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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
09kg | Let $0 \le a \le 1$. Prove that $m^2 + 2amn + an^2 \ge m + an$ holds for all integers $m$ and $n$, and determine the condition under which equality holds. | [
"Let us rewrite the inequality:\n\n$$\nm^2 + 2amn + an^2 \\ge m + an.\n$$\n\nBring all terms to one side:\n\n$$\nm^2 + 2amn + an^2 - m - an \\ge 0.\n$$\n\nGroup terms:\n\n$$\nm^2 - m + 2amn + an^2 - an = (m^2 - m) + (2amn + an^2 - an).\n$$\n\nFactor $a$ in the last three terms:\n\n$$\n(m^2 - m) + a(2mn + n^2 - n).\... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | The inequality holds for all integers m and n and all a between zero and one because (1 − a)(m^2 − m) + a((m + n)^2 − (m + n)) ≥ 0. Equality occurs as follows: if a = 0, then m is either zero or one (any n); if a = 1, then m + n is either zero or one; if 0 < a < 1, then (m, n) is one of (0, 0), (0, 1), (1, −1), (1, 0). | |
0b2n | Problem:
Consider all real numbers $c$ such that $|x-8| + \left|4 - x^{2}\right| = c$ has exactly three real solutions. The sum of all such $c$ can be expressed as a fraction $a / b$ in lowest terms. What is $a + b$? | [] | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 93 | |
0bkx | Problem:
Fie $a$, $b$, $c$ numere reale pozitive astfel încât $a + b + c = 3$. Determinaţi valoarea minimă a expresiei
$$
A = \frac{2 - a^{3}}{a} + \frac{2 - b^{3}}{b} + \frac{2 - c^{3}}{c}
$$ | [] | Romania | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 3 | |
016v | The polynomial $P(x) = 2x^3 - 30x^2 + cx$ takes consecutive integer values for three consecutive integers. Determine these values. | [
"Assume $P(m-1) = n-1$, $P(m) = n$ and $P(m+1) = n+1$. We have\n$$\n\\pm 1 = (n \\pm 1) - n = P(m \\pm 1) - P(m) = \\pm 6m^2 + 6m \\pm 2 \\mp 60m - 30 \\pm c\n$$\n\nand adding these two equations together we get\n$$\n0 = 12m - 60\n$$\nso $m = 5$. Put $m = 5$ into either of the equations in the first display to get ... | Baltic Way | BALTIC WAY | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 244, 245, 246 | |
0khr | Problem:
Two circles $\Gamma_{1}$ and $\Gamma_{2}$ of radius $1$ and $2$, respectively, are centered at the origin. A particle is placed at $(2,0)$ and is shot towards $\Gamma_{1}$. When it reaches $\Gamma_{1}$, it bounces off the circumference and heads back towards $\Gamma_{2}$. The particle continues bouncing off t... | [
"Solution:\n\nBy symmetry, the particle must bounce off of $\\Gamma_{2}$ at points that make angles of $60^{\\circ}, 120^{\\circ}, 180^{\\circ}, 240^{\\circ}$, and $300^{\\circ}$ with the positive $x$-axis. Similarly, the particle must bounce off of $\\Gamma_{1}$ at points that make angles of $30^{\\circ}, 90^{\\ci... | United States | HMMT Spring 2021 Guts Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 403 | |
02k7 | Problem:
Numa corrida de carros, um piloto percorreu três trechos: um de $240~\mathrm{km}$, um de $300~\mathrm{km}$ e um de $400~\mathrm{km}$. O piloto sabe que as velocidades médias nesses trechos foram $40~\mathrm{km}/\mathrm{h}$, $75~\mathrm{km}/\mathrm{h}$ e $80~\mathrm{km}/\mathrm{h}$, mas não se lembra qual dess... | [
"Solution:\n\nO menor tempo de percurso é obtido quando se percorre o maior trecho com a maior velocidade e o menor trecho com a menor velocidade. Já o maior tempo é obtido quando se percorre o maior trecho com a menor velocidade e o menor trecho com a maior velocidade. Assim, o tempo total gasto pelo piloto nos tr... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | MCQ | D | |
07am | Point $O$ is the center of the circumcircle $\omega$ of the acute-angled triangle $ABC$. A circle centered at $O$ tangent to side $BC$ of the triangle is drawn. Let $X$ and $Y$ be the intersection points of tangents from $A$ to this circle with side $BC$ in such a way that points $X$ and $B$ are on one side of the line... | [
"Let $M$ and $N$ be the feet of the perpendicular lines from $O$ to $BY$ and $AY$, respectively. Since $O$ is a point on the perpendicular bisector of $AB$, we have $OA = OB$ and hence\n$$\nAN^2 = AO^2 - ON^2 = BO^2 - OM^2 = BM^2.\n$$\nSo $BY = BM + MY = AN + NY = AY$ and hence\n$$\n\\angle AYS = \\angle BAY = \\an... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0ibo | Problem:
Andrea flips a fair coin repeatedly, continuing until she either flips two heads in a row (the sequence $H H$) or flips tails followed by heads (the sequence $T H$). What is the probability that she will stop after flipping $H H$? | [
"Solution:\nThe only way that Andrea can ever flip $H H$ is if she never flips $T$, in which case she must flip two heads immediately at the beginning. This happens with probability $\\frac{1}{4}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 1/4 | |
06k3 | Suppose $ABCD$ is a cyclic quadrilateral. Extend $DA$ and $DC$ to $P$ and $Q$ respectively such that $AP = BC$ and $CQ = AB$. Let $M$ be the midpoint of $PQ$. Show that $MA \perp MC$. | [
"Let $B'$ be the point such that $AB'BC$ is an isosceles trapezoid. Thus, $B'$ lies on the circle passing through $A$, $B$, $C$, $D$. Also, we have $AP = CB = AB'$ and $CQ = AB = CB'$. Let $E$ and $F$ be the midpoints of $B'P$ and $B'Q$ respectively. Then $AE \\perp B'P$ and $CF \\perp B'Q$.\n\n![](attached_image_1... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cvl | Let $b > a > 1$ be real numbers. Define
$$
x_n = 2^n \left( 2^n \sqrt{b} - 2^n \sqrt{a} \right).
$$
Prove that the sequence $x_1, x_2, \dots$ is strictly decreasing. | [
"Докажем, что $x_n > x_{n+1}$. Положим $A = 2^{n+1}\\sqrt{a}$ и $B = 2^{n+1}\\sqrt{b}$. Легко видеть, что $B > A > 1$, отсюда $\\frac{A+B}{2} > 1$. Тогда имеем\n$$\nx_{n+1} = 2^{n+1}(B-A) > 0, \\\\\nx_n = 2^n(B^2 - A^2) = 2^{n+1}(B-A) \\frac{A+B}{2} = x_{n+1} \\cdot \\frac{A+B}{2} > x_{n+1}, \\\\\n\\text{что и треб... | Russia | Regional round | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English; Russian | proof only | null | |
0g2i | Problem:
Sei $n \geq 2$ eine natürliche Zahl. Seien $d_{1}, \ldots, d_{r}$ alle verschiedenen positiven Teiler von $n$, die kleiner sind als $n$ selbst. Bestimme alle $n$, für die gilt:
$$
\operatorname{kgV}\left(d_{1}, \ldots, d_{r}\right) \neq n
$$ | [
"Solution:\n\nFalls $n$ eine Primpotenz ist, gilt $n=p^{\\alpha}$ für eine Primzahl $p$. Die Teiler von $n$ kleiner als $n$ sind dann $d_{1}=1, d_{2}=p, \\ldots, d_{r}=p^{\\alpha-1}$. Somit gilt $\\operatorname{kgV}\\left(1, p, \\ldots, p^{\\alpha-1}\\right)=p^{\\alpha-1} \\neq n$.\n\nFalls $n$ keine Primpotenz ist... | Switzerland | SMO - Vorrunde | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Exactly the prime powers | |
08ma | Problem:
Find all pairs $(x, y)$ of integers which satisfy the equation
$$
(x+y)^{2}\left(x^{2}+y^{2}\right)=2009^{2}
$$ | [
"Solution:\nLet $x+y=s$, $x y=p$ with $s \\in \\mathbb{Z}^{*}$ and $p \\in \\mathbb{Z}$. The given equation can be written in the form\n$$\ns^{2}\\left(s^{2}-2 p\\right)=2009^{2}\n$$\nor\n$$\ns^{2}-2 p=\\left(\\frac{2009}{s}\\right)^{2}\n$$\nSo, $s$ divides $2009=7^{2} \\times 41$ and it follows that $p \\neq 0$.\n... | JBMO | 2009 Shortlist JBMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (40, 9), (9, 40), (-40, -9), (-9, -40) | |
09ow | Find the number of couples of integers $(x, y)$ satisfying $1 \le x, y \le 45$ and $\frac{x^3+1}{xy+1}$ is an integer. | [] | Mongolia | MMO2025 Round 3 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 93 | |
0dxn | Problem:
Reši enačbo $\log_{3}(\log_{2} x + 12) + 2 = 4$ in rešitev zapiši v obliki ulomka. | [
"Solution:\n\nEnačbo najprej uredimo $\\log_{3}(\\log_{2} x + 12) = 2$. Upoštevamo definicijo logaritma in zapišemo $3^{2} = \\log_{2} x + 12$. Enačbo ponovno uredimo in dobimo $\\log_{2} x = -3$. Rešimo $2^{-3} = x$. Rešitev je $x = \\frac{1}{8}$."
