any positive ... | [
"Let $p$ be a prime with residue $3$ modulo $4$ and $p > 3$. Then $p - 1 = u \\cdot 2$ where $u > 1$ is odd. Choose $a_0 = 2$. The order of $2$ modulo $p$ (that is, the smallest positive integer $t$ such that $2^t \\equiv 1 \\pmod p$) is a divisor of $p-1 = u \\cdot 2$, but not a divisor of $2$ since $2^2 \\ne 1 \\... | Baltic Way | Baltic Way 2016 | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0hpy | Problem:
Count the number of permutations $a_{1} a_{2} \ldots a_{7}$ of $1234567$ with longest decreasing subsequence of length at most two (i.e. there does not exist $i<j<k$ such that $a_{i}>a_{j}>a_{k}$ ). | [
"Solution:\n\n$C(7) = 429$."
] | United States | null | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 429 | |
0iob | Problem:
Jack, Jill, and John play a game in which each randomly picks and then replaces a card from a standard 52 card deck, until a spades card is drawn. What is the probability that Jill draws the spade? (Jack, Jill, and John draw in that order, and the game repeats if no spade is drawn.) | [
"Solution:\n\n$\\boxed{\\dfrac{12}{37}}$"
] | United States | $10^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 12/37 | |
07xt | Suppose, for some pair of real numbers $x, y$, the real numbers $a, b, c$ satisfy the five equations
$$
\begin{aligned}
a+b+c &= 1 & bc-x-2ya &= 0 \\
a^2+b^2+c^2 &= 3 & ca-x-2yb &= 0 \\
ab-x-2yc &= 0.
\end{aligned}
$$
Prove that $abc$ is either $-1$ or $\frac{5}{27}$. | [
"Label the given equations as follows:\n$$\na + b + c = 1 \\tag{10}\n$$\n$$\na^2 + b^2 + c^2 = 3 \\tag{11}\n$$\n$$\nbc - x - 2ya = 0 \\tag{12}\n$$\n$$\nca - x - 2yb = 0 \\tag{13}\n$$\n$$\nab - x - 2yc = 0. \\tag{14}\n$$\n\n**Solution 1.** Eliminate $x$ by subtracting (13) from (12), and (14) from (13) to get $c(b-a... | Ireland | IRL_ABooklet_2025 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof only | null | |
0a6f | Problem:
Determine the least real number $L$ such that
$$
\frac{1}{a} + \frac{a}{b} + \frac{b}{c} + \frac{c}{d} \leqslant L
$$
for all quadruples $(a, b, c, d)$ of integers satisfying $1 < a < b < c < d$. | [
"Solution:\n\nAnswer: $3$. To solve this problem, two parts are required. Part A shows that $L = 3$ works. Part B shows that no $L' < 3$ works.\n\nPart A\n\nWe show that for all quadruples $(a, b, c, d)$ (with $1 < a < b < c < d$) we have\n$$\n\\frac{1}{a} + \\frac{a}{b} + \\frac{b}{c} + \\frac{c}{d} \\leqslant 3.\... | New Zealand | NZMO Round Two | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3 | |
0jux | Problem:
If $x + 2y - 3z = 7$ and $2x - y + 2z = 6$, determine $8x + y$. | [
"Solution:\n$8x + y = 2(x + 2y - 3z) + 3(2x - y + 2z) = 2(7) + 3(6) = 32$"
] | United States | HMMT November | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 32 | |
0e2m | Let $\triangle ABC$ be an acute triangle with $|AB| > |AC|$. Let $D$ be a point on the side $AB$, such that the angles $\angle ACD$ and $\angle CBD$ are equal. Let $E$ denote the midpoint of $BD$, and let $S$ be the circumcentre of the triangle $BCD$. Prove that the points $A, E, S$ and $C$ lie on the same circle. | [
"The central angle is twice the inscribed angle, so $\\angle CSD = 2\\angle CBD$. The triangle $CSD$ is isosceles with the apex at $S$, so\n$$\n\\angle DCS = \\frac{\\pi-\\angle CSD}{2} = \\frac{\\pi}{2} - \\angle CBD.\n$$\nThus,\n$$\n\\angle ACS = \\angle ACD + \\angle DCS = \\angle CBD + \\frac{\\pi}{2} - \\angle... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof only | null | |
0bld | Let $A$ be a ring. Prove that the following statements are equivalent:
a) For every elements $a$ and $b$ from $A$ there exists two odd integers $p$ and $q$ such that $a^2 + b^2 = (pa + qb)^2$;
b) The ring $A$ is commutative and $1 + 1 = 0$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Algebra > Abstract Algebra > Ring Theory"
] | null | proof only | null | |
0216 | Problem:
Determine all integers $k \geqslant 1$ with the following property: given $k$ different colours, if each integer is coloured in one of these $k$ colours, then there must exist integers $a_{1}<a_{2}<\cdots<a_{2023}$ of the same colour such that the differences $a_{2}-a_{1}, a_{3}-a_{2}, \ldots, a_{2023}-a_{2022... | [
