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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0fja | Problem:
Las longitudes de los lados de un triángulo están en progresión geométrica de razón $r$. Halla los valores de $r$ para los que el triángulo es, respectivamente, acutángulo, rectángulo u obtusángulo. | [
"Solution:\nPodemos suponer $r \\geq 1$, puesto que si no, basta con invertir el orden en que se consideran los lados. Si tomamos como unidad de longitud la del lado más corto, entonces las longitudes son $1$, $r$ y $r^{2}$. Llamemos $a$ al ángulo opuesto a $r^{2}$, que es el mayor. Por el teorema del coseno, se ti... | Spain | Viernes 19 de enero de 2001 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof and answer | Right: r = sqrt((1 + sqrt(5)) / 2). Acute: 1 ≤ r < sqrt((1 + sqrt(5)) / 2). Obtuse: sqrt((1 + sqrt(5)) / 2) < r < (1 + sqrt(5)) / 2. | |
036k | Problem:
Let $k$ be a circle with center $O$ and radius $\sqrt{5}$ and let $M$ and $N$ be points on a diameter of $k$ such that $MO = NO$. The chords $AB$ and $AC$, passing through $M$ and $N$, respectively, are such that
$$
\frac{1}{MB^{2}} + \frac{1}{NC^{2}} = \frac{3}{MN^{2}}
$$
Find the length of $MO$. | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 1 | |
07xy | The 2025 members of a football club run a tournament of “45-a-side” football matches, where each football match involves 90 people. After the tournament is concluded, it is noticed that no two football matches had four players in common. Prove that the number of football matches is less than 90. | [
"Denote the players by $P_1, P_2, \\dots, P_{2025}$ and let the number of matches be $N$. Also, denote the set of participants of match $i$ by $\\mathcal{T}_i$ for each $i = 1, 2, \\dots, N$. We will count, in two different ways, the total number of triples $(P_i, \\mathcal{T}_j, \\mathcal{T}_k)$ such that $P_i$ pa... | Ireland | IRL_ABooklet_2025 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0286 | Problem:
Sejam $O$ e $H$ o circuncentro e o ortocentro do triângulo $\triangle ABC$, respectivamente. Sejam $O_1$, $O_2$ tais que $AC$ é mediatriz de $OO_1$ e $BC$ é mediatriz de $OO_2$, respectivamente.
a) Verifique que $\angle BAH = \angle OAC$.
b) Se $M$ é o ponto médio de $BC$, mostre que $OM = \frac{AH}{2}$.
c... | [
"Solution:\n\na) Como o triângulo $ABF$ é retângulo, temos $\\angle BAH = 90^\\circ - \\angle ABC$. No triângulo isósceles $AOC$, o ângulo da base $\\angle OAC$ mede $\\frac{180^\\circ - \\angle AOC}{2} = 90^\\circ - \\angle ABC$. Portanto, $\\angle BAH = \\angle OAC$.\n\nb) Sejam $D$ e $G$ as interseções, diferent... | Brazil | NÍVEL 3 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof and answer | 80° | |
08qc | Problem:
Find all integers $x, y$ such that
$$
x^{3}(y+1)+y^{3}(x+1)=19
$$ | [
"Solution:\nSubstituting $s = x + y$ and $p = x y$ we get\n$$\n2 p^{2} - (s^{2} - 3 s) p + 19 - s^{3} = 0\n$$\nThis is a quadratic equation in $p$ with discriminant $D = s^{4} + 2 s^{3} + 9 s^{2} - 152$.\nFor each $s$ we have $D < (s^{2} + s + 5)^{2}$ as this is equivalent to $(2 s + 5)^{2} + 329 > 0$.\nFor $s \\ge... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (-1, -20), (-20, -1), (1, 2), (2, 1) | |
0ezh | Problem:
Given 5 segments such that any 3 can be used to form a triangle. Show that at least one of the triangles is acute-angled. | [
"Solution:\n\nLet the segments have lengths $a \\leq b \\leq c \\leq d \\leq e$.\n\nSuppose, for contradiction, that all triangles formed are obtuse-angled. Then, for any triangle with sides $x \\leq y \\leq z$, the triangle is obtuse only if $z^2 > x^2 + y^2$.\n\nConsider the three largest segments: $c$, $d$, $e$.... | Soviet Union | 4th ASU | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
06vf | There are 60 empty boxes $B_{1}, \ldots, B_{60}$ in a row on a table and an unlimited supply of pebbles. Given a positive integer $n$, Alice and Bob play the following game.
In the first round, Alice takes $n$ pebbles and distributes them into the 60 boxes as she wishes. Each subsequent round consists of two steps:
a... | [
"Solution 1 (Alice).\n\nAlice initially distributes pebbles according to $V_{\\left\\lceil\\frac{N}{2}\\right\\rceil}$. Suppose the current configuration of pebbles dominates $V_{i}$. If Bob makes a $k$-move with $k \\geqslant i$ then Alice picks the left group, which results in a configuration that dominates $V_{i... | IMO | IMO 2019 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 960 | |
0b3f | Problem:
Call a whole number ordinary if the product of its digits is less than or equal to the sum of its digits. How many numbers from the set $\{1,2, \ldots, 999\}$ are ordinary?
(a) 151
(b) 162
(c) 230
(d) 241 | [] | Philippines | 24th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | d | |
098y | Problem:
Pe laturile $BC$ și $BF$ ale paralelogramului $ABCD$ se iau respective punctele $E$ și $F$ astfel încât $\frac{EB}{EC}=p$ și $\frac{FC}{FD}=q$. Fie $M$ punctul de intersecție a dreptelor $AE$ și $BF$, iar $N$ punctul de intersecție a dreptelor $DE$ și $BF$. Determinați valorile raporturilor $\frac{AM}{ME}$ și... | [
"Solution:\n\nTrasăm $CL \\parallel BF$ ($L \\in AB$) și fie $K$ intersecția dreptelor $CL$ și $AE$.\n$$\n\\begin{aligned}\n& BM \\parallel CK \\Rightarrow \\triangle BEM \\sim \\triangle CEK \\Rightarrow \\frac{KE}{EM}=\\frac{CE}{EB}=\\frac{1}{p} \\\\\n& \\frac{AM}{ME}=\\frac{AM}{MK} \\cdot \\frac{MK}{ME} \\\\\n& ... | Moldova | OLIMPIADA REPUBLICANĂ LA MATEMATICĂ | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | AM/ME = ((p+1)(q+1))/(pq); DN/NE = (p+1)/(pq) | |
0jud | Problem:
Determine the remainder when
$$
\sum_{i=0}^{2015}\left\lfloor\frac{2^{i}}{25}\right\rfloor
$$
is divided by $100$, where $\lfloor x\rfloor$ denotes the largest integer not greater than $x$. | [
"Solution:\nLet $r_{i}$ denote the remainder when $2^{i}$ is divided by $25$. Note that because $2^{\\phi(25)} \\equiv 2^{20} \\equiv 1 \\pmod{25}$, $r$ is periodic with length $20$. In addition, we find that $20$ is the order of $2 \\bmod 25$. Since $2^{i}$ is never a multiple of $5$, all possible integers from $1... | United States | HMMT February 2016 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 14 | |
0aw9 | Problem:
In a certain school, there are $5000$ students. Each student is assigned an ID number from $0001$ to $5000$. No two students can have the same ID number. If a student is selected uniformly at random, what is the probability that the ID number of the student does not contain any $2$s among its digits? | [
"Solution:\n\nThe number of ID numbers in the range $0001$ to $5000$ which do not contain any $2$s is given by $4 \\times 9 \\times 9 + 1 - 1 = 2916$. For the thousands digit, we can use a number from $0$ to $4$ and for the rest we can use any digit except $2$. We add one since $5000$ was not counted and we subtrac... | Philippines | 18th PMO National Stage Oral Phase | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 729/1250 | |
08if | Problem:
The distinct points $M$ and $N$ lie on the hypotenuse ($AC$) of the right isosceles triangle $ABC$ so that $M \in (AN)$ and $MN^{2} = AM^{2} + CN^{2}$. Prove that $m(\angle MBN) = 45^{\circ}$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0hs9 | Problem:
There are seven green amoeba and three blue amoeba in a dish. Every minute, each amoeba splits into two identical copies; then, we randomly remove half the amoeba (thus there are always $10$ amoeba remaining). This process continues until all amoeba are the same color. What is the probability that this color ... | [
"Solution:\n\nThe answer is $70\\%$. Rather than considering the given colors, imagine the amoeba are colored with the following ten colors:\n- Forest green\n- Hunter green\n- Jungle green\n- Myrtle green\n- Mint green\n- Tea green\n- Aqua green\n- Sky blue\n- Navy blue\n- Midnight blue\n\nNow, we simply wait until... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 70% | |
08o8 | Problem:
Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 1$. Prove that
$$
\frac{7 + 2b}{1 + a} + \frac{7 + 2c}{1 + b} + \frac{7 + 2a}{1 + c} \geq \frac{69}{4}.