] | Slovenia | 7. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 1/8 | |
0gxj | A parallelogram $ABCD$ with $AC > BD$ is given. Let $k_1$ be the circle with diameter $AC$ and $k_2$ the circle with diameter $DC$. $k_1$ meets line $AB$ at the point $E$, $k_2$ meets the line $AC$ at the points $C$ and $O$, and line $AD$ at the point $F$. Suppose that $AO = a$, $FO = b$ and $\angle BAC = 45^\circ$. Fi... | [
"Finally, $EF$ and $AC$ are chords in the circle $k_1$, hence $EO \\cdot OF = AO \\cdot OC$, which implies:\n$$\n\\frac{S(AOE)}{S(COF)} = \\frac{AO \\cdot GE}{FO \\cdot CO} = \\left(\\frac{AO}{FO}\\right)^2 = \\left(\\frac{a}{b}\\right)^2\n$$\n\n\nFig.25"
] | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | (a/b)^2 | |
04g3 | Determine all complex numbers $z$ such that
$$
|z^2 - i| = 1 \quad \text{and} \quad |z| = \sqrt{2}.
$$ | [
"Let $z = x + iy$, where $x, y \\in \\mathbb{R}$.\n\nGiven $|z| = \\sqrt{2}$, so\n$$\n|z|^2 = x^2 + y^2 = 2.\n$$\n\nAlso, $|z^2 - i| = 1$.\nCompute $z^2$:\n$$\nz^2 = (x + iy)^2 = x^2 - y^2 + 2ixy.\n$$\nSo,\n$$\nz^2 - i = (x^2 - y^2) + 2ixy - i = (x^2 - y^2) + i(2xy - 1).\n$$\nTherefore,\n$$\n|z^2 - i|^2 = (x^2 - y^... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | 1 + i, -1 - i | |
081d | Problem:
In un torneo di pallacanestro ogni squadra affronta esattamente due volte tutte le altre squadre partecipanti. Il torneo viene vinto da una squadra sola in testa alla classifica con 26 punti, mentre esattamente due squadre arrivano ultime con 20 punti. Quante squadre hanno partecipato al torneo?
(Ricordiamo ... | [
"Solution:\n\nIndichiamo con $n$ il numero delle squadre partecipanti. Il numero di partite giocate durante tutto il torneo sarà uguale a $n(n-1)$, tante quante sono le coppie ordinate di elementi distinti di un insieme con $n$ elementi.\n\nIndichiamo con $M$ la media aritmetica dei punteggi totalizzati dalle squad... | Italy | Gara Nazionale di Matematica | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 12 | |
01ah | Consider a tetrahedron bounded by four right-angled triangles. It is known that three of its edges have the same length $s$. Compute its volume. | [
"The three equal edges clearly cannot bound a face by themselves, for then this triangle would be equilateral and not right-angled. Nor can they be incident to the same vertex, for then the opposite face would again be equilateral.\n\nHence we may name the tetrahedron $ABCD$ in such a way that $AB = BC = CD = s$. T... | Baltic Way | Baltic Way 2013 | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | s^3/6 | |
0h55 | Find all pairs of prime numbers $(p, q)$ with $p > q$, for which both numbers $p+q$ and $p-q$ are also prime. | [
"For the number $p+q$ to be prime the numbers $p$ and $q$ must be of different parity, which automatically means that $q=2$, since $p > q$. By the problem statement we then have that the numbers $p-2, p$, and $p+2$ should be prime. Since they obviously have different remainders in division by $3$, one of them must ... | Ukraine | Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (5, 2) | |
0bga | Problem:
Un şir de numere este numit complet dacă are termeni naturali nenuli şi orice număr natural nenul are cel puţin un multiplu printre termenii şirului.
Arătaţi că o progresie aritmetică este şir complet dacă şi numai dacă raţia sa divide primul termen.
Problem:
Egy számsorozatot nevezzünk teljesnek, ha elemei n... | [
"Solution:\nDacă raţia $r$ divide primul termen $a_{1}$, atunci $a_{1} = d r$, $d \\in \\mathbb{N}$ şi $a_{n} = (d + n - 1) r$, iar un multiplu al numărului natural nenul $k$ se obţine luând $d + n - 1$ multiplu de $k$.\n\nReciproc, dacă $r = 0$, atunci şirul nu poate fi complet.\n\nDeoarece $r \\neq 0$ şi, conform... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof only | null | |
07c8 | In this question, functions on natural numbers that are computable by a computer (which has a finite memory and output) for arbitrary large numbers are investigated. These functions are called Computable Functions. Since a computer has a finite number of outputs, the investigation is restricted only to those functions ... | [
"a) The idea here is that a computer (with sufficiently large but finite memory) can compute the remainder of arbitrary large natural numbers when divided by $m$. The following describes a Machine for computing the remainder of a given number modulo $m$ in base $k$:\n\n* **Inputs:** Set $\\{0, 1, \\dots, k-1\\}$.\n... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | null | proof only | null | |
0124 | Problem:
The points $A, B, C, D, E$ lie on the circle $c$ in this order and satisfy $AB \parallel EC$ and $AC \parallel ED$. The line tangent to the circle $c$ at $E$ meets the line $AB$ at $P$. The lines $BD$ and $EC$ meet at $Q$. Prove that $|AC| = |PQ|$. | [
"Solution:\n\nThe arcs $BC$ and $AE$ are of equal length (see Figure 1). Also, since $AB \\parallel EC$ and $ED \\parallel AC$, we have $\\angle CAB = \\angle DEC$ and the arcs $DC$ and $BC$ are of equal length. Since $PE$ is tangent to $c$ and $|AE| = |DC|$, then $\\angle PEA = \\angle DBC = \\angle QBC$. As $ABCD... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0d4p | Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that
$$
f(n+1) > \frac{f(n) + f(f(n))}{2}
$$
for all $n \in \mathbb{N}$, where $\mathbb{N}$ is the set of strictly positive integers. | [
"It is clear that $f(n) \\geq 1$ for all $n \\geq 1$. Assume that $f(n) \\geq m$ for all $n \\geq m$, for some $m \\geq 1$. Let $n \\geq m+1$. Because $n-1 \\geq m$, we have $f(n-1) \\geq m$ and therefore $f(f(n-1)) \\geq m$. We deduce that $f(n) > \\frac{f(n-1) + f(f(n-1))}{2} \\geq m$. Hence, $f(n) \\geq m+1$. In... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English, Arabic | proof and answer | All functions f: N -> N of the form: (i) f(n) = n for all n; (ii) f(n) = n + 1 for all n; (iii) for some integer m > 1, f(n) = n for n < m and f(n) = n + 1 for n >= m. | |
04r9 | Let $M$ be the midpoint of the side $AB$ of a triangle $ABC$. Prove that the equality $|\angle ABC| + |\angle ACM| = 90^\circ$ holds if and only if the triangle $ABC$ is isosceles or right-angled, with $AB$ as a base or a hypotenuse, respectively. (Pavel Novotný) | [
"Assume first that $|\\angle ABC| + |\\angle ACM| = 90^\\circ$. Using the notation $\\phi = |\\angle ACM|$ and $\\psi = |\\angle BCM|$ (Fig. 1), we conclude from our assumption that $|\\angle ABC| = 90^\\circ - \\phi$, and hence $|\\angle BAC| = 90^\\circ - \\psi$ as well, because of an easy angle computation in $\... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"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 ... | English | proof only | null | |
0gqj | Find all triples $(m, n, p)$ satisfying $m^3 + 7p^2 = 2^n$, where $p$ is prime and $m, n$ are positive integers. | [
"Note that $n \\ge 5$ and $m$ and $p$ have the same parity. If $m$ is even then $p = 2$ and we get $0 \\equiv 4 \\pmod{8}$. Therefore, $m$ and $p$ both are odd. $m^3 \\equiv 0, \\pm1 \\pmod{7}$ and $2^n \\equiv 2, 4, 1 \\pmod{7}$. Therefore, $2^n \\equiv 1 \\pmod{7}$ and consequently $n = 3k$: $2^{3k} - m^3 = 7p^2$... | Turkey | Team Selection Test | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest ... | null | proof and answer | (1, 6, 3) | |
0f4k | Problem:
$\{a\}_1, \{a\}_2, \ldots, \{a\}_{1982}$ is a permutation of $1, 2, \ldots, 1982$. If $\{a\}_1 > \{a\}_2$, we swap $\{a\}_1$ and $\{a\}_2$. Then if$\,$(the new)$\, \{a\}_2 > \{a\}_3$ we swap $\{a\}_2$ and $\{a\}_3$. And so on. After $1981$ potential swaps we have a new permutation $\{b\}_1, \{b\}_2, \ldots, \... | [] | Soviet Union | 16th ASU | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 100 | |
0e37 | Each point on the segment $\overline{D}$ is either red or blue. Show that there exist three different points $A$, $B$ and $C$ of the same colour such that $|AB| = |BC|$. | [
"Let us divide the segment into three parts of equal length. We can find two points of the same colour in the middle part. We may assume that these two points are red. Let us call them $A$ and $B$. Let $A'$ be the image of the point $A$ under reflection over the point $B$ and let $B'$ be the image of the point $B$ ... | Slovenia | National Math Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
00cw | Ignacio tiene una hoja de papel. La puede cortar en 6 pedazos o en 8 pedazos, a su elección. Luego, en cada etapa, puede elegir uno de los pedazos existentes y cortarlo en 6 pedazos o cortarlo en 8 pedazos.