"Solution:\nWe claim that only $k=1$ and $k=2$ satisfy the required property. First, if $k \\geqslant 3$, we colour each integer with its residue class modulo $3$, so that, whenever two integers have the same colour, their difference is divisible by $3$, so is not a power of $2$. This shows that no $k \\geqslant 3$... | Benelux Mathematical Olympiad | 15th Benelux Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1 and 2 | |
0c8k | Determine all functions $f: \mathbb{R} \to \mathbb{R}$, integrable over any bounded interval, satisfying the condition
$$
\int_{x-y}^{x+y} f(t) \, dt = y(f(x+y) + f(x-y)), \quad \text{for all real numbers } x \text{ and } y.
$$ | [
"It is readily checked that any affine function satisfies the condition in the statement. To prove the converse, set $y = x$ to write\n$$\n\\int_{0}^{2x} f(t) \\, dt = x(f(2x) + f(0)). \\quad (*)\n$$\nSince $f$ is integrable, $(*)$ shows that the function $g: \\mathbb{R} \\to \\mathbb{R}$, $g(x) = x(f(2x) + f(0))$,... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All affine functions: f(x) = a x + b for real constants a and b. | |
0i3n | Problem:
Evaluate $\sum_{n=0}^{\infty} \cot^{-1}\left(n^{2}+n+1\right)$. | [
"Solution:\n\n$\\sum_{n=0}^{\\infty} \\cot^{-1} \\left(n^{2}+n+1\\right) = \\sum_{n=0}^{\\infty} \\arctan \\left(\\frac{1}{n^{2}+n+1}\\right) = \\sum_{n=0}^{\\infty} \\arctan (n+1) - \\arctan (n)$ by the sum/difference formula for tangent. This sum, taken out to $n=N$, telescopes to $-\\arctan (0) + \\arctan (N+1)$... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | pi/2 | |
04v4 | Consider 26 letters $A, \dots, Z$. A *string* is a finite sequence consisting of those letters. We say that a string $s$ is *nice* if it contains each of the 26 letters at least once, and each permutation of letters $A, \dots, Z$ occurs in $s$ as a subsequence the same number of times. Prove that:
a. There exists a ni... | [] | Czech Republic | Czech-Polish-Slovak Match | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2022 | |
019y | Players $A$ and $B$ play a game: given a positive integer $n$, one has to choose a divisor $m$ of $n$ which is greater than $1$ and smaller than $n$ and replace $n$ with $n - m$. Player $A$ makes the first move, players move alternately. The player who can't make a move loses the game. For which starting positive numbe... | [
"Firstly note that for given $n$ exactly one player has a winning strategy. We'll show by induction that $B$ has a winning strategy if $n$ is odd.\n\nFirst step of induction is clear. Assume $n$ is odd and $B$ has a winning strategy for all odd integers smaller than $n$. If player $A$ can't make a move, $B$ wins. I... | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All odd n, and n = 2·4^k for k ≥ 0 (i.e., powers of two with odd exponent). | |
0dks | Given a set $S$ of $n$ elements, Ali wants to color all $2^n$ subsets of $S$ using $m$ colors such that if two different subsets $X$ and $Y$ are colored the same, the subset $X \cup Y$ must be colored in a color different from the color of $X$ and $Y$. What is the smallest $m$ for which Ali can achieve this? | [] | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n+1 | |
02m0 | Prove that there exists a positive integer $n_0$ with the following property: for each integer $n \ge n_0$ it is possible to partition a cube into $n$ smaller cubes. | [
"Consider the following two operations, A and B: A consists in cutting a cube in 8 equal cubes of half its dimensions; B consists in cutting a cube in 27 equal cubes of one third of its dimensions. A and B increases the total number of cubes in 7 and 26, respectively, so one can obtain $1 + 7x + 26y$ cubes by perfo... | Brazil | XXXI Brazilian Math Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
09ne | Let $d_n = \sum_{k=0}^{2n} \frac{(-1)^k}{n+k} = \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \dots + \frac{1}{3n}$ for $n \ge 1$.