$$
When does equality hold? | [
"Solution:\n\nThe inequality can be written as:\n$$\n\\frac{5 + 2(1 + b)}{1 + a} + \\frac{5 + 2(1 + c)}{1 + b} + \\frac{5 + 2(1 + a)}{1 + c} \\geq \\frac{69}{4}.\n$$\nWe substitute $1 + a = x$, $1 + b = y$, $1 + c = z$.\nSo, we have to prove the inequality\n$$\n\\frac{5 + 2y}{x} + \\frac{5 + 2z}{y} + \\frac{5 + 2x}... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds when a = b = c = 1/3. | |
0irn | Problem:
Pirate ships Somy and Lia are having a tough time. At the end of the year, they are both one pillage short of the minimum required for maintaining membership in the Pirate Guild, so they decide to pillage each other to bring their counts up. Somy by tradition only pillages $28 \cdot 3^{k}$ coins for integers ... | [
"Solution:\n\nAnswer: 2\n\nClearly, $n=1$ cannot be achieved, because $28 \\cdot 3^{k}$ is never a multiple of $82$. However, two pillages is enough: Somy pillages $28$ and $28 \\cdot 81$ from Lia, and Lia pillages $81$ and $81 \\cdot 27$ from Somy. As is easily checked, both pillage $28 \\cdot 82$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 2 | |
03jl | Problem:
Suppose that $S$ is a finite set of at least five points in the plane; some are coloured red, the others are coloured blue. No subset of three or more similarly coloured points is collinear. Show that there is a triangle
(i) whose vertices are all the same colour, and
(ii) at least one side of the triangle do... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0a5n | Problem:
Let $S$ be a set of 10 positive integers. Prove that one can find two disjoint subsets $A = \{a_{1}, \ldots , a_{k}\}$ and $B = \{b_{1}, \ldots , b_{k}\}$ of $S$ with $|A| = |B|$ such that the sums
$$x = \frac{1}{a_{1}} +\dots +\frac{1}{a_{k}}$$
and
$$y = \frac{1}{b_{1}} +\dots +\frac{1}{b_{k}}$$
diffe... | [
"Solution:\nPartition the interval $(0.00, 2.50]$ into 250 intervals each of size 0.01. \n$$(0.00, 2.50] = (0.00, 0.01] \\cup (0.01, 0.02] \\cup (0.02, 0.03] \\cup \\dots \\cup (2.49, 2.50].$$ \nNow consider all possible sets, $S$, we can choose from the given 10 positive integers with $|S| = 5$. Because each of ... | New Zealand | New Zealand Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
08a5 | Problem:
Per ogni numero naturale $n$ di 3 cifre decimali (quindi con la prima cifra diversa da zero), consideriamo il numero $n_{0}$ ottenuto da $n$ eliminando le sue eventuali cifre uguali a zero. Per esempio, se $n=205$ allora $n_{0}=25$.
Determinare il numero degli interi $n$ di tre cifre per i quali $n_{0}$ è un ... | [
"Solution:\n\nEscludiamo subito il caso in cui $n$ abbia tutte le cifre diverse da zero, in quanto in questo caso si ha $n_{0}=n$. Restano quindi due possibilità:\n\n1) $n$ ha la terza cifra uguale a zero; la scrittura decimale di $n$ è del tipo $n=ab0$, $n_{0}=ab$ e $n_{0} \\mid n$ per ogni scelta di $a \\in \\{1,... | Italy | Cesenatico | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 93 | |
0l1a | A graph is symmetric about a line if the graph remains unchanged after reflection in that line. For how many quadruples of integers $(a, b, c, d)$, where $|a|, |b|, |c|, |d| \le 5$ and $c$ and $d$ are not both $0$, is the graph of
$$
y = \frac{ax + b}{cx + d}
$$
symmetric about the line $y = x$? | [
"**Answer (B):** There are two cases, depending on whether the denominator of the fraction is constant.\n\n**CASE 1.** Suppose that $c = 0$. Then the graph of $y = \\frac{ax+b}{d}$ is a line. It is symmetric with respect to the line $y = x$ if the slope of the line is $1$ and the line passes through the origin, or ... | United States | AMC 12 A | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | B | |
0hml | Problem:
Circles $k_{1}$ and $k_{2}$ intersect at points $A$ and $B$. Line $l$ is the common tangent to these circles and it touches $k_{1}$ at $C$ and $k_{2}$ at $D$ such that $B$ belongs to the interior of the triangle $ACD$. Prove that $\angle CAD + \angle CBD = 180^{\circ}$. | [
"Solution:\n\nHere we use the fact that the angle between the tangent and the chord is equal to the peripheral angle corresponding to the chord (this fact is easy to verify and can be found in any book on geometry). We have $\\angle CAB = \\angle DCB$ and $\\angle DAB = \\angle DCB$. Hence $\\angle CAD = \\angle CA... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a9z | Problem:
Let $n > 1$ and $p(x) = x^{n} + a_{n-1} x^{n-1} + \cdots + a_{0}$ be a polynomial with $n$ real roots (counted with multiplicity). Let the polynomial $q$ be defined by
$$
q(x) = \prod_{j=1}^{2015} p(x + j)
$$
We know that $p(2015) = 2015$. Prove that $q$ has at least 1970 different roots $r_{1}, \ldots, r_{197... | [
"Solution:\nLet $h_{j}(x) = p(x + j)$. Consider $h_{2015}$. Like $p$, it has $n$ real roots $s_{1}, s_{2}, \\ldots, s_{n}$, and $h_{2015}(0) = p(2015) = 2015$. By Viète, the product $\\left| s_{1} s_{2} \\cdots s_{n} \\right|$ equals $2015$. Since $n \\geq 2$, there is at least one $s_{j}$ such that $\\left| s_{j} ... | Nordic Mathematical Olympiad | The 29th Nordic Mathematical Contest | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
00b6 | A grid rectangle that is not a square is cut into 8 different (non-congruent) grid polygons along the grid lines. What is its minimal possible area? | [
"There is one grid polygon of area $1$ ($1\\times 1$ square), one such polygon of area $2$ ($1\\times 2$ rectangle), $2$ such polygons of area $3$ ($1\\times 3$ rectangle and an angle of $3$ squares). To satisfy the condition one must then use at least $4$ grid polygons of area $4$ or greater. Hence the area of the... | Argentina | Argentina_2018 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 26 | |
0gjk | 令 $k$ 為一正整數。天橋上的魔術師和小不點玩一場遊戲。一開始，小不典將 $N = 2^k$ 顆相異的球在桌面上排成一橫排，並各自用一個杯子罩住。在每一回合，小不點可以指定兩個杯子，然後魔術師可以交換這兩個杯子內的球，或著選擇做假動作讓兩個杯子內的球維持不變。小不點無法透過魔術師的動作來判別是否為假動作，也無法看到動作前後杯子內的球。
在 $M = k \times 2^{k-1}$ 個回合後，魔術師會打開所有杯子，讓小不點確認每個杯子裡的球。若此時小不點可以確知魔術師在每一回合中是否有做假動作，則小不點獲勝。證明小不點存在必勝策略。
Let $k$ be a positive integer. The little one a... | [
"將所有杯子依序編號為 1 到 $N$。我們將用歸納法建構策略。$k=1$ 時顯然。現在假設 $k-1$ 時存在一個必勝策略，其在第 $i=1,2,\\dots,(k-1)2^{k-2}$ 回合選取編號 $a_i$ 與 $b_i$ 的杯子。則在 $k$ 時，令 $M_1 = 2^{k-1}$ 與 $M_2 = (k+1)2^{k-2}$，並考慮以下策略：\n\nStage 1: 在第 $i = 1,2,\\dots,M_1$ 回合選取編號 $i$ 與 $i + M_1$\n(注意到 $i + M_1 \\le 2M_1 = 2^k$)；\n\nStage 2: 在第 $i = M_1 + 1, \\dots, M_2$ 回合... | Taiwan | IMO 2J, Independent Study 2 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Chinese; English | proof only | null | |
0gy2 | Given the sequence of polynomials $P_0(x), P_1(x), \dots, P_n(x)$, $n \ge 2$. It is known that for every integer $i$ ($0 \le i \le n$): $\deg(P_i(x)) = n-i$, and $P_n(x) \ne 0$. It is also known that for every integer $i$ ($2 \le i \le n$) there exists polynomial $Q_i(x)$ such that $P_i(x) = P_{i-2}(x) + P_{i-1}(x)Q_i(... | [
"It is possible that the degree of the sum of two polynomials is less than the degree of one summand only if their degrees equal. For shortening we pull down $(x)$ in the notation. $\\deg P_i < \\deg P_{i-2}$ and $P_i = P_{i-2} + P_{i-1}Q_i$, therefore $\\deg(P_{i-1}Q_i) = n-i+2$, hence $\\deg(Q_i) = 1$ for all int... | Ukraine | The Problems of Ukrainian Authors | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0iwj | Problem:
$$
F(x) = \sum_{n=1}^{105^{m}} \frac{d(n)}{n^{x}}
$$
Define the numbers $a(n)$ to be the positive integers for which
$$
F(x)^{2} = \sum_{n=1}^{105^{2m}} \frac{a(n)}{n^{x}}
$$
for all real $x$. Express $a\left(105^{m}\right)$ in terms of $m$. | [
"Solution:\n\nThe denominator of a term in the expansion of $F(x)^{2}$ is equal to $n^{x}$ if and only if it is a product of two terms of $F$ of the form $\\frac{d(n / k)}{(n / k)^{x}}$ and $\\frac{d(k)}{k^{x}}$ for some divisor $k$ of $n$. Thus $a\\left(105^{m}\\right) = \\sum_{k \\mid 105^{m}} d(k) d\\left(\\frac... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | binom(m+3, 3)^3 | |
0j9i | Problem:
Purineqa is making a pizza for Arno. There are five toppings that she can put on the pizza. However, Arno is very picky and only likes some subset of the five toppings. Purineqa makes five pizzas, each with some subset of the five toppings. For each pizza, Arno states (with either a "yes" or a "no") if the pi... | [
"Solution:\n\nWe claim the only way for Purineqa to deduce Arno's preferences is for each pizza to contain exactly one topping, with no topping be repeated. It is obvious that she can deduce the toppings in this case.\n\nWe now claim that this is not possible with any other combination. Suppose that Arno tells Puri... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Logic"
] | null | proof and answer | The five initial pizzas must each have exactly one topping, with all five toppings appearing exactly once; no other configuration works. | |
0jdn | Problem:
Triangle $ABC$ is inscribed in a circle $\omega$. Let the bisector of angle $A$ meet $\omega$ at $D$ and $BC$ at $E$. Let the reflections of $A$ across $D$ and $C$ be $D'$ and $C'$, respectively. Suppose that $\angle A = 60^\circ$, $AB = 3$, and $AE = 4$. If the tangent to $\omega$ at $A$ meets line $BC$ at $... | [
"Solution:\n\nAnswer: $2 \\sqrt{13-6 \\sqrt{3}}$\n\nFirst observe that by angle chasing, $\\angle PAE = 180 - \\frac{1}{2} \\angle BAC - \\angle ABC = \\angle AEP$, so by the cyclic quadrilateral $APD'F$, $\\angle EFD' = \\angle PAE = \\angle PEA = \\angle D'EF$. Thus, $ED'F$ is isosceles.\n\nDefine $B'$ to be the ... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2 sqrt(13-6 sqrt 3) | |
08b8 | Problem:
Dato un triangolo $ABC$, sia $A'$ il simmetrico di $A$ rispetto a $C$, $A''$ il simmetrico di $A$ rispetto a $B$, $B'$ il simmetrico di $B$ rispetto a $A$, $B''$ il simmetrico di $B$ rispetto a $C$, $C'$ il simmetrico di $C$ rispetto a $B$ e $C''$ il simmetrico di $C$ rispetto a $A$. Determinare il rapporto t... | [
"Solution:\n\nLa risposta è (B). Il segmento $A'B''$ è il simmetrico del segmento $AB$ rispetto al punto $C$. Quindi $A'B''$ e $AB$ sono paralleli e congruenti. Inoltre $AB' = AB$ perché $B'$ è il simmetrico di $B$ rispetto a $A$. Quindi il quadrilatero $AB'B''A'$ è un parallelogramma perché ha una coppia di lati o... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | B | |
039m | Find all values of the real parameter $p$ such that the equation $x^2 + (p^2 + 1)x + p = 2$ has two real distinct roots $x_1$ and $x_2$ such that
$$
\frac{2x_1 - 1}{x_2} + \frac{2x_2 - 1}{x_1} = x_1x_2 + \frac{55}{x_1x_2}.
$$ | [] | Bulgaria | Winter Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | -2 | |
0ach | There were 25 birds on two branches. After a while 5 birds from the first branch flew over to the second and after a while longer 7 birds from the second flew away. In that moment on the first branch there were twice as many birds than on the second. How many birds were there on each of the two branches in the beginnin... | [
"When the 7 birds flew away from the second branch the total number of birds on both branches was $25-7=18$. Because in that moment on the first branch there were twice as many birds than on the second, there were $18:3=6$ birds on the second branch, and on the first one were $2 \\cdot 6=12$ birds. Hence, before th... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | Initially 17 birds on the first branch and 8 birds on the second branch. | |
0cjd | We will call *smalls* the positive integers not larger than $2025$.
a) Find the number of smalls which are perfect squares.
b) Find the number of smalls which are perfect squares and leave remainder $0$ when divided by $45$.
c) Find the number of smalls which neither are perfect squares, nor leave remainder $0$ when... | [
"a) Since $45^2 = 2025$, the smalls which are perfect squares are $1^2, 2^2, 3^2, \\dots, 45^2$, that is $45$ such numbers.\n\n\nb) The smalls which are perfect squares and are divisible by $45$ are $15^2, 4 \\cdot 15^2$ and $9 \\cdot 15^2$ – there are $3$ of them.\n\n\nc) The smalls divisible by $45$ are $45 \\cdo... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | a) 45, b) 3, c) 1938 | |
01b0 | Real numbers $x$ and $y$ are such that $x^4 y^2 + y^4 + 2x^3 y + 6x^2 y + x^2 + 8 \le 0$. Prove that $x \ge -1/6$. | [
"By removing $y^4$ from the left hand side of our inequality, we see that the inequality $x^4 y^2 + 2x^2(x+3)y + x^2 + 8 \\le 0$ holds. This is a quadratic inequality with respect to $y$ whose discriminant is equal to $4x^4(6x+1)$. If $x < -1/6$, then the discriminant is negative, so that the left hand side must be... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
03sx | Given a right triangular prism $A_1B_1C_1 - ABC$ with $\angle BAC = \frac{\pi}{2}$ and $AB = AC = AA_1 = 1$, let $G, E$ be the midpoints of $A_1B_1$, $CC_1$ respectively; and $D, F$ be variable points lying on segments $AC, AB$ (not including endpoints) respectively. If $GD \perp EF$, the range of the length of $DF$ is... | [
"We establish a coordinate system with point $A$ as the origin, line $AB$ as the $x$-axis, $AC$ the $y$-axis and $AA_1$ the $z$-axis. Then we have $F(t_1, 0, 0)$ ($0 < t_1 < 1$), $E(0, 1, \\frac{1}{2})$,\n$G(\\frac{1}{2}, 0, 1)$, $D(0, t_2, 0)$ ($0 < t_2 < 1$). Therefore $\\vec{EF} = (t_1, -1, -\\frac{1}{2})$, $\\v... | China | China Mathematical Competition | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Linear Algebra > Vectors"
] | English | MCQ | A | |
0bf5 | Let $n$ be a positive integer. Show that $2\sqrt{2^n} \cos\left(n \arccos \frac{\sqrt{2}}{4}\right)$ is an odd integer number. | [
"Let $\\alpha = \\arccos \\frac{\\sqrt{2}}{4}$ and $z = \\cos \\alpha + i \\sin \\alpha$. Since $\\cos n\\alpha = \\frac{1}{2}(z^n + \\bar{z}^n)$, we have $2\\sqrt{2^n} \\cos n\\alpha = \\sqrt{2^n}(z^n + \\bar{z}^n) = S_n$.\n\nNotice that $z^n + \\bar{z}^n = (z + \\bar{z})(z^{n-1} + \\bar{z}^{n-1}) - z\\bar{z}(z^{n... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | null | proof only | null | |
0aod | Problem:
Let $\{JOHN\}$ be a rhombus with $JH = 16$ and $ON = 12$. Let $G$ and $P$ be points in $JN$ and $HN$ respectively such that $JG : GN : NP = 2 : 2 : 1$. What is the length $GP$?