a) Decidir si de esta manera Ignacio puede tener, después de alguna etapa, exactamente 24 pedazos de papel.
b) ... | [
"Comenzamos con una hoja y en cada paso agregamos $5$ o $7$ trozos. Luego queremos que\n$$\na) 1+5n+7m=24 \\leftrightarrow 5n+7m=23. \\text{ Notemos que } n \\le 4 \\text{ y } m \\le 3.\n$$\nConsideramos la igualdad módulo $7$ y tenemos $5n \\equiv 23 \\equiv 2 \\ (\\text{mod } 7)$, pero ninguno de los posibles val... | Argentina | Nacional OMA | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Spanish | proof and answer | a) No. b) Yes: perform two cuts into six pieces and three cuts into eight pieces to reach thirty two pieces. | |
0cxw | Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a strictly increasing function such that $f(f(n))=3 n$, for all $n \in \mathbb{N}$. Find $f(2010)$.
(Note: $\mathbb{N}=\{0,1,2, \ldots\}$ ). | [
"It follows $f(f(f(n)))=f(3 n)$, hence for all $n \\in \\mathbb{N}$ we have\n$$\nf(3 n)=3 f(n) \\tag{1}\n$$\nFor $n=0$ we find $f(0)=0$. We have $f(1) \\neq 1$. Indeed, if $f(1)=1$, then we get $3=f(f(1))=f(1)=1$, not possible. Hence $f(1)>1$, so $3=f(f(1))>f(1)>1$. It follows $f(1)=2$, and consequently $f(2)=f(f(1... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 3843 | |
02fl | Does there exist a set of $n > 2$ points in the plane such that no three are collinear and the circumcenter of any three points of the set is also in the set? | [
"No, it's not possible. Let $S$ be the set of $n$ points. Consider the smallest circumcircle $\\omega$ of all triangles determined by the $n$ points. Let $A, B, C \\in S$ be three points on $\\omega$ and let $O$ be its center. If one of the central angles $\\angle AOB$, $\\angle BOC$, $\\angle COA$ is less than $12... | Brazil | XVIII OBM | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08t3 | Suppose $2008$ real numbers $x_1, x_2, \dots, x_{2008}$ are given and suppose that $|x_1| = 999$ and $|x_n| = |x_{n-1} + 1|$ for all $n$ with $2 \le n \le 2008$ are satisfied. Determine the smallest possible value that $x_1 + x_2 + \dots + x_{2008}$ can have under these conditions. | [
"Let $S = x_1 + x_2 + \\dots + x_{2008}$. Since $x_1^2 = |x_1|^2 = 999^2$, and $x_n^2 = |x_{n-1} + 1|^2 = (x_{n-1} + 1)^2$ for $2 \\le n \\le 2008$, we have\n$$\n\\begin{aligned}\nx_1^2 + x_2^2 + \\dots + x_{2008}^2 &= 999^2 + (x_1 + 1)^2 + \\dots + (x_{2007} + 1)^2 \\\\\n&= (x_1^2 + \\dots + x_{2007}^2) + 2(x_1 + ... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | -500004 | |
0c21 | Let $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ be two quadratic functions with the following property: for any real number $r$, if $f(r)$ is an integer number, then $g(r)$ is also an integer number.
Prove that there are two integers $m$ and $n$ such that $g(x) = m f(x) + n$, for any real number $... | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0gtp | There are $n$ empty red boxes numbered $1, 2, \dots, n$ and $n$ empty white boxes numbered $1, 2, \dots, n$ on the table. At each step we choose one red and one white box and put one ball into each chosen box. After finite number of steps it turns out that for any two identically numbered red and white boxes either the... | [
"Answer: $n = 11k$, where $k$ is a positive integer.\n\nSuppose that in $m$ identically numbered pairs of red and white boxes each red box contains more balls than the white box. Then the difference between the total number of balls in red boxes and the total number of balls in white boxes is $6m - 16(n - m)$. On t... | Turkey | 31st Junior Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | n = 11k for positive integer k | |
0jp1 | Problem:
A parallelogram has 2 sides of length $20$ and $15$. Given that its area is a positive integer, find the minimum possible area of the parallelogram. | [
"Solution:\nAnswer: $1$\nThe area of the parallelogram can be made arbitrarily small, so the smallest positive integer area is $1$."
] | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | final answer only | 1 | |
0efm | Problem:
Kenguruju Pitagori so všeč le tista naravna števila, ki so deljiva s $4$ in imajo vsoto števk enako $3$ ter imajo v zapisu natanko $5$ števk enakih $0$. Koliko naravnih števil je všeč kenguruju Pitagori?
(A) $15$
(B) $16$
(C) $19$
(D) $20$
(E) $25$ | [
"Solution:\n\nŠtevke, ki so enake $0$, vsoti števk ne prispevajo ničesar. Iz neničelnih števk pa lahko dobimo vsoto $3$ le na tri načine: $3=3$, $1+2=3$ ali $1+1+1=3$. Ker je natanko $5$ števk števila, ki je všeč kenguruju Pitagori, enakih $0$, imamo torej tri možnosti.\n\nNaravno število, ki ima števke $3,0,0,0,0,... | Slovenia | Slovenian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | MCQ | D | |
0d9m | Denote $S$ as the set of prime divisors of all integers of form $2^{n^{2}+1}-3^{n}$, $n \in \mathbb{Z}^{+}$. Prove that $S$ and $\mathscr{P} \backslash S$ both contain infinitely many elements ($\mathscr{P}$ is the set of prime numbers). | [
"First, suppose on the contrary that $S$ is finite, then $S=\\{p_{1}, p_{2}, \\ldots, p_{k}\\}$ for $k=|S|$. It is easy to check that $2,3 \\notin S$.\n\nConsider number $N=(p_{1}-1)(p_{2}-1) \\cdots (p_{k}-1)$ and $M=2^{N^{2}+1}-3^{N}$. For some $p \\in S$, by Euler's theorem, we have\n$$\n2^{p-1} \\equiv 1 \\pmod... | Saudi Arabia | Team selection tests for IMO 2018 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity"
] | English | proof only | null | |
095r | Problem:
Rezolvați ecuația $\sqrt{y \sqrt{5}}-\sqrt{x \sqrt{5}}=\sqrt{3 \sqrt{5}-5}$ în numere raționale. | [
"Solution:\nDeterminăm DVA al acestei ecuații:\n$$\n\\left\\{\n\\begin{array}{c}\n\\sqrt{y \\sqrt{5}}-\\sqrt{x \\sqrt{5}} \\geq 0 \\\\\ny \\geq 0, \\quad x \\geq 0\n\\end{array} \\Leftrightarrow 0 \\leq x \\leq y\\right.\n$$\nFie $x, y \\in \\mathbb{Q}$. Ridicăm ambele părți ale ecuației la pătrat:\n$$\n\\begin{ali... | Moldova | A 62 - A OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | (1/2, 5/2) | |
08gv | Problem:
Let $a, b, c, d$ be real numbers such that $0 \leq a \leq b \leq c \leq d$. Prove the inequality
$$
a b^{3}+b c^{3}+c d^{3}+d a^{3} \geq a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}
$$ | [
"Solution:\nThe inequality is equivalent to\n$$\n\\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\\right)^{2} \\geq\\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}\\right)^{2}\n$$\nBy the Cauchy-Schwarz inequality,\n$$\n\\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\\right)\\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\\right) \\geq\\... | JBMO | null | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
03yn | Given an integer $n \ge 3$, find the maximum real number $M$, such that for any positive numbers $x_1, x_2, \dots, x_n$, there exists a permutation $y_1, y_2, \dots, y_n$ of $x_1, x_2, \dots, x_n$ that satisfies
$$
\sum_{i=1}^{n} \frac{y_i^2}{y_{i+1}^2 - y_{i+1} y_{i+2} + y_{i+2}^2} \ge M,
$$
where $y_{n+1} = y_1$, $y_... | [
"Let\n$$\nF(x_1, \\dots, x_n) = \\sum_{i=1}^{n} \\frac{x_i^2}{x_{i+1}^2 - x_{i+1}x_{i+2} + x_{i+2}^2}.\n$$\nFirst, take $x_1 = x_2 = \\cdots = x_{n-1} = 1$, $x_n = \\epsilon$, then all permutations are the same in the sense of circulation. In this case, we have\n$$\nF(x_1, \\dots, x_n) = n - 3 + \\frac{2}{1 - \\eps... | China | China National Team Selection Test | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | M = n - 1 | |
080n | Problem:
Qual è il più piccolo numero intero positivo che possiede esattamente 15 divisori? | [
"Solution:\nLa risposta è $144$. Supponiamo che $n = p_{1}^{a_{1}} \\cdots p_{k}^{a_{k}}$ sia la scomposizione di $n$ in fattori primi, con $p_{1} < p_{2} < \\cdots < p_{k}$ tutti distinti e $a_{i} > 0$ per $i = 1, \\ldots, k$. Un intero positivo $d$ è un divisore di $n$ se e solo se la sua scomposizione in fattori... | Italy | Progetto Olimpiadi di Matematica | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 144 | |
0535 | a) There are three numbers $a$, $b$, $c$ such that $a \le b \le c$. Let $p$, $q$, $r$ be the pairwise sums $a+b$, $b+c$, $c+a$ in the order such that $p \le q \le r$. Given that $r-q = q-p$, is it certainly true that $c-b = b-a$?