Prove that
$$
\frac{2}{3n+1} < d_n < \frac{2}{3n-2}.
$$
(Otgonbayar Uuye) | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
03zs | Consider 12 figures on the clock face as 12 points. Color them in four colors: red, yellow, blue and green. Each color is used for three points. Configure $n$ convex quadrilaterals with vertices in these points, such that
(1) there are no same color of vertices for each quadrilateral.
(2) among any three of these quadr... | [
"We use $A, B, C, D$ to represent these four colors, respectively, and the points in the same color by lower letters as $a_1, a_2, a_3; b_1, b_2, b_3; c_1, c_2, c_3$ and $d_1, d_2, d_3$ respectively.\n\nNow consider color $A$. If, in $n$ quadrilaterals, the number of points $a_1, a_2, a_3$ in color $A$ are $n_1, n_... | China | China Southeastern Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 9 | |
03ng | Problem:
Let $n$ be an integer greater than $1$, and let $a_{0}, a_{1}, \ldots, a_{n}$ be real numbers with $a_{1} = a_{n-1} = 0$. Prove that for any real number $k$,
$$
\left|a_{0}\right| - \left|a_{n}\right| \leq \sum_{i=0}^{n-2} \left|a_{i} - k a_{i+1} - a_{i+2}\right|
$$ | [
"Solution:\nLet $Q(x) = x^{2} - k x - 1$ and let $P(x) = a_{0} + a_{1} x + \\cdots + a_{n} x^{n}$. Note that the product of the two roots of $Q(x)$ is $-1$ and thus one of the two roots has magnitude at most $1$. Let $z$ be this root. Now note that since $a_{1} = a_{n-1} = 0$, we have that\n$$\n\\begin{aligned}\n0 ... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
00jg | Let $a$ and $b$ be positive real numbers with $a \le 2b \le 4a$.
Prove that
$$
4ab \le 2(a^2 + b^2) \le 5ab.
$$ | [
"The terms in the left inequality can be rewritten as a square:\n$$\n2(a^2 + b^2) \\ge 4ab \\Leftrightarrow (a - b)^2 \\ge 0.\n$$\nThe square in the last inequality is clearly weakly positive.\n\nIn the right inequality, we multiply with $8$ and complete the square to obtain\n$$\n16a^2 - 40ab + 16b^2 \\le 0 \\Leftr... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0lgn | Problem:
On a party with 99 guests, hosts Ann and Bob play a game (the hosts are not regarded as guests). There are 99 chairs arranged in a circle; initially, all guests hang around those chairs. The hosts take turns alternately. By a turn, a host orders any standing guest to sit on an unoccupied chair $c$. If some ch... | [
"Solution:\n\nAnswer. $k=34$.\n\nPreliminary notes. Let $F$ denote the number of occupied chairs at the current position in the game. Notice that, on any turn, $F$ does not decrease. Thus, we need to determine the maximal value of $F$ Ann can guarantee after an arbitrary move (either hers or her opponent's).\n\nSay... | Zhautykov Olympiad | Zhautykov Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 34 | |
01pe | Find the smallest positive integer $n$ such that the number $2013n$ can be presented as the difference of two cubes of positive integer numbers. | [
"Answer: $n=39$. (Solution of A. Semchankau, A. Zhuk.) Let\n$$\n2013n = a^3 - b^3. \\qquad (1)\n$$\nThen\n$$\n(1) \\Leftrightarrow 61 \\cdot 11 \\cdot 3n = 2013n = (a-b)^3 + 3ab(a-b) \\Rightarrow (a-b) \\ge 3,\n$$\ni.e. $(a-b) = 3k, k \\in \\mathbb{N}$. So $61 \\cdot 11 \\cdot 3n = 3^3 k^3 + 3^2 abk$, whence $n \\g... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 39 | |
07nr | Suppose $a$, $b$, and $c$ are positive real numbers. Prove that,
$$
24abc \leq |a^3 + b^3 + c^3 - (a+b+c)^3| \leq \frac{8}{9}(a+b+c)^3,
$$
and that for both inequalities, equality occurs if and only if $a = b = c$. | [