(a) $\frac{2 \sqrt{19}}{5}$
(b) $\frac{2 \sqrt{13}}{3}$
(c) $\frac{3 \sqrt{17}}{2}$
(d) $\frac{3 \sqrt{15}}{2}$ | [] | Philippines | Qualifying Round | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | (c) | |
07na | The incircle $C_1$ of triangle $ABC$ touches the sides $AB$ and $AC$ at the points $D$ and $E$, respectively. The incircle $C_2$ of the triangle $ADE$ touches the sides $AB$ and $AC$ at the points $P$ and $Q$, and intersects the circle $C_1$ at the points $M$ and $N$. Prove that
(a) the centre of the circle $C_2$ lies ... | [
"(a) Let $I$ be the incentre of $\\triangle ABC$, and let $J$ be the point of intersection of $AI$ and $C_1$. Then, the line $AJ$ is the angle bisector of $\\angle DAE$ and the triangles $ADI$ and $AEI$ are congruent. In particular, $\\angle EIA = \\angle AID$. We have $\\angle EIJ = 2\\angle EDJ$ as both are subte... | Ireland | Ireland | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",... | English | proof only | null | |
080j | Problem:
Per ogni numero reale $x$, indichiamo con $[x]$ la "parte intera di $x$", definita come il più grande intero $\leq x$. Così ad esempio abbiamo che $[3/2]=1$, $[\pi]=3$, $[8]=8$. Determinare quante sono le soluzioni reali positive $(>0)$ dell'equazione $32^{x}=64^{[x]}$. | [
"Solution:\n\nLa risposta è 4. L'equazione si può scrivere nella forma $2^{5x}=2^{6[x]}$, che è equivalente a $5x=6[x]$. Poiché $x<[x]+1$, si ha che $6[x]=5x<5[x]+5$, da cui $[x]<5$. Poiché per ogni $x>0$ la parte intera $[x]$ è un intero $\\geq 0$, rimangono solo le possibilità $[x]=0,1,2,3,4$. Sostituendo tali va... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 4 | |
08kh | Problem:
Let $A$ and $P$ are the points of intersection of the circles $k_{1}$ and $k_{2}$ with centers $O$ and $K$, respectively. Let also $B$ and $C$ be the symmetric points of $A$ with respect to $O$ and $K$, respectively. A line through $A$ intersects the circles $k_{1}$ and $k_{2}$ at the points $D$ and $E$, respe... | [
"Solution:\nThe points $B, P, C$ are collinear, and\n$$\n\\angle A P C = \\angle A P B = 90^\\circ\n$$\nLet $N$ be the midpoint of $D P$.\nSo we have:\n$$\n\\begin{aligned}\n& \\angle N O P = \\angle D A P \\\\\n& = \\angle E C P = \\angle E C A + \\angle A C P\n\\end{aligned}\n$$\nSince $O K // B C$ and $O K$ is t... | JBMO | OJBM | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03id | Problem:
A dog standing at the centre of a circular arena sees a rabbit at the wall. The rabbit runs around the wall and the dog pursues it along a unique path which is determined by running at the same speed and staying on the radial line joining the centre of the arena to the rabbit. Show that the dog overtakes the r... | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
08hn | Problem:
For every positive integer $n \geq 2$ the affirmation $P_{n}^{\prime}$: "If the derivative $P^{\prime}(X)$ of a polynomial $P(X)$ of degree $n$ with real coefficients has $n-1$ real distinct roots, then there exists a real constant $C$ such that the equation $P(x) = C$ has $n$ real distinct solutions" is cons... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | P4 is true; P5 is false. | |
0dq9 | Let $A$ be a set of positive integers and let $k$ be a positive integer. Set $A$ is said to be $k$-thin if there are $k$ positive integers $n_1, \dots, n_k$ so that for all $1 \le i < j \le k$ and all $a, a' \in A$, $n_i + a \ne n_j + a'$. Suppose that $A_r$ is a $k_r$-thin set for $1 \le r \le m$ and that $\cup_{r=1}^... | [
"For any positive integer $n$, let $[n] = \\{1, 2, \\dots, n\\}$. Consider a $k$-thin set $A$ with the associated integers $n_1, \\dots, n_k$. Let $N$ be the maximum of these integers. We let\n$$\nA + n = \\{a + n : a \\in A\\}.\n$$\nFor any fixed integer $L$, let $B = A \\cap [L]$. Now $B + n_i$, $i = 1, \\dots, k... | Singapore | Singapore International Mathematical Olympiad Committee National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
06dl | Considering $\pmod{4}$ or otherwise, show that if $a, b, c, d$ are integers such that
$$
(3a + 5b)(7b + 11c)(13c + 17d)(19d + 23a) = 2001^{2001},
$$
then $a$ must be even. | [
"Taking modulo $4$, we obtain\n$$\n(-a + b)(-b - c)(c + d)(-d - a) \\equiv 1 \\pmod{4}.\n$$\nThis implies all factors on the left are odd, and an even number of them is congruent to $3$ modulo $4$ and the others are congruent to $1$ modulo $4$. It follows that their sum is a multiple of $4$, i.e.\n$$\n4 \\mid (-a +... | Hong Kong | IMO HK TST | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
07az | Find the maximum number of permutations of the set $\{1, 2, \ldots, 2014\}$ such that for each two distinct numbers $a$ and $b$ of this set, one could find at most one permutation in which $b$ has appeared exactly after $a$. | [
"There are $2014 \\times 2013$ ordered pairs $(a, b)$ such that $1 \\le a \\ne b \\le 2014$. On the other hand, there are $2013$ consecutive pairs in each permutation. Therefore, the number of such permutations is at most $2014$.\n\nConsider the $2014$ permutations of the form\n$$\n\\sigma_i = (2014 + i, 1 + i, 201... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2014 | |
0jcq | Problem:
Let $n$ be the 200th smallest positive real solution to the equation $x-\frac{\pi}{2}=\tan x$. Find the greatest integer that does not exceed $\frac{n}{2}$. | [
"Solution:\n\nDrawing the graphs of the functions $y=x-\\frac{\\pi}{2}$ and $y=\\tan x$, we may observe that the graphs intersect exactly once in each of the intervals $\\left(\\frac{(2 k-1) \\pi}{2}, \\frac{(2 k+1) \\pi}{2}\\right)$ for each $k=1,2, \\cdots$. Hence, the 200th intersection has $x$ in the range $\\l... | United States | HMMT November 2012 | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | 314 | |
0hyw | Problem:
For what single digit $n$ does 91 divide the 9-digit number $12345 n 789$? | [
"Solution 1:\n\n$123450789$ leaves a remainder of $7$ when divided by $91$, and $1000$ leaves a remainder of $90$, or $-1$, so adding $7$ multiples of $1000$ will give us a multiple of $91$.\n\n\nSolution 2:\n\nFor those who don't like long division, there is a quicker way. First notice that $91 = 7 \\cdot 13$, and... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | final answer only | 7 | |
0ihf | Problem:
Farmer Bill's 1000 animals - ducks, cows, and rabbits - are standing in a circle. In order to feel safe, every duck must either be standing next to at least one cow or between two rabbits. If there are 600 ducks, what is the least number of cows there can be for this to be possible? | [
"Solution:\nSuppose Bill has $r$ rabbits and $c$ cows. At most $r-1$ ducks can be between two rabbits: each rabbit can serve up to two such ducks, so at most $2r/2 = r$ ducks will each be served by two rabbits, but we cannot have equality, since this would require alternating between rabbits and ducks all the way a... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 201 | |
0ehd | Problem:
Pokaži, da je izraz $2^{2 n+3}+3^{n+2} \cdot 7^{n}$ deljiv s 17 za vsako naravno število $n$. | [
"Solution:\n\nNaj bo $n$ poljubno naravno število. Dani izraz preoblikujemo\n$$\n2^{2 n+3}+3^{n+2} \\cdot 7^{n}=2^{3} \\cdot 2^{2 n}+3^{2} \\cdot 3^{n} \\cdot 7^{n}=8 \\cdot 4^{n}+9 \\cdot 21^{n}=17 \\cdot 4^{n}+9 \\cdot\\left(21^{n}-4^{n}\\right)\n$$\nKer je prvi člen večkratnik števila 17, zadošča dokazati, da je... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
01uq | A village $M$ is on the road between the villages $A$ and $B$. The distance between $A$ and $M$ is twice as long as the distance between $B$ and $M$. Ann, Bob and Tom live in $A$, $B$, and $M$ respectively. One time Tom invites Ann and Bob to a game of chess. Ann and Bob walk along the road with constant and equal spee... | [
"**1.** Answer: 2 hours.\nLet the distance between $B$ and $M$ be $S$ (km), then the distance between $A$ and $M$ is $2S$. Denote by $v$ (km/h) the speed of Ann and Bob, then the speed of Tom on a motorbike is $9v$.\n\nConsider the case in which Tom first drives Ann. Since the distance between $A$ and $M$ is $2S$, ... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 2 hours | |
078l | * $a^{2n} - a$ is divisible by $n$,
* $\sum_{k=1}^{n} k^{2024} a^{2k}$ is not divisible by $n$.
Prove that $n$ has a prime factor *smaller* than $2024$. | [
"Let $n$ be a positive integer. Call a positive integer $a$ *good* for $n$ if $n \\mid a^{2n} - a$.\nDefine $d_n(a) = d$ to be the smallest positive integer such that $n \\mid a^{2d} - a$. Then $a^{2d+k} \\equiv a^{2k} \\pmod{n}$ holds for $k \\ge 0 \\implies a^{2i} \\pmod{n}$ is periodic with period $d$. The minim... | India | IMO TST | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Numb... | null | proof only | null | |
0c2e | Find the natural number $n$ for which $\sqrt{\frac{20^n - 18^n}{19}}$ is a rational number. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2 | |
0f3g | Problem:
There is a flu epidemic in elf city. The course of the disease is always the same. An elf is infected one day, he is sick the next, recovered and immune the third, recovered but not immune thereafter. Every day every elf who is not sick visits all his sick friends. If he is not immune he is sure to catch flu ... | [
"Solution:\n\nThis is curiously easy. Write $S$ for sick, $N$ for not sick and not immune, and $I$ for immune. Suppose group $A$ are $S$ on day 1 and group $B$ are $N$ on day 1. Then on day 2, $B$ are $S$, and $A$ are $I$. So on day 3 no one is sick, $A$ are $N$ and $B$ are $I$. Thereafter no one can get sick, so t... | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Other"
] | null | proof only | null | |
08v8 | Determine all the quintets $(a, n, p, q, r)$ of positive integers for which the identity
$$a^n - 1 = (a^p - 1)(a^q - 1)(a^r - 1)$$
is satisfied. | [
"If $a = 1$, the identity of the problem holds regardless of how other positive integers $n, p, q, r$ are chosen. So, we assume that $a \\ge 2$ in the sequel. Since the given identity is symmetric in $p, q, r$, we may assume that $p \\le q \\le r$ holds.\n\nWe can rewrite the given identity in the form\n$$\na^n = a... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | All solutions are:
- a = 1 with arbitrary positive integers n, p, q, r.