b) There are four numbers $e$, $f$, $g$, $h$ such that $e \le f \le g \le h$. Let $u$, $v... | [
"a) If $a \\le b \\le c$, then $a+b \\le a+c \\le b+c$, due to which $p = a+b$, $q = a+c$ and $r = b+c$. Equality $r-q = q-p$ can now be written as $(b+c) - (a+c) = (a+c) - (a+b)$, simplifying to $b-a = c-b$.\n\nb) Let $e = 0$, $f = 1$, $g = 2$ and $h = 4$. Their pairwise sums in increasing order are $u = 0+1=1$, $... | Estonia | Open Contests | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) Yes. b) No; for example, zero, one, two, and four. | |
0gtt | In the interior of a trapezoid $ABCD$ with $AB \parallel CD$ a point $T$ is chosen such that $\angle ATD = \angle CTB$. The second intersection point of the line $AT$ with the circumcircle of $ACD$ is $K$. The second intersection point of the line $BT$ with the circumcircle of $BCD$ is $L$. Show that $KL \parallel AB$. | [
"Let the lines $AT$ and $BT$ intersect the line $CD$ at the points $A_1$ and $B_1$, and intersect the circle $(TCD)$ at the points $A_2$ and $B_2$ (beside $T$). Now one has $\\overline{A_2D} = \\overline{CB_2}$ on the circle $(TCD)$, which implies $A_2B_2 \\parallel CD$. On the other hand $A_1T \\cdot A_1A_2 = A_1C... | Turkey | Team Selection Test for IMO 2023 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00bw | Let $ABC$ be an acute-angled and scalene triangle. Consider the altitudes $BE$ and $CD$, that intersect in $H$. The bisector of the angle $B\hat{A}C$ intersects the altitudes $BE$ and $CD$ in $P$ and $Q$ respectively. Let $T$ be the orthocenter of the triangle $HPQ$. Prove that the triangles $TDA$ and $TEA$ have the sa... | [
"As $T$ is the orthocenter of the triangle $HPQ$, we have that $QT$ is perpendicular to $BE$; then, since $BE$ is perpendicular to $AC$, it follows that $QT$ is parallel to $AC$. Similarly, $TP$ is parallel to $AB$. Assume $QT$ intersects $AB$ at the point $F$ and $TP$ intersects $AC$ at a point $G$.\n\n![](attache... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06c1 | The general term of a sequence $\{a_n\}$ is given by $a_{n+2} = a_{n+1} - a_n$. If the sum of the first 1997 terms is 1879 and the sum of the first 1879 terms is 1997, what is the sum of the first 2000 terms? | [
"The answer is 5873.\nLet $a_1 = x$ and $a_2 = y$. Then the sequence is\n$$\nx, y, y-x, -x, -y, x-y, x, y, \\dots\n$$\nNote that the sequence repeats every 6 terms. Also, the sum of every 6 consecutive terms is 0. Thus, the sum of the first 1997 = $6 \\times 332 + 5$ terms is $x + y + (y-x) + (-x) + (-y) = y-x$, an... | Hong Kong | HKG TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 5873 | |
09aa | Let us say that an integer number is a power number, if there exist positive integers $a$ and $n$ such that the number equals to $a^n$ and $n > 1$.
a) Prove that there exist $2010$ positive integers such that every sum of the integers selected from them is not a power number.
b) Prove that there exist $2010$ positive... | [
"a) Let $p_i$ be the $i$-th prime number. Consider the numbers\n$$\np_1,\\ p_1^2 p_2,\\ p_1^2 p_2^2 p_3,\\ \\dots,\\ p_1^2 p_2^2 \\dots p_{2009} p_{2010}.\n$$\nIf $p_1^2 p_2^2 \\dots p_k^2 p_{k+1}$ is the least number in the sum, then the sum is divisible by $p_{k+1}$, but not divisible by $p_{k+1}^2$. Hence each s... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
092w | Problem:
Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying
$$
f\left(x^{2}+f(x) f(y)\right)=x f(x+y)
$$
for all real numbers $x$ and $y$. | [
"Solution:\nPut $x:=0$. Then $f(f(0) f(y))=0$, so there is at least one real number $a$ such that $f(a)=0$.\n\nLet $z$ be an arbitrary real number, let's put $x:=a, y:=z-a$. Then we get $f\\left(a^{2}\\right)=a \\cdot f(z)$. If $a \\neq 0$, we'll get that $f$ is constant, i.e. $f(x) \\equiv c$. The equation (1) is ... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0, f(x) = x, or f(x) = -x | |
04xt | Positive real numbers $a$, $b$, $c$, $d$ satisfy the relations
$$
abcd = 4, \quad a^2 + b^2 + c^2 + d^2 = 10.
$$
Determine the largest possible value of the expression $ab + bc + cd + da$. | [
"Let $V = ab + bc + cd + da$. We will find the maximum value of\n$$\nV^2 = (a+c)^2(b+d)^2 = (a^2+c^2+2ac)(b^2+d^2+2bd). \\quad (1)\n$$\nAll the given expressions do not change under the simultaneous replacement of $a$ by $b$, $b$ by $c$, $c$ by $d$ and $d$ by $a$. Since $ac \\cdot bd = 4$, at least one of the numbe... | Czech-Polish-Slovak Mathematical Match | 12th Czech-Polish-Slovak Mathematics Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | sqrt(82) | |
0gt9 | Points $A$, $B$, $C$ are given on a semicircle. The line which is tangent to the semicircle at $A$ intersects the line of the diameter at $M$, and the line which is tangent to the semicircle at $B$ intersects the line of the diameter at $N$. The line which passes through $A$ and is perpendicular to the diameter interse... | [
"In the solution, we use directed lengths on the line of the diameter. Let $A'$, $B'$, $C'$ be the feet of the perpendiculars dropped onto the diameter from $A$, $B$, $C$. We have\n$$\n\\frac{ZB'}{ZA'} = \\frac{|SB'|}{|RA'|} = \\frac{|SB'|}{|CC'|} \\cdot \\frac{|CC'|}{|RA'|} = \\frac{MB'}{MC'} \\cdot \\frac{NC'}{NA... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
0amp | Problem:
The length $d$ of a tangent, drawn from a point $A$ to a circle, is $\frac{4}{3}$ of the radius $r$. What is the shortest distance from $A$ to the circle? | [] | Philippines | Area Stage | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2r/3 | |
0ccu | Determine the positive integers $a$, $b$, $c$ with the following properties:
(i) $(a^2 + b^2)(c^2 + 2023^2) = (ab + 2023c)^2$;
(ii) $(a^2 + 2023^2)(b^2 + c^2) = (2023a + bc)^2$;
(iii) the greatest common divisor of $a$, $b$, $c$ and $2023$ equals $1$. | [
"By subtracting the equality (i) from (ii) we obtain $(a^2 - c^2) \\cdot (b^2 - 2023^2) = 0$. Because $a$, $b$, $c$ are positive integers, we deduce that $a = c$ or $b = 2023$.\n\nIf $a = c$, using (i) we find $(a^2 - 2023b)^2 = 0$, that is $a^2 = 2023b = 7 \\cdot 17^2 \\cdot b$. Consequently $7 \\mid b$ and $7 \\m... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (a, b, c) in {(1, 2023, 2023^2), (7^2, 2023, 17^4), (17^4, 2023, 7^2), (2023^2, 2023, 1)} | |
0a9r | Problem:
Define a sequence $\left(n_{k}\right)_{k \geq 0}$ by $n_{0}=n_{1}=1$, and $n_{2k}=n_{k}+n_{k-1}$ and $n_{2k+1}=n_{k}$ for $k \geq 1$. Let further $q_{k}=n_{k} / n_{k-1}$ for each $k \geq 1$. Show that every positive rational number is present exactly once in the sequence $\left(q_{k}\right)_{k \geq 1}$. | [
"Solution:\n\nClearly, all the numbers $n_{k}$ are positive integers. Moreover,\n$$\nq_{2k} = \\frac{n_{2k}}{n_{2k-1}} = \\frac{n_{k} + n_{k-1}}{n_{k-1}} = q_{k} + 1\n$$\nand similarly,\n$$\n\\frac{1}{q_{2k+1}} = \\frac{n_{2k}}{n_{2k+1}} = \\frac{n_{k} + n_{k-1}}{n_{k}} = \\frac{1}{q_{k}} + 1\n$$\nIn particular, $q... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
08hj | Problem:
The sequences $\left(a_{n}\right)_{n \geq 0}$ and $\left(b_{n}\right)_{n \geq 0}$ satisfy the conditions $(1+\sqrt{3})^{2 n+1}=a_{n}+b_{n} \sqrt{3}$ and $a_{n}, b_{n} \in \mathbb{Z}$. Find the recurrent relation for each of the sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | Coupled first-order system: a_{n+1} = 4 a_n + 6 b_n, b_{n+1} = 2 a_n + 4 b_n, with a_0 = 1, b_0 = 1. Decoupled second-order recurrences: a_{n+2} = 8 a_{n+1} − 4 a_n with a_0 = 1, a_1 = 10; and b_{n+2} = 8 b_{n+1} − 4 b_n with b_0 = 1, b_1 = 6. | |
070w | Problem:
Label the vertices of a regular $n$-gon from $1$ to $n > 3$. Draw all the diagonals. Show that if $n$ is odd then we can label each side and diagonal with a number from $1$ to $n$ different from the labels of its endpoints so that at each vertex the sides and diagonals all have different labels. | [
"Solution:\n\nLabel the diagonal/side between $i$ and $j$ as $i + j$ (reduced if necessary mod $n$) almost works. The labels for all the lines at a given vertex will be different. But the line between $i$ and $n$ will have label $i$, the same as one endpoint. However, we are not using the label $2i$ for the lines f... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
01kx | Bob has the collection of $n$ coins. Each of them weights an integer number of grams, and their total weight is equal to 300 grams.