"We factorise $a^3 + b^3 + c^3 - (a+b+c)^3$ as follows:\n$$\n\\begin{aligned}\n& a^3 + b^3 + c^3 - ((a+b)^3 + 3(a+b)^2c + 3(a+b)c^2 + c^3) \\\\\n&= a^3 + b^3 - (a^3 + b^3 + 3a^2b + 3ab^2 + b^3 + 3(a+b)^2c + 3(a+b)c^2) \\\\\n&= -3(a+b)(ab + (a+b)c + c^2) \\\\\n&= -3(a+b)(b+c)(c+a).\n\\end{aligned}\n$$\nHence $|a^3+b... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0hct | In the company there are no three people who know each other, but for any group of five there are three who have the common acquaintance (not necessarily from that group). Prove that this company can be split into two groups of pairwise unfamiliar people.
(ViktorBogdanskyi) | [
"We are going to show that the acquaintances graph has no odd cycles. For the sake of contradiction, let us take the smallest odd cycle. Clearly, there are no edges connecting vertices of this cycle (as it would divide the cycle into two smaller cycles, at least one of which would be odd). Next, let us enumerate pe... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
01gg | A real sequence $(a_n)_{n=0}^{\infty}$ is defined recursively by $a_0 = 2$ and the recursion formula
$$
a_n = \begin{cases} a_{n-1}^2 & \text{if } a_{n-1} < \sqrt{3} \\ \frac{a_{n-1}^2}{3} & \text{if } a_{n-1} \ge \sqrt{3}. \end{cases}
$$
Another real sequence $(b_n)_{n=1}^{\infty}$ is defined in terms of the first by ... | [
"The first step is to prove, using induction, the formula\n$$\na_n = \\frac{2^{2^n}}{3^{2^n}(b_1+b_2+\\cdots+b_n)}.\n$$\nThe base case $n = 0$ is trivial. Assume the formula is valid for $a_{n-1}$, that is,\n$$\na_{n-1} = \\frac{2^{2^{n-1}}}{3^{2^{n-1}}(b_1+b_2+\\cdots+b_{n-1})}.\n$$\nIf now $a_{n-1} < \\sqrt{3}$, ... | Baltic Way | Baltic Way 2020 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Exponential functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ipy | Problem:
Find the coefficient of $x^{6}$ in the expansion of
$$
(x+1)^{6} \cdot \sum_{i=0}^{6} x^{i}
$$ | [
"Solution:\nAnswer: 64\nEach term of $(x+1)^{6}$ can be multiplied by a unique power $x^{i}$, $0 \\leq i \\leq 6$, to get a sixth degree term. So the answer is the sum of the coefficients of the terms of $(x+1)^{6}$, which is the same as substituting $x=1$ into this power to get $2^{6}=64$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 64 | |
0cow | Seven skiers, numbered from $1$ to $7$, participated in a race. They started one by one, and each skier passed the distance with a constant speed (which could be different for different skiers). It happened that each skier participated in an overtaking exactly two times (in each overtaking, exactly two skiers participa... | [
"Заметим, что чётность места каждого лыжника менялась при любом обгоне; значит, его место на финише — той же чётности, что и на старте.\nТак как скорости постоянны, каждые два лыжника встречались не более одного раза. Будем обозначать лыжников их стартовыми номерами.\nПобедителя никто не мог обогнать, значит, он са... | Russia | Regional round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof only | null | |
03tq | Consider five points $A$, $B$, $C$, $D$ and $E$ such that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral. Let $l$ be a line passing through $A$. Suppose that $l$ intersects the interior of the segment $DC$ at $F$ and intersects line $BC$ at $G$. Suppose also that $EF = EG = EC$. Prove that $l$ is the bi... | [