- (a, n, p, q, r) = (3, 2, 1, 1, 1).
- (a, n, p, q, r) = (2, 6, 2, 2, 3) and all permutations of (p, q, r).
- (a, n, p, q, r) = (2, k, 1, 1, k) for any positive integer k, and all permutations of (p, q, r). | |
006p | Determinar si es posible distribuir $60$ ceros y $61$ unos en las casillas de un tablero de $11 \times 11$, un número en cada casilla, de modo que la suma de los números de cada fila sea impar, la suma de los números de cada columna sea impar y la suma de los números de cada una de las dos diagonales sea impar. ¿Y si e... | [] | Argentina | Argentina 2009 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | Yes for 11×11; Yes for 12×12. | |
02es | Given $k > 0$, the sequence $a_1, a_2, a_3, \dots$ is defined by its first two members and $a_{n+2} = a_{n+1} + \frac{k}{n} a_n$. For which $k$ can we write an $a_n$ as a polynomial in $n$? For which $k$ can we write $\frac{a_{n+1}}{a_n} = \frac{p(n)}{q(n)}$, where $p$ and $q$ are polynomials? | [
"Rewrite the equation as $\\frac{n(a_{n+2}-a_{n+1})}{a_n} = k$. If $a_n$ is a polynomial in $n$, say $a_n = r(n) = \\sum_{i=0}^{d} r_i n^i$, then the limit $\\lim_{n \\to \\infty} \\frac{n(a_{n+2}-a_{n+1})}{a_n}$ must exist and be equal to $k$. But\n$$\n\\begin{align*}\n& \\lim_{n \\to \\infty} \\frac{n(a_{n+2} - a... | Brazil | XIII OBM | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | First: k must be a nonnegative integer. Second: k must be an integer. | |
0l2q | Problem:
An underground burrow consists of an infinite sequence of rooms labeled by the integers $\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots$. Initially, some of the rooms are occupied by one or more rabbits.
Each rabbit wants to be alone. Thus, if there are two or more rabbits in the same room (say, room $m$), half the ... | [
"Solution:\n\nFirst, we show that the process eventually stops.\n\nCall a room interior if it is occupied, or if at least one room somewhere to its left and at least one room somewhere to its right are occupied.\n\nWe claim that it is not possible for a gap of two or more consecutive unoccupied interior rooms to ev... | United States | 25th Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | The process eventually stops. In the final configuration there is exactly one rabbit in every room from minus twenty six through forty, except that room twenty nine is empty. | |
0lfn | For a positive integer $n$, let $\tau(n)$ be the number of positive divisors of $n$.
a) Find all positive integers $n$ such that $\tau(n) + 2023 = n$.
b) Prove that there exist infinitely many positive integers $k$ such that there are exactly two positive integers $n$ satisfying $\tau(kn) + 2023 = n$. | [
"a) First, we will prove the following lemma.\n\n*Lemma.* For any positive integer $n$, then $\\tau(n) \\leq 2\\sqrt{n}$.\n\n*Proof.* Consider any positive integer $n$, let $d_1, d_2, \\dots, d_s$ be all positive divisors not exceeding $\\sqrt{n}$ of $n$. Then, obviously we have $s \\leq \\sqrt{n}$.\n\nNote that, i... | Vietnam | Vietnamese MO | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | a) No positive integer n satisfies the equation. b) For any prime k greater than 6996, there are exactly two solutions n, namely 2027 and 2031. | |
0d5b | Given 2015 subsets $A_{1}, A_{2}, \ldots, A_{2015}$ of the set $\{1,2, \ldots, 1000\}$ such that $|A_{i}| \geq 2$ for every $i \geq 1$ and $|A_{i} \cap A_{j}| \geq 1$ for every $1 \leq i < j \leq 2015$. Prove that $k=3$ is the smallest number of colors such that we can always color the elements of the set $\{1,2, \ldot... | [
"Consider the collection: $A_{1} = \\{1,2\\}$, $A_{2} = \\{2,3\\}$, $A_{3} = \\{1,3\\}$, $A_{4}, \\ldots, A_{2015}$ be any 2011 subsets of $\\{1,2, \\ldots, 1000\\}$ that contain all $1,2,3$. One can check that this collection satisfies the given condition. For any $2$-coloring of the set $\\{1,2, \\ldots, 1000\\}$... | Saudi Arabia | SAMC 2015 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English, Arabic | proof and answer | 3 | |
03am | A number $r_X > 0$ is assigned to any point $X$ in the plane such that $2|r_X - r_Y| \le |XY|$ for any two points $X$ and $Y$. A cricket can jump from $X$ to $Y$ if $r_X = |XY|$. Prove that for any two points $X$ and $Y$ the cricket can move from $X$ to $Y$ by finite number of jumps. | [
"Denote by $D(Z, r)$ and $S(Z, r)$ the open disc and the circle with center $Z$ and radius $r$, respectively.\nLet $A = A_X$ be the set of points that can be reached from $X$ by finite number of jumps (possibly zero). We have to show that $A = \\mathbb{R}^2$.\n\nFirst, we shall prove that $D(X, 2r_X) \\subset A$. W... | Bulgaria | Team selection test for 50. IMO | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0c9j | Problem:
Arătaţi că pentru orice $a, b, c > 0$ cu $a + b + c = 1$ are loc inegalitatea:
$$
\frac{1}{a + b c} + \frac{1}{b + c a} + \frac{1}{c + a b} \geqslant \frac{7}{1 + a b c}
$$ | [] | Romania | Al treilea test de selecţie pentru OBMJ Bucureşti | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0769 | Problem:
Let $n$ be a natural number and $X = \{1, 2, \ldots, n\}$. For subsets $A$ and $B$ of $X$ we define $A \Delta B$ to be the set of all those elements of $X$ which belong to exactly one of $A$ and $B$. Let $\mathcal{F}$ be a collection of subsets of $X$ such that for any two distinct elements $A$ and $B$ in $\ma... | [
"Solution:\nFor each subset $A$ of $\\{1, 2, \\ldots, n-1\\}$, we pair it with $A \\cup \\{n\\}$. Note that for any such pair $(A, B)$ not both $A$ and $B$ can be in $\\mathcal{F}$. Since there are $2^{n-1}$ such pairs it follows that $\\mathcal{F}$ can have at most $2^{n-1}$ elements.\n\nWe shall show by induction... | India | Indian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Maximum size is 2^{n-1}. The extremal families are exactly the collections of all even-cardinality subsets of X or all odd-cardinality subsets of X. | |
0l2d | The number $2024$ is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum? | [
"In order to minimize the number of terms in the sum, the greatest two-digit number, $99$, should be used as many times as possible. Because $20 \\cdot 99 = 1980$, the minimum number of terms is greater than $20$. On the other hand, $2024 = 20 \\cdot 99 + 44$, so the least number of two-digit numbers needed is $21$... | United States | AMC 10 A | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | final answer only | 21 | |
0cbv | A positive integer is called *special* if all of its digits are nonzero and any two adjacent digits are consecutive (not necessarily in ascending order).
a) Find the largest special number for which the sum of its digits is $2023$.
b) Find the smallest special number for which the sum of its digits is $2022$. | [
"a) The largest number $m$ will have as many digits as possible, so we will choose the smallest possible digits. As next to a $1$ we can only put a $2$, and $2023 = 3 \\cdot 674 + 1$, we choose $m = \\underbrace{12121 \\dots 121}_{1349 \\text{ digits}}$, with $674$ digits equal to $2$ and $675$ digits equal to $1$.... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) The largest special number is the 1349-digit alternating sequence starting with one, i.e., 121212…121, with 675 ones and 674 twos.
b) The smallest special number is 56787 followed by 117 repetitions of 89, i.e., 56787 8989…89 (117 copies of 89). | |
0ba6 | Given a set $\mathcal{L}$ of lines in general position in the plane (no two lines in $\mathcal{L}$ are parallel and no three lines are concurrent) and another line $\ell$, show that the total number of edges of all faces in the corresponding arrangement intersected by $\ell$ is at most $6|\mathcal{L}|$.
Chazelle et al.... | [
"Assume without loss of generality that $\\ell$ is horizontal and does not pass through any vertex of the arrangement of lines in $\\mathcal{L}$.\n\nFirst, we shall bound the total number of edges of the upper parts of all faces intersected by $\\ell$, that is, those parts that lie above $\\ell$. The boundary of th... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
03jz | Problem:
In the figure, the side length of the large equilateral triangle is $3$ and $f(3)$, the number of parallelograms bounded by sides in the grid, is $15$. For the general analogous situation, find a formula for $f(n)$, the number of parallelograms, for a triangle of side length $n$.
 | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | f(n) = 3 * C(n+2, 4) = n(n-1)(n+1)(n+2)/8 | |
03ra | Let $l$ be the perimeter of an acute triangle $\triangle ABC$ which is not equilateral, $P$ a variable point inside $\triangle ABC$, and $D, E$ and $F$ be projections of $P$ on $BC, CA$ and $AB$ respectively.