Find the smallest possible value of $n$ for which Bob always (independently of coin's weights) can partition his collection into three groups so that the total weight the coins in each of ... | [
"Answer: 201.\nLet $a_1 \\le a_2 \\le \\dots \\le a_n$ be the weights of the coins in Bob's collection. If there exists an $i$ such that $a_i \\ge 101$, and $k = l = m = 100$, then obviously Bob\n\nThen $S_j - S_i \\doteq 100$, and, on the other hand, $S_j - S_i < S_{100} + a_{201} = 200$. It follows that $S_j - S_... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 201 | |
04yj | The plane was divided by vertical and horizontal lines into unit squares. Determine whether it is possible to write integers into cells of this infinite grid so that:
(i) every cell contains exactly one integer
(ii) every integer appears exactly once
(iii) for every two cells $A$ and $B$ sharing exactly one vertex, if ... | [
"Yes, this is possible. Consider the spiral depicted below and write consecutive integers along the spiral:\n\n\nWe claim that this works. Consider any two cells $A$ and $B$ sharing exactly one vertex. Consider the $2 \\times 2$ square containing $A$ and $B$. If the $2 \\times 2$ square con... | Czech-Polish-Slovak Mathematical Match | CAPS Match 2025 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes, it is possible. | |
0k7o | Problem:
Let $x$ and $y$ be positive real numbers. Define $a=1+\frac{x}{y}$ and $b=1+\frac{y}{x}$. If $a^{2}+b^{2}=15$, compute $a^{3}+b^{3}$. | [
"Solution:\n\nNote that $a-1=\\frac{x}{y}$ and $b-1=\\frac{y}{x}$ are reciprocals. That is,\n$$\n(a-1)(b-1)=1 \\Longrightarrow a b-a-b+1=1 \\Longrightarrow a b=a+b\n$$\nLet $t=a b=a+b$. Then we can write\n$$\na^{2}+b^{2}=(a+b)^{2}-2 a b=t^{2}-2 t\n$$\nso $t^{2}-2 t=15$, which factors as $(t-5)(t+3)=0$. Since $a, b>... | United States | HMMT February 2019 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 50 | |
00ge | Prove that for every irrational real number $a$, there are irrational real numbers $b$ and $b^{\prime}$ so that $a+b$ and $a b^{\prime}$ are both rational while $a b$ and $a+b^{\prime}$ are both irrational. | [
"Let $a$ be an irrational number. If $a^{2}$ is irrational, we let $b = -a$. Then, $a + b = 0$ is rational and $a b = -a^{2}$ is irrational.\n\nIf $a^{2}$ is rational, we let $b = a^{2} - a$. Then, $a + b = a^{2}$ is rational and $a b = a^{2}(a - 1)$. Since\n$$\na = \\frac{a b}{a^{2}} + 1\n$$\nis irrational, so is ... | Asia Pacific Mathematics Olympiad (APMO) | XVII APMO - March, 2005 | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Number Theory > Other"
] | null | proof only | null | |
0bkc | Let $f: \mathbb{N} \to \mathbb{N}^*$ be a strictly increasing function. Prove that:
a) there exists a decreasing sequence of positive real numbers, $(y_n)_{n \in \mathbb{N}}$, converging to $0$, such that $y_n \le 2y_{f(n)}$, for all $n \in \mathbb{N}$;
b) if $(x_n)_{n \in \mathbb{N}}$ is a decreasing sequence of real ... | [
"a) Since $f(0) > 0$ and $f$ is strictly increasing, it follows that $f(n) > n$, for all $n \\in \\mathbb{N}$. Consider the sequence of non-negative integers $(n_k)_{k \\in \\mathbb{N}}$, defined by $n_0 = 0$ and $n_k = f(n_{k-1})$, $k \\in \\mathbb{N}^*$. The properties of $f$ imply that the sequence is strictly i... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0eke | Problem:
Naj bosta $p$ in $q$ različni praštevili. Za koliko različnih vrednosti $a$ sta obe rešitvi kvadratne enačbe $x^{2}+a x+p q=0$ celoštevilski?
(A) 4
(B) 3
(C) 2
(D) 1
(E) 0 | [
"Solution:\nKer je koeficient pri $x^{2}$ enak 1, je po Vietovih pravilih produkt $x_{1} x_{2}$ obeh rešitev enačbe enak $p q$. Ker morata biti obe rešitvi celoštevilski in sta $p$ in $q$ praštevili, imamo za množico rešitev $\\{x_{1}, x_{2}\\}$ le možnosti $\\{1, p q\\},\\{-1,-p q\\},\\{p, q\\}$ ali $\\{-p,-q\\}$.... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | MCQ | A | |
0dqv | Find all pairs of positive integers $(m, n)$ such that
$$
m+n - \frac{3mn}{m+n} = \frac{2011}{3}.
$$ | [
"Answer: $(m, n) = (1144, 377)$ or $(377, 1144)$.\n\nLet $m$ and $n$ be positive integers satisfying the given equation. That is $2011(m + n) = 3(m^2 - mn + n^2)$. Since the equation is symmetric in $m$ and $n$, we may assume $m \\ge n$. If $m = n$, then $m = n = 4022/3$ which is not an integer. So we may further a... | Singapore | Singapore Mathematical Olympiad (SMO) 2011 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (1144, 377) and (377, 1144) | |
06b8 | Prove that there exists integer $n \ge 1$, such that the number of the pairs $(a, b)$ of positive integers satisfying the equation
$$
\frac{1}{a-b} - \frac{1}{a} + \frac{1}{b} = \frac{1}{n}
$$
is greater than 2024. | [
"We seek solutions $(a, b)$ with $a - b = k b$, that is $a = (k+1) b$ for some integer $k > 0$. Then the left part of the given equation is equal to\n$$\n\\frac{1}{k b} - \\frac{1}{(k+1) b} + \\frac{1}{b} = \\frac{1}{b} \\left( \\frac{1}{k} - \\frac{1}{k+1} + 1 \\right) = \\frac{1}{b} \\cdot \\frac{k(k+1)+1}{k(k+1)... | Greece | Hellenic Mathematical Olympiad | [
"Number Theory > Diophantine Equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0465 | Determine the smallest real number $\lambda$ with the following property: every positive integer $n$ can be written as a product $n = x_1x_2\cdots x_{2023}$, with each $x_i$ either a prime number or a positive integer that is less than or equal to $n^\lambda$. | [
"The number $\\lambda$ is $\\frac{1}{1012}$.\nThe minimal such $\\lambda$ is $\\frac{1}{1012}$.\nTo see that $\\lambda < \\frac{1}{1012}$ does not work, we take $n = p^{2024}$ for some prime number $p$. Then when writing $n$ as the product of $2023$ integers, at least one of the integers is $p^\\alpha$ for some $\\... | China | Chinese Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics >... | English | proof and answer | 1/1012 | |
048t | Determine all the positive integers $n$ such that $\frac{n-1}{n-5}$ is an integer. | [] | Croatia | Hrvatska 2011 | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 1, 3, 4, 6, 7, 9 | |
0lee | Consider the real sequence $(x_n)$ such that $x_1 \in (0, \frac{1}{2})$ and
$$
x_{n+1} = 3x_n^2 - 2n x_n^3, \quad \forall n \ge 1.
$$
a) Prove that $\lim_{n \to \infty} x_n = 0$.