"Draw the altitudes of two isosceles triangles $EGC$ and $ECF$ as in the figure.\nIn view of the given condition, it is easy to see that $\\triangle ADF \\sim \\triangle GCF$. Hence\n\n\n$$\n\\begin{aligned}\n\\frac{AD}{GC} &= \\frac{DF}{CF} \\Rightarrow \\frac{BC}{CG} = \\frac{DF}{CF} \\Ri... | China | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0i03 | Problem:
You are somewhere on a ladder with 5 rungs. You have a fair coin and an envelope that contains either a double-headed coin or a double-tailed coin, each with probability $1/2$. Every minute you flip a coin. If it lands heads you go up a rung, if it lands tails you go down a rung. If you move up from the top ru... | [
"Solution:\nFirst consider the probability of winning if you never open the envelope. Let $q(n)$ be the probability of winning from the $n$\\text{th}$ rung with just the fair coin, then $q(n) = \\frac{q(n-1) + q(n+1)}{2}$, so it is not hard to calculate that $q(n) = n/6$.\n\nIf we open the envelope, then there's a ... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | Open the envelope only on the bottom rung. | |
0ayd | Problem:
How many distinct prime factors does $5^{14} - 30 + 5^{13}$ have? | [] | Philippines | 21st PMO Area Stage | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | 7 | |
04k4 | If $a$ and $b$ are positive integers, then $\overline\{a.b\}$ is a decimal number obtained by writing the number $a$, then the decimal point and then the number $b$. For example, if $a = 20$ and $b = 17$, then $\overline\{a.b\} = 20.17$ and $\overline\{b.a\} = 17.2$.
Determine all pairs $(a, b)$ of positive integers su... | [
"If $(a, b)$ is a solution, then so is $(b, a)$ – and vice versa. Therefore, we may assume that $a \\ge b$.\nIf $a \\ge 10$, then necessarily $b = 1$, and the only possibilities for $a$ are $10$, $11$ and $12$.\nWe can verify that in those cases the product is not $13$.\nTherefore, numbers $a$ and $b$ are both smal... | Croatia | Mathematical competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (a, b) = (2, 5) and (5, 2) | |
059z | Find the value of
$$
\frac{1 \cdot 3}{3 \cdot 5} + \frac{2 \cdot 4}{5 \cdot 7} + \frac{3 \cdot 5}{7 \cdot 9} + \frac{4 \cdot 6}{9 \cdot 11} + \dots + \frac{1009 \cdot 1011}{2019 \cdot 2021}
$$ | [
"**Answer:** $\\frac{505 \\cdot 1009}{2021} = \\frac{509545}{2021} = 252 + \\frac{253}{2021}$.\n\nThe sum consists of $1009$ terms, where the $i$-th term is of the form $\\frac{i(i+2)}{(2i+1)(2i+3)}$. Let $s$ be the desired sum. Notice that\n$$\n\\frac{i(i+2)}{(2i+1)(2i+3)} = \\frac{1}{4} - \\frac{3}{4} \\cdot \\fr... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | final answer only | 509545/2021 | |
05tq | Problem:
Soit $ABC$ un triangle non isocèle en $B$. Soit $D$ le pied de la bissectrice de l'angle $\widehat{ABC}$. Soit $M$ le milieu de l'arc $\widehat{AC}$ contenant $B$ sur le cercle circonscrit au triangle $ABC$. Le cercle circonscrit au triangle $BDM$ coupe le segment $[AB]$ en un point $K$ distinct de $B$. Soit ... | [
"Solution:\n\n\n\nTout d'abord, on reconnaît dans la figure quelques points connus : le point $M$ est le milieu de l'arc $AC$ contenant $B$, il s'agit donc du pôle Nord du point $B$ dans le triangle $ABC$, c'est-à-dire qu'il est le point d'intersection de la bissectrice extérieure de l'angl... | France | Envoi 5: Pot Pourri | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07nv | The integer $N$ has eight positive divisors $1 = d_1 < d_2 < d_3 < \dots < d_8 = N$. Find the two smallest integers $N$ for which $2d_2d_5 = d_3d_4 + 3$. | [