Prove that
$$
2(AF + BD + CE) = l,
$$
if and only if $P$ is collinear with the incenter and circumcenter of $\t... | [
"Denote the lengths of three sides of $\\triangle ABC$ by $BC = a$, $CA = b$ and $AB = c$ respectively. No loss of generality, we can suppose $b \\neq c$. We choose a rectangular coordinate system (see the figure), then we have $A(m, n)$, $B(0, 0)$, $C(a, 0)$ and $P(x, y)$.\n\n\n\nSince $AF... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Concurr... | English | proof only | null | |
0al3 | Let $ABC$ be a scalene acute triangle with orthocenter $H$. The circle with center $A$ and radius $AH$ meets the circumcircle of $\triangle BHC$ at the point $T_a \neq H$. Points $T_b$ and $T_c$ are defined similarly. Prove that $H$ lies on the circumcircle of $\triangle T_a T_b T_c$. | [] | North Macedonia | Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Advanced Config... | English | proof only | null | |
06za | Problem:
Show that there are infinitely many solutions in positive integers to $2 a^{2} - 3 a + 1 = 3 b^{2} + b$. | [
"Solution:\nPut $A = a - 1$ and the equation becomes $A (2A + 1) = b (3b + 1)$. Let $d$ be the greatest common divisor of $A$ and $b$. Put $A = d x$, $b = d y$. Then $x (2 d x + 1) = y (3 d y + 1)$. Since $x$ and $y$ are coprime, $x$ must divide $3 d y + 1$. So put $3 d y + 1 = n x$. Then $2 d x + 1 = n y$. Solving... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0kra | Let $w = \frac{\sqrt{3}+i}{2}$ and $z = \frac{-1+i\sqrt{3}}{2}$, where $i = \sqrt{-1}$. Find the number of ordered pairs $(r, s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s$. | [
"Note that $w = \\cos \\frac{\\pi}{6} + i \\cdot \\sin \\frac{\\pi}{6}$ is a primitive twelfth root of unity, so $w^3 = i$ and $w^r = w^{r+12m}$ for all integers $m$ and $r$. Furthermore, $z = \\cos \\frac{2\\pi}{3} + i \\cdot \\sin \\frac{2\\pi}{3}$ is a primitive cube root of unity, so $w^4 = z$ and\n$$\ni \\cdot... | United States | 2022 AIME I | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 834 | |
09tv | Problem:
Bepaal alle paren $(a, b)$ van positieve gehele getallen zodat $(a+b)^3-2 a^3-2 b^3$ een tweemacht is. | [
"Solution:\n\nWe bepalen eerst de paren $(a, b)$ met $\\operatorname{ggd}(a, b)=1$. Er geldt\n$$\n\\begin{aligned}\n(a+b)^3-2 a^3-2 b^3 & =a^3+3 a^2 b+3 a b^2+b^3-2 a^3-2 b^3 \\\\\n& =-a^3-b^3+3 a b(a+b) \\\\\n& =-(a+b)\\left(a^2-a b+b^2\\right)+3 a b(a+b) \\\\\n& =(a+b)\\left(-a^2+a b-b^2+3 a b\\right) \\\\\n& =(a... | Netherlands | IMO-selectietoets III | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (a, b) = (2^k, 2^k), (2^k, 3*2^k), or (3*2^k, 2^k) for any nonnegative integer k | |
0c8d | Let $ABC$ be an acute triangle with $AB = AC$, let $D$ be the midpoint of the side $AC$, and let $\gamma$ be the circumcircle of the triangle $ABD$. The tangent of $\gamma$ at $A$ crosses the line $BC$ at $E$. Let $O$ be the circumcentre of the triangle $ABE$. Prove that the midpoint of the segment $AO$ lies on $\gamma... | [
"Let $\\Gamma$ be the image of $\\gamma$ under the homothety of centre $A$ and factor $2$. Clearly, $\\Gamma$ is also tangent to $AE$ at $A$, and the conclusion is equivalent to $\\Gamma$ passing through $O$, which is the same as $AE$ being tangent to the circle $ACO$.\n\nAlternatively, but equivalently, this amoun... | Romania | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Ge... | English | proof only | null | |
08ne | Problem:
Decipher the equality
$$
(\overline{V E R}-\overline{I A}):(\overline{G R E}+\overline{E C E})=G^{R^{E}}
$$
assuming that the number $\overline{G R E E C E}$ has a maximum value. It is supposed that each letter corresponds to a unique digit from 0 to 9 and different letters correspond to different digits, and... | [
"Solution:\n\nDenote\n$$\nx=\\overline{V E R}-\\overline{I A}, \\quad y=\\overline{G R E}+\\overline{E C E}, \\quad z=G^{R^{E}}\n$$\nThen obviously, we have\n$$\n\\begin{aligned}\n& (201+131 \\text{ or } 231+101) \\leq y \\leq (879+969 \\text{ or } 869+979 \\text{ or. } 769+989) \\\\\n\\Rightarrow \\quad & 332 \\le... | JBMO | JBMO Shortlist | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | G=7, R=0, E=1, C=6, V=9, I=4, A=8; the maximal six-digit number is 701161 | |
02yf | Problem:
Seja $ABCD$ um trapézio, com $AD \parallel BC$, tal que o lado $CD$ é tangente ao círculo com diâmetro $AB$. Se $G$ é o ponto médio de $CD$ e $CD = 8\ \mathrm{cm}$, determine a medida da altura $GF$.
 | [
"Solution:\n\nSeja $O$ o centro do círculo de diâmetro $AB$. Como $O$ é ponto médio de $AB$ e $G$ é ponto médio de $CD$, segue que $GO$ é base média do trapézio $ABCD$. Daí, $GO$ é paralelo aos lados $AD$ e $BC$. Consequentemente, temos as seguintes igualdades de áreas: $A_{DGO} = A_{AGO}$ e $A_{COG} = A_{BGO}$.\n\... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4 cm | |
03um | Suppose that $f(x)$ is defined on $\mathbb{R}$, satisfying $f(0) = 2008$, and for any $x \in \mathbb{R}$
$$
\begin{aligned}
f(x+2) - f(x) &\le 3 \times 2^x, \\
f(x+6) - f(x) &\ge 63 \times 2^x.
\end{aligned}
$$
Then $f(2008) = \underline{\hspace{2cm}}$. | [
"**Solution I** We have\n$$\n\\begin{aligned}\nf(x+2) - f(x) &= -(f(x+4) - f(x+2)) - (f(x+6) - f(x+4)) + (f(x+6) - f(x)) \\\\\n&\\geq -3 \\times 2^{x+2} - 3 \\times 2^{x+4} + 63 \\times 2^x = 3 \\times 2^x.\n\\end{aligned}\n$$\nThis means that $f(x+2) - f(x) = 3 \\times 2^x$. So we have\n$$\n\\begin{aligned}\nf(200... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | 2^{2008} + 2007 | |
0962 | Problem:
Determinaţi toate valorile parametrului real $a$ pentru care ecuaţia
$$
\left|a x^{2}-6\right|=|2 a x|+|3 a| \text{ are cel puţin o soluţie reală . }
$$ | [
"Solution:\n1. Fie $a=0$, atunci ecuaţia nu are soluţii.\n\n2. Fie $a \\neq 0$, atunci ecuaţia este echivalentă cu\n$$\n\\left|x^{2}-\\frac{6}{a}\\right|=2|x|+3 \\Leftrightarrow\n\\left[\\begin{array}{c}\nx^{2}-\\frac{6}{a}=2|x|+3, \\\\\n\\frac{6}{a}-x^{2}=2|x|+3\n\\end{array}\\right. \\Leftrightarrow\n\\left[\\beg... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (-∞, -3/2] ∪ (0, +∞) | |
03fy | Let $\mathcal{F}$ be a family of 4-element subsets of a set of size $5^m$, where $m$ is a fixed positive integer. If the intersection of any two sets in $\mathcal{F}$ does not have size exactly $2$, find the maximal value of $|\mathcal{F}|$. | [] | Bulgaria | 5 Bulgarian National Olympiad - Final Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 5^m(5^m - 1)/4 | |
0ar0 | Problem:
Find all real numbers $a$ such that $x^{3} + a x^{2} - 3 x - 2$ has two distinct real zeros. | [] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a = 0 | |
05gy | Problem:
Soit $n$ l'entier $4 \times 201420142014 \ldots 2014$ (où $2014$ est écrit $117819$ fois). Montrer que $2014^{3}$ divise $n$. | [
"Solution:\n\nOn peut écrire\n$$\n4 \\times 201420142014 \\ldots 2014 = 4 \\cdot 2014 \\cdot \\frac{10^{4 \\cdot 117819}-1}{10^{4}-1}\n$$\nComme $2014 = 2 \\cdot 19 \\cdot 53$ et $10^{4}-1 = 3^{2} \\cdot 11 \\cdot 101$ sont premiers entre eux, il suffit de montrer que $19^{2}$ et $53^{2}$ divisent $10^{4 \\cdot 117... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES, ENVOI No. 3 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0fiy | Problem:
Sea $P$ un punto, en el interior del triángulo $ABC$, de modo que el triángulo $ABP$ es isósceles. Sobre cada uno de los otros dos lados de $ABC$ se construyen exteriormente triángulos $BCQ$ y $CAR$, ambos semejantes al triángulo $ABP$.