b) For each $n \ge 1$, let $y_n = x_1 + 2x_2 + \dots + n x_n$. Prove that $(y_n)$ converges. | [
"a. First, we prove by induction that $0 < x_n < \\frac{3}{2n}$, $\\forall n \\ge 1$.\nThe base case $n = 1$ is trivial. For $n = 2$, we have\n$$\n0 < x_1^2(3 - 2x_1) < 3x_1^2 < \\frac{3}{4} \\Rightarrow 0 < x_2 < \\frac{3}{4}.\n$$\nAssume that $0 < x_k < \\frac{3}{2k}$ for some $k \\ge 2$, by the AM-GM inequality,... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0gkv | Determine a five-digit positive integer $n$ (in base 10) whose digit sum is least and $n^3 - 1$ is divisible by $2556$. | [
"We first show that for $n \\in \\mathbb{N}$, the integer $n^3-1$ is divisible by $2556 = 2^2 \\cdot 3^2 \\cdot 71$ if and only if it is of the form $n = 852k + 1$ ($k \\in \\mathbb{N}$).\n(⇒) If $2556 \\mid (n^3 - 1)$, then\n$$\nn^3 \\equiv 1 \\pmod{2^2 \\cdot 3^2 \\cdot 71} \\quad (1)\n$$\nand so $n^3 \\equiv 1 \... | Thailand | The 10th Thailand Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 21301 | |
0d80 | Determine all positive integers $n \geq 3$ such that we can divide a convex $n$-polygon into triangles by using some diagonals of this polygon such that the number of the used diagonals of every vertex is even. | [
"The answer is $3 \\mid n$.\n\n1) We show that if $n=3k$ then we can divide every convex $n$-polygon (without overlapping) into triangles by using some diagonals of this polygon such that the number of the used diagonals of every vertex is an even integer. The proof is by induction in $k$. The figure shows how to r... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | n is divisible by 3 | |
021n | Problem:
Arquimedes possui uma balança de dois pratos com braços de comprimentos diferentes. Objetivando pesar dois quilos de açúcar, ele procedeu da seguinte forma: colocou um peso de um quilo no prato da esquerda e açúcar no outro lado até que a balança ficasse equilibrada. Em seguida, ele esvaziou os dois pratos, c... | [
"Solution:\n\nDenotando o comprimento do braço da esquerda por $p$, o da direita por $q$ e por $x$ e $y$ as quantidades pesadas de açúcar na primeira e na segunda pesagem, respectivamente. Em virtude do equilíbrio, podemos escrever\n$$\n\\begin{aligned}\n& 1 \\cdot p = x \\cdot q \\\\\n& y \\cdot p = 1 \\cdot q\n\\... | Brazil | NÍVEL 3 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | more than two kilograms | |
0dht | Let $a_1, a_2, \dots, a_n$ be non-zero integers such that
$$
a_1 a_2 \dots a_n \left( \frac{1}{a_1^2} + \frac{1}{a_2^2} + \dots + \frac{1}{a_n^2} \right)
$$
is an integer. Prove that $a_k^2 \mid a_1 a_2 \dots a_n$ for all $k = 1, 2, \dots, n$. | [
"For each $k = 1, 2, \\dots, n$, denote $b_k = \\frac{a_1 a_2 \\dots a_n}{a_k^2}$ and consider the following polynomial\n$$\nP(x) = (x - b_1)(x - b_2) \\dots (x - b_n) = x^n + c_{n-1} x^{n-1} + \\dots + c_1 x + c_0.\n$$\nBased on Vieta's theorem, one can see that $c_{n-1}$ is an integer, and for any $k$ integers $i... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | English | proof only | null | |
0ae9 | Нека $a$ и $b$ се природни броеви. Докажи дека $a^2 + ab + b^2$ е делител на бројот $(a+b)^6 - a^6$. | [
"Ќе ги искористиме идентитетите $A^6 - B^6 = (A^3 - B^3)(A^3 + B^3)$ и $A^3 + B^3 = (A+B)(A^2 - AB + B^2)$. Навистина, добиваме\n$$\n\\begin{aligned} (a+b)^6 - a^6 &= [(a+b)^3 - a^3][(a+b)^3 + a^3] = [(a+b)-a][(a+b)^2 + a(a+b) + a^2] \\\\\n&[(a+b)+a][(a+b)^2 - (a+b)a + a^2] = b(2a+b)(a^2 + ab + b^2)(3a^2 + 3ab + b^... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Macedonian, English | proof only | null | |
0bs8 | Let $a$, $b$, $c \in \mathbb{C}^*$, be distinct complex numbers with equal moduli, such that $a^2 + b^2 + c^2 + ab + ac + bc = 0$. Prove that $a$, $b$, $c$ are the complex coordinates of the vertices of a triangle which is either right angled or equilateral.
Marian Ionescu | [
"As usual, we could suppose that $|a| = |b| = |c| = 1$. The equality in the hypothesis is equivalent with $(a + b + c)^2 = ab + bc + ca$, and thus $(a + b + c)^2 = abc\\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right)$, which is the same with $(a + b + c)^2 = abc(a + b + c)$. It is clear now that $|a + b + c... | Romania | 67th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Algebra > Algebraic E... | English | proof only | null | |
010g | Problem:
We say that an integer $m$ covers the number 1998 if $1,9,9,8$ appear in this order as digits of $m$. (For instance, 1998 is covered by 215993698 but not by 213326798.) Let $k(n)$ be the number of positive integers that cover 1998 and have exactly $n$ digits ($n \geqslant 5$), all different from 0. What is th... | [
"Solution:\n\nLet $1 \\leqslant g < h < i < j \\leqslant n$ be fixed integers. Consider all $n$-digit numbers $a = \\overline{a_{1} a_{2} \\ldots a_{n}}$ with all digits non-zero, such that $a_{g} = 1$, $a_{h} = 9$, $a_{i} = 9$, $a_{j} = 8$ and this quadruple 1998 is the leftmost one in $a$; that is,\n$$\n\\begin{c... | Baltic Way | Baltic Way 1998 | [
"Discrete Mathematics > Combinatorics",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 1 | |
0goy | The incircle of a triangle $ABC$ touches the sides $[BC], [CA], [AB]$ at points $D, E, F$, respectively. The circle passing through point $A$ and touches the line $BC$ at $D$ intersects the line segments $[BF]$ and $[CE]$ at the points $K$ and $L$, respectively. The line passing through $E$ and parallel to $DL$ and the... | [
"Let $M$ be the intersection of the lines $PE$ and $BC$, $N$ be the intersection of the lines $PF$ and $BC$. We will prove that $MD = ND$.\n\nThe power of $B$ with respect to the circumcircle of the triangle $AKD$ gives\n$$\nBK \\cdot BA = BD^2 = BF^2,\ni.e.\\ BK^2 + BK \\cdot KF + BK \\cdot AF = (BK + KF)^2.\n$$\n... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
0294 | Problem:
Seja $ABCD$ um paralelogramo, e $ABF$ e $ADE$ triângulos equiláteros construídos exteriormente ao paralelogramo. Prove que $FCE$ também é equilátero.
 | [
"Solution:\n\nSeja $\\alpha = \\angle ABC$. Como $ABCD$ é um paralelogramo, ele tem ângulos opostos iguais, logo $\\angle ADC = \\alpha$, e ângulos adjacentes suplementares, logo $\\angle BAD = 180^\\circ - \\angle ABC = 180^\\circ - \\alpha$. Portanto conseguimos calcular o ângulo $\\angle FAE$, usando o fato de q... | Brazil | null | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06ur | Determine all functions $f:(0, \infty) \rightarrow \mathbb{R}$ satisfying
$$
\left(x+\frac{1}{x}\right) f(y)=f(x y)+f\left(\frac{y}{x}\right)
$$
for all $x, y>0$. | [
"Answer: $f(x)=C_{1} x+\\frac{C_{2}}{x}$ with arbitrary constants $C_{1}$ and $C_{2}$.\n\nSolution 1. Fix a real number $a>1$, and take a new variable $t$. For the values $f(t), f\\left(t^{2}\\right)$, $f(a t)$ and $f\\left(a^{2} t^{2}\\right)$, the relation (1) provides a system of linear equations:\n$$\n\\begin{a... | IMO | IMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | f(x) = C1*x + C2/x for arbitrary real constants C1 and C2 | |
0csv | Трапеция $ABCD$ с основаниями $AB$ и $CD$ вписана в окружность $\Omega$. Окружность $\omega$ проходит через точки $C, D$ и пересекает отрезки $CA, CB$ в точках $A_1, B_1$ соответственно. Точки $A_2$ и $B_2$ симметричны точкам $A_1$ и $B_1$ относительно середин отрезков $CA$ и $CB$ соответственно. Докажите, что точки $A... | [
"**Первое решение.** Утверждение задачи эквивалентно равенству $CA_2 \\cdot CA = CB_2 \\cdot CB$. Поскольку $AA_1 = CA_2$ и $BB_1 = CB_2$, достаточно доказать, что $AA_1 \\cdot AC = BB_1 \\cdot BC$.\nПусть $D_1$ — вторая точка пересечения $\\omega$ с $AD$ (см. рис. 9). Из симметрии имеем $AD = BC$ и $AD_1 = BB_1$. ... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
08ah | Problem:
Tre amici entrano nella pizzeria di Giorgio e siedono ciascuno a un lato di un tavolo rettangolare; il primo è seduto a un lato di lunghezza $70~\mathrm{cm}$, il secondo e il terzo siedono uno di fronte all'altro, su lati di lunghezza $l$. Le pizze hanno un diametro di $30~\mathrm{cm}$; Giorgio serve la pizza... | [
"Solution:\n\nLa risposta è (D). Consideriamo un tavolo rettangolare in cui la lunghezza $l$ sia la minima tale che le tre pizze sono interamente contenute sulla sua superficie, e cioè tale che la seconda e la terza pizza siano tangenti al lato al quale non siede nessuno dei tre amici. Chiamiamo $O_{1}, O_{2}$ i ce... | Italy | Progetto Olimpiadi della Matematica - Gara di Febbraio | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | MCQ | D | |
05sa | Problem:
Soit $n \geqslant 2$ un entier. Clara dispose d'un plateau de taille $3 n \times 3 n$, semblable à un échiquier. Elle vient d'inventer une nouvelle pièce, le léopard, qu'elle peut mouvoir comme suit : en le déplaçant d'une case vers le haut, d'une case vers la droite, ou bien d'une case en diagonale, vers le ... | [
"Solution:\n\nTout d'abord, Clara peut effectuer $9 n^2 - 3$ déplacements en procédant comme illustré ci-dessous dans le cas où $n = 3$.\n\n\n\nMontrons maintenant qu'elle ne peut pas mieux faire. Tout d'abord, on identifie chaque case en ligne $i$ (en partant du bas) et en colonne $j$ (en ... | France | Préparation Olympique Française de Mathématiques - Test du 15 Mai 2019 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 9n^2 - 3 | |
06m4 | If $x^2 + y^2 = \frac{3961xy}{1980}$ where $x > y > 0$, find the value of $\frac{x+y}{x-y}$. | [
"Answer: 89\n\nWe have\n$$\n\\left(\\frac{x+y}{x-y}\\right)^2 = \\frac{x^2+y^2+2xy}{x^2+y^2-2xy} = \\frac{\\left(\\frac{3961}{1980}+2\\right)xy}{\\left(\\frac{3961}{1980}-2\\right)xy} = 7921 = 89^2\n$$\nand so the answer is 89. (Note that $\\frac{x+y}{x-y} > 0$ as $x > y > 0$.)"
] | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 89 | |
00ro | Find all natural numbers $n$ for which $1^{\phi(n)} + 2^{\phi(n)} + \dots + n^{\phi(n)}$ is coprime with $n$. | [
"Consider the given expression (mod $p$) where $p \\mid n$ is a prime number. $p \\mid n \\Rightarrow p-1 \\mid \\phi(n)$, thus for any $k$ that is not divisible by $p$, one has $k^{\\phi(n)} \\equiv 1 \\pmod p$. There are $n - \\frac{n}{p}$ numbers among $1, 2, \\dots, n$ that are not divisible by $p$. Therefore\n... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All square-free natural numbers | |
0kke | Problem:
Triangle $A B C$ has a right angle at $C$, and $D$ is the foot of the altitude from $C$ to $A B$. Points $L$, $M$, and $N$ are the midpoints of segments $A D$, $D C$, and $C A$, respectively. If $C L = 7$ and $B M = 12$, compute $B N^{2}$. | [
"Solution:\n\nNote that $C L$, $B M$, and $B N$ are corresponding segments in the similar triangles $\\triangle A C D \\sim \\triangle C B D \\sim \\triangle A B C$. So, we have\n$$\nC L : B M : B N = A D : C D : A C\n$$\nSince $A D^{2} + C D^{2} = A C^{2}$, we also have $C L^{2} + B M^{2} = B N^{2}$, giving an ans... | United States | HMMT Spring 2021 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 193 | |
06gw | Find a positive integer $m$ such that $x^3 \equiv 1 \pmod m$ has more than 2012 solutions. (Note: If $x \equiv y \pmod m$, then $x$ and $y$ are considered as the same solution.) | [
"A possible $m$ is $5201836549$.\nLet $p_1, p_2, \\dots, p_7$ be the primes $7, 13, 19, 31, 37, 43, 61$ which are congruent to $1$ modulo $3$, and let $m = p_1p_2\\cdots p_7$. Since $3 \\mid p_j - 1$, the equation $x^3 \\equiv 1 \\pmod{p_j}$ has three distinct solutions modulo $p_j$ by the cubing lemma (or primitiv... | Hong Kong | IMO HK TST | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | 5201836549 | |
0gd5 | 給定凸五邊形 $ABCDE$, 令點 $A_1$ 為直線 $BD$ 與 $CE$ 的交點, 點 $B_1$ 為直線 $CE$ 與 $DA$ 的交點, 並依此類推定義 $C_1, D_1, E_1$ 等點。又令點 $A_2$ 為三角形 $ABD_1$ 的外接圓與三角形 $AEC_1$ 的外接圓的另一個交點, 點 $B_2$ 為三角形 $BCE_1$ 的外接圓與三角形 $BAD_1$ 的外接圓的另一個交點, 並依此類推定義 $C_2, D_2, E_2$ 等點。證明直線 $AA_2, BB_2, CC_2, DD_2, EE_2$ 共點。 | [
"Perform an inversion with center $A$ and denote inverse point with prime.\n\n\n\nLet $P \\equiv AC'_1 \\cap \\odot(AB'B'_2)$, $Q \\equiv AD'_1 \\cap \\odot(AE'E'_2)$, $R \\equiv B'D'_1 \\cap \\odot(AB'D')$, $S \\equiv E'C'_1 \\cap \\odot(AC'E')$. Clearly $E'_1 \\in \\odot(AB'D')$, $B'_2 \\... | Taiwan | 二〇一九數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0765 | Problem:
Let $n$ be a natural number. Prove that
$$
\left[\frac{n}{1}\right]+\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\cdots+\left[\frac{n}{n}\right]+[\sqrt{n}]
$$
is even. (Here $[x]$ denotes the largest integer smaller than or equal to $x$.) | [
"Solution:\nLet $f(n)$ denote the given expression. Then $f(1)=2$ which is even. Now suppose that $f(n)$ is even for some $n \\geq 1$. Then\n$$\n\\begin{aligned}\nf(n+1) & =\\left[\\frac{n+1}{1}\\right]+\\left[\\frac{n+1}{2}\\right]+\\left[\\frac{n+1}{3}\\right]+\\cdots+\\left[\\frac{n+1}{n+1}\\right]+[\\sqrt{n+1}]... | India | Indian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof only | null | |
0gov | Let the line segment $[AB]$ be a chord of the circle $\Gamma$ not passing through the center of it and $M$ be the midpoint of $[AB]$. Let $C$ be a variable point on $\Gamma$ different from $A$ and $B$, and let $P$ be the point where the tangent line to the circumcircle of the triangle $CAM$ at the point $A$ meets the t... | [
"Let $Q$ be the point where the tangent lines to $\\Gamma$ at $A$ and $B$ meet. We will show that $Q$ is the point.\nSince $\\angle MCA = \\angle MAP$ and $\\angle MCB = \\angle MBP$, we have $\\angle ACB + \\angle APB = 180^\\circ$ and $P$ lies on $\\Gamma$. Therefore, if $P'$ is the point where $QC$ intersect the... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ioq | Problem:
Find all positive integers $n$ such that
$$
\sum_{k=1}^{n} \phi(k)=\frac{3 n^{2}+5}{8}
$$ | [
"Solution:\nAnswer: $1$, $3$, $5$.\n\nWe contend that the proper relation is\n$$\n\\sum_{k=1}^{n} \\phi(k) \\leq \\frac{3 n^{2}+5}{8}\n$$\nLet $\\Phi(k)$ denote the left hand side of $(*)$. It is trivial to see that for $n \\leq 7$ the posed inequality holds, has equality where $n=1,3,5$, and holds strictly for $n=... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1, 3, 5 | |
0ef2 | Problem:
a) Reši neenačbo $x^{4}-2 x^{3}-4 x^{2}>6-5 x$.
b) Reši enačbo $x^{6}-5 x^{3}=14$. | [
"Solution:\n\na) Uredimo neenačbo $x^{4}-2 x^{3}-4 x^{2}+5 x-6>0$. Rešimo enačbo $x^{4}-2 x^{3}-4 x^{2}+5 x-6=0$ in zapišemo realni rešitvi enačbe $x_{1}=-2$ in $x_{2}=3$. Zapišemo rešitev neenačbe $x \\in (-\\infty,-2) \\cup (3, \\infty)$.\n\nb) Uredimo enačbo $x^{6}-5 x^{3}-14=0$ in jo razcepimo $\\left(x^{3}-7\\... | Slovenia | 17. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Odbirno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | a) (-∞, -2) ∪ (3, ∞)
b) x = ∛7 or x = -∛2 | |
0blj | Triangle $ABC$ is isosceles with basis $[BC]$ and points $E, G \in (AC)$, $F \in (BC)$, $H \in (AB)$ are such that $(BE$ is the bisector of angle $ABC$, $(EF$ is the bisector of angle $BEC$, $(FG$ is the bisector of angle $EFC$ and $(GH$ is the bisector of angle $EGF$. It is known that $GH \parallel BC$.
a) Prove that... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | ∠A = 90°, ∠B = 45°, ∠C = 45° | |
04vp | Martin is arranging $n^2$ equal dice in the shape of an $n \times n$ square such that whenever two faces touch, the same number is on both of them. What is the maximum number of different numbers that can appear on the top faces of the dice? (Martin Panák)
 | [] | Czech Republic | First Round (take-home) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Solid Geometry > 3D Shapes"
] | English | proof and answer | 1 if n = 1; 4 if n ≥ 2 | |
0aib | In each vertex of a regular $n$-gon $A_1, A_2, \ldots, A_n$ there is a unique pawn. In each step it is allowed:
1. to move all pawns one step in the clockwise direction or
2. to swap the pawns at vertices $A_1$ and $A_2$.