"The number of distinct positive divisors of an integer $N = \\prod p_i^{e_i}$ with distinct primes $p_i$ is equal to $\\prod (e_i + 1)$. Because $8$ has only the factorizations $8$, $4 \\cdot 2$ and $2 \\cdot 2 \\cdot 2$, we have to consider three cases:\n\nCase A: $N = p^7$\nCase B: $N = p^3q$\nCase C: $N = pqr$,... | Ireland | Ireland | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1107 and 2013 | |
0bit | Let $S$ be a set of positive integers such that $\lfloor \sqrt{x} \rfloor = \lfloor \sqrt{y} \rfloor$ for all $x, y \in S$. Show that the products $xy$, where $x, y \in S$, are pairwise distinct. | [
"We first show that if $x_1, x_2, x_3, x_4$ are members of $S$ such that $x_1 x_2 \\le x_3 x_4$, then $x_1 + x_2 \\le x_3 + x_4$. Suppose, if possible, that $x_1 + x_2 > x_3 + x_4$. Let $n = \\lfloor \\sqrt{x} \\rfloor$, $x \\in S$, and write $x_k = n^2 + w_k$, where the $w_k$ are non-negative integers less than $2... | Romania | THE 2014 DANUBE MATHEMATICAL COMPETITION | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
05vt | Problem:
Aurélien écrit 11 entiers naturels au tableau. Montrer qu'il peut choisir certains de ces entiers et placer des signes + et - entre eux de telle sorte que le résultat soit divisible par 2021. | [
"Solution:\n\nOn peut voir un choix de certains entiers avec des signes $+$ et $-$ comme un choix de certains entiers que l'on va compter positivement et de certains entiers que l'on va compter négativement. Les nombres qu'Aurélien peut obtenir sont des nombres qui s'écrivent comme la différence entre deux sommes d... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
010m | Problem:
Determine all real numbers $a, b, c, d$ that satisfy the following system of equations.
$$
\left\{\begin{array}{r}
a b c+a b+b c+c a+a+b+c=1 \\
b c d+b c+c d+d b+b+c+d=9 \\
c d a+c d+d a+a c+c+d+a=9 \\
d a b+d a+a b+b d+d+a+b=9
\end{array}\right.
$$ | [
"Solution:\nAnswer: $a = b = c = \\sqrt[3]{2} - 1$, $d = 5 \\sqrt[3]{2} - 1$.\n\nSubstituting $A = a + 1$, $B = b + 1$, $C = c + 1$, $D = d + 1$, we obtain\n$$\n\\begin{aligned}\n& A B C = 2 \\\\\n& B C D = 10 \\\\\n& C D A = 10 \\\\\n& D A B = 10\n\\end{aligned}\n$$\nMultiplying (1), (2), (3) gives $C^{3}(A B D)^{... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a = b = c = 2^(1/3) - 1, d = 5·2^(1/3) - 1 | |
0aas | For which values of $x$ and $y$, the difference of $\frac{2x+15}{8}$ and $1\frac{1}{3} \cdot (y-1)$ will be 3 times smaller than $2 \cdot (5-2y)$ and $\frac{x+5}{2}$ will be for $0.125$ bigger than $3y$? | [
"From the condition in the problem we get the equations\n$$ \\frac{2x+15}{8} - 1\\frac{1}{3} \\cdot (y-1) = \\frac{2(5-2y)}{3} $$\nand\n$$ \\frac{x+5}{2} = 3y + 0.125. $$\nThey are equivalent to\n$$ 6x = 3 $$\nand\n$$ 4x - 24y = -22. $$\nThe solution to this system is\n$$ x = \\frac{1}{2} $$\nand\n$$ y = 1. $$"
] | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | x = 1/2, y = 7/8 | |
09mj | Let $P$ and $Q$ be distinct intersection points of circles $\omega_1$ and $\omega_2$. A line passing through point $P$ intersects $\omega_1$ again at point $A$ and $\omega_2$ again at point $B$. Line $l$ intersects $\omega_1$ at points $E$ and $D$, and intersects $\omega_2$ at points $C$ and $F$, where $C$ and $D$ are ... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
02me | Problem:
Sete números inteiros positivos e consecutivos estão escritos em ordem crescente numa mesma linha. Determine se é possível colocar entre esses números cinco sinais de "+" e só um de "=" de tal modo que resulte uma igualdade. | [] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Yes. The unique possibility is 9 + 10 + 11 + 12 = 13 + 14 + 15. | |