Probar que los puntos $P$, $Q$, $C$ y $R$ o están alineados o son los vért... | [
"Solution:\nLos triángulos $ABC$ y $PBQ$ son semejantes pues tienen un ángulo igual $\\angle ABC = \\angle PBQ$ y los lados que lo forman proporcionales:\n$$\n\\frac{c}{a} = \\frac{BP}{BQ}\n$$\n\nDe modo análogo, $ABC$ es semejante a $APR$, por tanto $PBQ$ y $APR$ son semejantes (y al ser $... | Spain | XXXVII Olimpiada Española de Matemáticas | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08x3 | Suppose two $20 \times 13$ rectangular grids consisting of $260$ small squares are given. We insert into each square box of the two grids, numbers $1, 2, \ldots, 260$ in the following way:
* For the first grid, we start inserting numbers $1, 2, \ldots, 13$ into the boxes on the top row from left to right. Continue to i... | [
"$87$, $174$\n\nThe number inserted into the box located on the $i$-th row from the top and $j$-th column from the left is given by $13(i - 1) + j$ for the case of the first grid, and by $20(13 - j) + i$ for the case of the second grid. If the same number goes into the boxes located at the same position in the two ... | Japan | Japan 2013 Initial Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 87, 174 | |
0jsj | Problem:
Let $ABCD$ be an isosceles trapezoid with parallel bases $AB = 1$ and $CD = 2$ and height $1$. Find the area of the region containing all points inside $ABCD$ whose projections onto the four sides of the trapezoid lie on the segments formed by $AB$, $BC$, $CD$ and $DA$. | [
"Solution:\nAnswer: $\\frac{5}{8}$\nLet $E, F$ be the projections of $A, B$ on $CD$. A point whose projections lie on the sides must be contained in the square $ABFE$. Furthermore, the point must lie under the perpendicular to $AD$ at $A$ and the perpendicular to $BC$ at $B$, which have slopes $\\frac{1}{2}$ and $-... | United States | HMMT November | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 5/8 | |
07g0 | Given a triangle $\triangle ABC$ with circumcircle $\Gamma$. Points $E$ and $F$ are the feet of angle bisectors of $B$ and $C$, let $I$ be incenter and $K$ be the intersection point of $AI$ and $EF$. Suppose that $T$ is the midpoint of arc $\overarc\{BAC\}$. Circle $\Gamma$ intersects the $A$-median and circumcircle of... | [
"Let $P$ be the $A$-mixtilinear touchpoint with $\\Gamma$.\n\n---\n\n\nWe know that $P, I$ and $T$ are collinear. First suppose that we have $IJ \\parallel AP$. We have\n$$\n\\angle ITX = \\angle PTX = \\angle PAX = \\angle IJX \\implies \\angle ITX = \\angle IJX.\n$$\nSo, $IJTX$ is cyclic ... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations... | English | proof only | null | |
04sw | Let $ABC$ be an acute triangle, which is not equilateral. Denote by $O$ and $H$ its circumcenter and orthocenter, respectively. The circle $k$ passes through $B$ and touches the line $AC$ at $A$. The circle $l$ with center on the ray $BH$ touches the line $AB$ at $A$. The circles $k$ and $l$ meet in $X$ ($X \neq A$). S... | [
"Let $E$ be the intersection point of the circle $l$ and the line $AC$ ($E \\neq A$). Since $k$ lies in the half-plane $ACB$ and $l$ lies in the half-plane $ABC$, the point $X$ lies inside of the angle $BAC$ (Fig. 4). Using the well known fact about the angle between a tangent and a chord of a given circle we get $... | Czech Republic | 15th Czech-Polish-Slovak Mathematics Competition | [
"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 > Transformations > Spiral similarity... | English | proof only | null | |
010i | Problem:
Let $a$ be an odd digit and $b$ an even digit. Prove that for every positive integer $n$ there exists a positive integer, divisible by $2^{n}$, whose decimal representation contains no digits other than $a$ and $b$. | [
"Solution:\nIf $b=0$, then $N=10^{n} a$ meets the demands. For the sequel, suppose $b \\neq 0$.\n\nLet $n$ be fixed. We prove that if $1 \\leqslant k \\leqslant n$, then we can find a positive integer $m_{k}<5^{k}$ such that the last $k$ digits of $m_{k} 2^{n}$ are all $a$ or $b$.\n\nClearly, for $k=1$ we can find ... | Baltic Way | Baltic Way 1998 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
02sb | Problem:
Um número natural $N$ maior que 10 é chamado "superquadrado" se o número formado por cada dois algarismos consecutivos do número $N$ (considerados na mesma ordem) é sempre um quadrado perfeito. Por exemplo, $8164$ é "superquadrado" porque os números $81$, $16$ e $64$ são quadrados perfeitos. Outros exemplos d... | [
"Solution:\n\na) Será útil adiante que observemos a lista de todos os quadrados perfeitos de dois dígitos:\n$$\n16, \\quad 25, \\quad 36, \\quad 49, \\quad 64, \\quad 81\n$$\nO número formado pelos primeiros dois algarismos de um superquadrado deve ser algum número da lista (.1). Assim, começemos contando os superq... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Other",
"Number Theory > Other"
] | null | proof and answer | a) 14; b) 81649 | |
0e2e | Problem:
Za katera naravna števila $n$ obstaja večkratnik števila 11, ki ima vsoto števk enako $n$? | [
"Solution:\n\nOznačimo vsoto števk naravnega števila $m$ z $S(m)$. Velja $S(11)=2$. Opazimo, da ima prvih nekaj večkratnikov sodo vsoto števk. Prvi večkratnik, ki ima liho vsoto števk, je $209=19 \\cdot 11$, vsota števk je 11. S pomočjo teh dveh primerov lahko konstruiramo večkratnike z vsoto števk $n$ za skoraj vs... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All natural numbers except 1, 3, 5, 7, and 9 | |
0e0y | Problem:
Poišči vsa naravna števila $n$, za katera obstaja praštevilo $p$, da je število $p^{2}+7^{n}$ popoln kvadrat. | [
"Solution:\n\nNaj bo $p^{2}+7^{n}=m^{2}$. Potem velja $7^{n}=m^{2}-p^{2}=(m-p)(m+p)$. Ločimo dve možnosti.\n\nČe je $m-p=1$ in $m+p=7^{n}$, sledi $2p=7^{n}-1$. Denimo, da je $n \\geq 2$. Tedaj je $2p=7^{n}-1=(7-1)\\left(7^{n-1}+7^{n-2}+\\ldots+7+1\\right)$, zato dobimo $p=3\\left(7^{n-1}+7^{n-2}+\\ldots+7+1\\right)... | Slovenia | Slovenian Secondary School Mathematical Competition | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1 | |
0k06 | Problem:
Find the number of quadruples $(a, b, c, d)$ of integers with absolute value at most $5$ such that
$$
\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}=(a+b+c+d)(a-b+c-d)\left((a-c)^{2}+(b-d)^{2}\right)
$$ | [
"Solution:\nLet $x = a + c$, $y = a - c$, $w = b + d$, and $z = b - d$. Then\n$$\n\\left(w^{2} + x^{2} + y^{2} + z^{2}\\right)^{2} = 4\\left(x^{2} - w^{2}\\right)\\left(y^{2} + z^{2}\\right)\n$$\nand since $\\left|x^{2} + w^{2}\\right| \\geq \\left|x^{2} - w^{2}\\right|$ it follows that $w = 0$ or $y = z = 0$.\n\nN... | United States | HMMT November 2017 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 49 | |
01sw | Given real numbers $x, y, z$, with
$$
x + y + z = xy + yz + zx = -1.