Prove that by a finite series of such steps it is possible to swap the pawns at vertices:
a) $A_i$... | [
"We denote a pawn that was initially at point $A_i$ as $i$. We will prove that a) and then use it to show part b).\n\na) We apply first operation $i-1$ times which will bring $i$ and $i+1$ as they are at points $A_1$ and $A_2$ and move every other pawn $i-1$ steps in clockwise direction.\nWe can now apply second op... | North Macedonia | European Mathematical Cup | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0ehr | Problem:
Izračunaj vrednost izraza $\left(-4 x y z^{-1}\right)^{-2}$, če so $x$, $y$ in $z$ neznanke v sistemu
$$
\begin{aligned}
& \frac{2}{x}-\frac{1}{y}+\frac{1}{z}=-3 \\
& \frac{1}{x}-\frac{1}{y}-\frac{1}{z}=3 \\
& \frac{3}{x}+\frac{2}{y}+\frac{2}{z}=4
\end{aligned}
$$ | [
"Solution:\n\nUvedemo nove neznanke $a=\\frac{1}{x}$, $b=\\frac{1}{y}$, $c=\\frac{1}{z}$ in dobimo sistem enačb $2a-b-c=-3$, $a-b-c=3$, $3a+2b+2c=4$, ki ga rešimo. Rešitev sistema je $a=2$, $b=3$, $c=-4$ in dobimo $x=\\frac{1}{2}$, $y=\\frac{1}{3}$, $z=-\\frac{1}{4}$.\n\nIzračunamo\n$$\n\\left(-4 x y z^{-1}\\right)... | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških i strokovnih šol, Odbirno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 9/64 | |
0k7t | Problem:
Katie has a fair 2019-sided die with sides labeled $1, 2, \ldots, 2019$. After each roll, she replaces her $n$-sided die with an $(n+1)$-sided die having the $n$ sides of her previous die and an additional side with the number she just rolled. What is the probability that Katie's $2019^{\text{th}}$ roll is a ... | [
"Solution:\n\nSince Katie's original die is fair, the problem is perfectly symmetric. So on the 2019th roll, each number is equally probable as any other. Therefore, the probability of rolling a 2019 is just $\\frac{1}{2019}$."
] | United States | HMMT November 2019 | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 1/2019 | |
0l5v | Problem:
Let $f$ be a function from nonnegative integers to nonnegative integers such that $f(0) = 0$ and
$$
f(m) = f\left(\left\lfloor \frac{m}{2} \right\rfloor\right) + \left\lceil \frac{m}{2} \right\rceil^2
$$
for all positive integers $m$. Compute
$$
\frac{f(1)}{1 \cdot 2} + \frac{f(2)}{2 \cdot 3} + \frac{f(3)}... | [
"Solution:\nFor all positive integers $n$, let $\\omega(n) = f(n) - f(n - 1)$. We claim that $\\omega(n)$ is the largest odd divisor of $n$ for all $n > 0$. Indeed, for all positive integers $k$, we have\n$$\n \\omega(2k) = f(2k) - f(2k - 1) = f(k) + k^2 - (f(k - 1) + k^2) = f(k) - f(k - 1) = \\omega(k)\n$$\nand\n... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization"
] | null | final answer only | 341/32 | |
0lcb | Let $ABCD$ be a cyclic quadrilateral. Let $M$, $N$ respectively be the midpoints of the segment $AC$, $BD$; $E$, $F$ respectively be the intersection points of $AB$ and $CD$, $AD$ and $BC$. Prove that $\frac{2MN}{EF} = \left|\frac{AC}{BD} - \frac{BD}{AC}\right|$. | [] | Vietnam | Vietnamese Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
075b | Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D$ be a point on the segment $BC$ such that $BD = 2DC$. Let $P$ be a point on the segment $AD$ such that $\angle BAC = \angle BPD$. Prove that $\angle BAC = 2\angle DPC$. | [
"Extend $AD$ to $E$ such that $PE = PB$. Join $EB$ and $EC$.\n\n\n\n$$\n\\angle BPE = \\angle BAC \\text{ and } \\frac{PB}{PE} = \\frac{AB}{AC} = 1.\n$$\n\nHence it follows that $CAB$ is similar to $EFB$. Thus $\\angle PEB$. This shows that $A$, $B$, $E$, $C$ are concyclic. In turn we obtai... | India | Indija TS 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06xm | Elisa has 2023 treasure chests, all of which are unlocked and empty at first. Each day, Elisa adds a new gem to one of the unlocked chests of her choice, and afterwards, a fairy acts according to the following rules:
- if more than one chests are unlocked, it locks one of them, or
- if there is only one unlocked chest,... | [
"We will prove that such a constant $C$ exists when there are $n$ chests for $n$ an odd positive integer. In fact we can take $C = n-1$. Elisa's strategy is simple: place a gem in the chest with the fewest gems (in case there are more than one such chests, pick one arbitrarily).\n\nFor each integer $t \\geqslant 0$... | IMO | International Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
012k | Problem:
Does there exist an infinite non-constant arithmetic progression, each term of which is of the form $a^{b}$, where $a$ and $b$ are positive integers with $b \geqslant 2$? | [
"Solution:\n\nFor an arithmetic progression $a_{1}, a_{2}, \\ldots$ with difference $d$ the following holds:\n$$\n\\begin{aligned}\nS_{n} & =\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\ldots+\\frac{1}{a_{n+1}}=\\frac{1}{a_{1}}+\\frac{1}{a_{1}+d}+\\ldots+\\frac{1}{a_{1}+n d} \\geqslant \\\\\n& \\geqslant \\frac{1}{m}\\left(... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | proof and answer | No | |
01zy | Problem:
An ordered pair of integers $(m, n)$ with $1 < m < n$ is said to be a Benelux couple if the following two conditions hold: $m$ has the same prime divisors as $n$, and $m+1$ has the same prime divisors as $n+1$.
a. Find three Benelux couples $(m, n)$ with $m \leqslant 14$.
b. Prove that there exist infinitel... | [
"Solution:\n\na.\nIt is possible to see that $(2, 8)$, $(6, 48)$ and $(14, 224)$ are Benelux couples.\n\nb.\nLet $k \\geqslant 2$ be an integer and $m = 2^{k} - 2$. Define $n = m(m+2) = 2^{k}(2^{k} - 2)$. Since $m$ is even, $m$ and $n$ have the same prime factors. Also, $n+1 = m(m+2) + 1 = (m+1)^{2}$, so $m+1$ and ... | Benelux Mathematical Olympiad | THIRD BENELUX MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Examples: (2, 8), (6, 48), (14, 224). Infinite family: (2^k − 2, 2^k(2^k − 2)) for all integers k ≥ 2. | |
015f | Nonnegative integers $a_1, \dots, a_{100}$ satisfy the inequality
$$
a_1 \cdot (a_1-1) \cdot \dots \cdot (a_1-20) + a_2 \cdot (a_2-1) \cdot \dots \cdot (a_2-20) + \dots + a_{100} \cdot (a_{100}-1) \cdot \dots \cdot (a_{100}-20) \le 100 \cdot 99 \cdot 98 \cdot \dots \cdot 79.
$$
Prove that $a_1 + \dots + a_{100} \le 990... | [
"Consider a function\n$$\nf(x) = \\begin{cases} 0, & x \\in [0, 20] \\\\ x(x-1) \\dots (x-20), & x \\ge 20. \\end{cases}\n$$\nThen for any nonnegative integer $a$ we have the equality $f(a) = a \\cdot (a-1) \\cdots (a-20)$ and we can write the given inequality as follows:\n$$\nf(a_1) + \\cdots + f(a_{100}) \\le 100... | Baltic Way | Baltic Way SHL | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0fbg | Problem:
En un disco metálico se quita un sector circular, de modo que con la parte restante se pueda formar un vaso cónico de volumen máximo. Calcular, en radianes, el ángulo del sector que se quita. | [
"Solution:\n\nSea $r$ el radio del disco y $\\alpha$ la medida en radianes del ángulo del sector. El arco $AB$ de dicho sector tiene una longitud igual a $\\alpha r$.\n\nLa longitud de la circunferencia de la base del cono es $2\\pi r - r\\alpha = r(2\\pi - \\alpha)$, así que el radio $\\rho$ de esta circunferencia... | Spain | OME 11 | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 2π − (2√6/3)π | |
0e5a | Twelve balls are numbered by the numbers $1$, $2$, $3$, $\ldots$, $12$. Each ball is coloured either red or green, so that the following two conditions are satisfied:
a. If two balls marked by different numbers $a$ and $b$ are coloured red and $a+b < 13$, then the ball marked by the number $a+b$ is coloured red, too.
... | [
"Assume that the ball denoted by $1$ is red. If the ball denoted by $2$ is also red, then $1+2=3$, $1+3=4$, $\\ldots$, $1+11=12$ are also red. So in this case all balls are red.\n\nIf the ball denoted by $2$ is green, we have to consider two more cases. If the ball number $3$ is red, then $1+3=4$, $1+4=5$, $\\ldots... | Slovenia | National Math Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 6 |
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