0227 | Problem:
1. Potências de 10 - O valor de $\frac{0,00001 \times (0,01)^{2} \times 1000}{0,001}$ é:
(a) $10^{-1}$
(b) $10^{-2}$
(c) $10^{-3}$
(d) $10^{-4}$
(e) 1 | [
"Solution:\n\n1. Potências de 10 - Temos:\n$$\n\\begin{aligned}\n\\frac{0,00001 \\times (0,01)^{2} \\times 1000}{0,001} & = \\frac{10^{-5} \\times \\left(10^{-2}\\right)^{2} \\times 10^{3}}{10^{-3}} = \\frac{10^{-5} \\times 10^{-4} \\times 10^{3}}{10^{-3}} = \\\\\n& = \\frac{10^{-5+(-4)+3}}{10^{-3}} = \\frac{10^{-6... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | c | |
09zy | Problem:
Vind alle functies $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ waarvoor geldt dat $f(n) \mid f(m)-n$ dan en slechts dan als $n \mid m$ voor alle natuurlijke getallen $m$ en $n$. | [
"Solution:\nAls we $m=n$ invullen, dan vinden we dat $f(n) \\mid f(n)-n$, dus voor alle natuurlijke getallen geldt $f(n) \\mid n$. Als we dat weer gebruiken op de originele functievoorwaarde vinden we dat $f(n) \\mid f(m)$ dan en slechts dan als $n \\mid m$.\nNu bewijzen we dat $f(n)=n$ via inductie naar het aantal... | Netherlands | Selectietoets | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | f(n)=n | |
01qy | Determine all functions $f: \mathbb{Z} \to \mathbb{Z}$ satisfying
$$
f(f(m) + n) + f(m) = f(n) + f(3m) + 2014
$$
for all integers $m$ and $n$. | [
"(Solution by I. Voronovich.) Rewrite the given equation as\n$$\nf(f(m) + n) = f(n) - f(m) + f(3m) + c, \\quad (1)\n$$\nwhere $c = 2014$; all we need in this solution is that $c$ is even and coprime with 3. Let $f(0) = a$.\n\n1) Setting $m = 0$ we get $f(n+a) = f(n)+c$, then by standard induction to both sides $f(n... | Belarus | SELECTION and TRAINING SESSION | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | f(n) = 2n + 1007 for all integers n | |
05fw | Problem:
Sept élèves d'une classe comparent leurs notes dans 12 épreuves, et remarquent qu'il n'existe pas deux élèves ayant des notes identiques dans chacune des 12 épreuves.
Montrer que l'on peut choisir 6 épreuves telles que deux élèves quelconques aient des notes différentes à au moins l'une de ces épreuves. | [
"Solution:\n\nSoit $m$ le plus petit entier tel que l'on peut choisir $m$ épreuves $E_{1}, \\ldots, E_{m}$ de sorte que deux élèves quelconques aient des notes différentes à au moins l'une de ces $m$ épreuves. On doit montrer que $m \\leqslant 6$.\n\nComme $m$ est minimal, pour tout $i=1, \\ldots, m$ il existe une ... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0ld5 | For positive integers $a$, $n$; consider the following equation:
$$
a^2x + 6a y + 36z = n
$$
where $x$, $y$, $z$ are non-negative natural numbers.
a) Find all integers $a$ such that for all $n \ge 250$, the given equation always has natural roots $(x, y, z)$.
b) Given that $a > 1$ and $\gcd(a, 6) = 1$. Find the great... | [
"We first state a well-known lemma\n**Lemma.** (Sylvester's theorem) For two positive integers $a$ and $b$ such that $\\gcd(a, b) = 1$, the largest integer that could not be written in the form $a x + b y$ where $x$ and $y$ are non-negative integers is $N_0 = ab - a - b$.\n\na) Let $a$ be the satisfying value. A na... | Vietnam | VMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | a) a = 1 or a = 5.
b) The greatest n with no solution is 5a^2 + 30a − 36. |
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