$$
Prove that $(xy - z^2)(yz - x^2)(zx - y^2) = xyz - 1$. | [
"Let $xyz = a$. Then\n$$\n\\begin{aligned}\n(xy - z^2)(yz - x^2)(zx - y^2) &= \\\\\n&= \\left(\\frac{a}{z} - z^2\\right) \\left(\\frac{a}{x} - x^2\\right) \\left(\\frac{a}{y} - y^2\\right) = \\frac{1}{a}(a - z^3)(a - x^3)(a - y^3) = \\\\\n&= \\frac{1}{a}(a^3 - (x^3 + y^3 + z^3)a^2 + (x^3y^3 + y^3z^3 + z^3x^3)a - x^... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
0ett | Determine which of the following is larger:
$$
\sqrt{2 + \sqrt[3]{5}} \quad \text{or} \quad \sqrt[3]{5 + \sqrt{2}}
$$
Fully explain your reasoning. | [
"Let us denote $x = \\sqrt{2 + \\sqrt[3]{5}}$ and $y = \\sqrt[3]{5 + \\sqrt{2}}$. Since both are positive, $x - y$ has the same sign as $x^6 - y^6$. We have\n$$\nx^6 = (2 + \\sqrt[3]{5})^3 = 8 + 12\\sqrt[3]{5} + 6\\sqrt[3]{25} + 5 = 13 + 12\\sqrt[3]{5} + 6\\sqrt[3]{25}\n$$\nand\n$$\ny^6 = (5 + \\sqrt{2})^2 = 25 + 1... | South Africa | The South African Mathematical Olympiad Third Round | [
"Algebra > Equations and Inequalities",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | sqrt(2 + cube_root(5)) is larger than cube_root(5 + sqrt(2)). | |
03b7 | Let $f_1 \in \mathbb{R}[x]$ be a quadratic polynomial with positive leading coefficient. Set $f_{n+1} = f_1 \circ f_n$ for $n \ge 1$. It is known that the polynomial $f_2$ has four non-positive different zeroes. Prove that the polynomial $f_n$ has $2^n$ different real zeroes. | [
"Note that if $x_1, \\dots, x_{2^n}$ are the zeroes of $f_n$, then the zeroes of $f_{n+1}$ are the solutions of the equations $f_1(x) = x_k$, $1 \\le k \\le 2^n$. Moreover, the equation $f_1(x) = a$ has two different real roots if and only if $a > m := \\min f_1$.\n\nAssume that $f_2$ has four non-positive differen... | Bulgaria | Bulgaria | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
002w | Sea $ABC$ un triángulo tal que al construir exteriormente al triángulo los cuadrados $ABB_1A_2$, $BCC_1B_2$ y $CAA_1C_2$, los puntos $A$, $B$ y $C$ quedan en el interior de los triángulos $A_1B_1C_1$ y $A_2B_2C_2$. Demostrar que los triángulos $A_1B_1C_1$ y $A_2B_2C_2$ tienen la misma área. | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | Español | proof only | null | |
0j7j | Problem:
Let $a$, $b$, and $c$ be complex numbers such that $|a|=|b|=|c|=1$. If
$$
\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}=1
$$
as well, determine the product of all possible values of $|a+b+c|$. | [
"Solution:\nLet $s=a+b+c$. Then\n$$\n\\begin{aligned}\ns^{3} & =a^{3}+b^{3}+c^{3}+3\\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\\right)+6 a b c \\\\\n& =a b c\\left(\\frac{a^{2}}{b c}+\\frac{b^{2}}{c a}+\\frac{c^{2}}{a b}+3\\left(\\frac{a}{b}+\\frac{b}{a}+\\frac{b}{c}+\\frac{c}{b}+\\frac{c}{a}+\\frac{a}{c... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Descartes' Rule of Signs"
] | null | proof and answer | 2 | |
0cvx | The points $D$ and $E$ are chosen on the sides $AB$ and $AC$ of a triangle $ABC$ so that $DB = BC = CE$. The segments $BE$ and $CD$ meet at $P$. Show that the circumcircles of the triangles $BDP$ and $CEP$ meet at the incenter of $\triangle ABC$. | [
"Обозначим через $I$ центр вписанной окружности треугольника $ABC$, точка $I$ является точкой пересечения биссектрис. Доказем, что точки $B$, $D$, $P$, $I$ лежат на одной окружности. Аналогично покажем, что точки $C$, $E$, $P$, $I$ лежат на одной окружности, и задача будет решена.\n\nДостаточно установить равенство... | Russia | Regional round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
02ly | Solve, in real numbers, the system
$$
\begin{aligned}
x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} \\
xyz = 1
\end{aligned}
$$ | [
"Since $xyz = 1$,\n$$\n\\begin{aligned}\nx + \\frac{1}{y} &= y + \\frac{1}{z} \\iff xyz + z = y^2z + y \\\\\n\\iff 1 - y = -z(1 - y^2) \\iff y = 1 \\text{ or } z = -\\frac{1}{1+y}\n\\end{aligned}\n$$\nIf $y = 1$, the system reduces to $xz = 1 \\iff z = \\frac{1}{x}$ and\n$$\nx + 1 = 1 + \\frac{1}{z} = z + \\frac{1}... | Brazil | XXXI Brazilian Math Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | (x,y,z) = (1,1,1) or (x,y,z) = (-(1+t)/t, t, -1/(1+t)) for any real t ≠ 0, −1, together with all cyclic permutations of this triple. | |
0bdn | Given a positive integer $n$, consider a triangular array with entries $a_{ij}$ where $i$ ranges from $1$ to $n$ and $j$ ranges from $1$ to $n - i + 1$. The entries of the array are all either $0$ or $1$, and, for all $i > 1$ and any associated $j$, $a_{ij}$ is $0$ if $a_{i-1,j} = a_{i-1,j+1}$, and $a_{ij}$ is $1$ othe... | [
"$$(a_{11}, a_{12}, \\dots, a_{1n}) \\mapsto (a_{11}, a_{21}, \\dots, a_{n1})$$\nis bijective, on one hand, and that $(a_{11}, a_{12}, \\dots, a_{1n})$ is fixed by $f$ if and only if $(a_{11}, a_{21}, \\dots, a_{n1})$ is a palindrome $(a_{i1} = a_{n-i+1,1}, i = 1, \\dots, n)$, on the other.\nTo begin with, notice t... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 2^{floor((n+1)/2)} | |
0ina | Problem:
Compute
$$
\frac{2^{3}-1}{2^{3}+1} \cdot \frac{3^{3}-1}{3^{3}+1} \cdot \frac{4^{3}-1}{4^{3}+1} \cdot \frac{5^{3}-1}{5^{3}+1} \cdot \frac{6^{3}-1}{6^{3}+1}
$$ | [
"Solution:\nAnswer: $\\frac{43}{63}$\n\nUse the factorizations $n^{3}-1=(n-1)\\left(n^{2}+n+1\\right)$ and $n^{3}+1=(n+1)\\left(n^{2}-n+1\\right)$ to write\n$$\n\\frac{1 \\cdot 7}{3 \\cdot 3} \\cdot \\frac{2 \\cdot 13}{4 \\cdot 7} \\cdot \\frac{3 \\cdot 21}{5 \\cdot 13} \\cdot \\frac{4 \\cdot 31}{6 \\cdot 21} \\cdo... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 43/63 | |
05xn | Problem:
Thanima possède un magnifique collier constitué de rubis, d'émeraudes et de saphirs, que l'on représente par une suite de $R$, $E$ et $S$. En une opération magique, elle peut faire l'une des actions suivantes :
- Remplacer un motif $RR$ dans le collier par un motif $ES$ (ou $ES$ par $RR$).
- Remplacer un moti... | [
"Solution:\n\nSupposons que Thanima dispose de trois objets numérotés $1$, $2$ et $3$ sur trois supports numérotés $1$, $2$ et $3$. Au début, chaque objet se trouve sur le support qui porte le même numéro. Thanima peut lire les lettres du collier une par une de gauche à droite et faire les opérations suivantes :\n-... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0d79 | Let $a, b, c$ be positive numbers such that $a^{2}+b^{2}+c^{2}+a b c=4$. Prove that
$$
\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} .
$$ | [
"Without loss of generality, assume that $a \\geq b \\geq c$. From the condition, we get $a \\geq 1 \\geq c > 0$. Write the inequality as\n$$\n\\frac{a+b+c}{c}+\\frac{a+b+c}{b}+\\frac{a+b+c}{c} \\geq a+b+c+\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+3\n$$\nwhich is equivalent to\n$$\n(a+b+c-1)\\left(\\frac{1}{a}+\\frac{... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0322 | Problem:
Find the least positive integer $m$ such that $2^{2000}$ divides $2003^{m} - 1$. | [
"Solution:\nSet $m = 2^{k} p$, where $p > 1$ is an odd integer. Then\n$$\n2003^{m} - 1 = 2003^{2^{k} p} - 1 = \\left(2003^{2^{k}} - 1\\right) K\n$$\nwhere $K$ is a sum of $p$ even integers and $1$; in particular, $K$ is odd. Hence $m$ must be of the form $m = 2^{k}$. In this case we have that\n$$\n\\begin{aligned}\... | Bulgaria | Bulgarian Mathematical Competitions | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2^1998 | |
08s5 | There is a grid of $5 \times 5$. And write the integers $1,2,\ldots,16$ in all the grids (each number can be written only once) in the upper left grid of $4 \times 4$ (♣). About the 4 rows, write the sum of four numbers that are written in each row at the right end of each row. Similarly, about the 4 columns, write the... | [
"Define the numbers written to each grid as shown in the table 1. I can assume that $A_1$ is the minimum and $A_4$ is the maximum in $A_1, A_2, A_3, A_4$, and that $B_1$ is the minimum and $B_4$ is the maximum in $B_1, B_2, B_3, B_4$, by rearranging rows and columns appropriately. Now $m \\leq A_4 - A_1$, $m \\leq ... | Japan | Japan 2007 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 35 |
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