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09hh | Let $a$ and $b$ be real numbers such that $a + b = 1$. Prove the following inequality.
$$
\sqrt{1 + 5a^2} + 5\sqrt{2 + b^2} \ge 9
$$ | [
"By the Cauchy-Schwarz inequality, we have\n$$\n3\\sqrt{1 + 5a^2} = \\sqrt{2^2 + (\\sqrt{5})^2} \\cdot \\sqrt{1 + 5a^2} \\geq 2 + 5a\n$$\nand\n$$\n3\\sqrt{2 + b^2} = \\sqrt{(2\\sqrt{2})^2 + 1} \\cdot \\sqrt{2 + b^2} \\geq 4 + b.\n$$\nHence\n$$\n\\sqrt{1 + 5a^2} + 5\\sqrt{2 + b^2} \\geq \\frac{2 + 5a}{3} + \\frac{5(... | Mongolia | Mongolian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0630 | Problem:
Für neun verschiedene positive ganze Zahlen $d_{1}, d_{2}, \ldots, d_{9}$ betrachten wir das Polynom $P(n)=\left(n+d_{1}\right)\left(n+d_{2}\right) \cdot \ldots \cdot\left(n+d_{9}\right)$.
Man zeige, dass eine ganze Zahl $N$ mit folgender Eigenschaft existiert:
Für alle ganzen Zahlen $n \geq N$ ist die Zahl $... | [
"Solution:\n\nWir können der Aufgabenstellung gemäß $N>0$ und $d_{1}<d_{2}<\\ldots<d_{9}$ annehmen. Dann ist $k=d_{9}-d_{1}=\\max _{1 \\leq q<r \\leq 9}\\left(d_{r}-d_{q}\\right)$. Es sei $z \\in \\mathbb{N}$ so gewählt, dass $2^{z}>k$.\n\nBehauptung: Mit $N=2^{z} \\cdot 3^{z} \\cdot 5^{z} \\cdot 7^{z} \\cdot 11^{z... | Germany | 2. Auswahlklausur | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
034d | Problem:
Let $k$ be the incircle of $\triangle ABC$ with $AC \neq BC$, $I$ be the center of $k$ and let $D$, $E$ and $F$ be the tangent points of $k$ to the sides $AB$, $BC$ and $AC$, respectively.
a) If $S = CI \cap EF$, prove that $\triangle CDI \sim \triangle DSI$.
b) Let $M$ be the second intersection point of $k... | [
"Solution:\n\na) From the right $\\triangle CEI$ we have $EI^2 = SI \\cdot CI = DI^2$. Then we get $\\frac{DI}{SI} = \\frac{CI}{DI}$ and therefore $\\triangle CDI \\sim \\triangle DSI$.\n\n\n\nb) The quadrilateral $DIM G$ is cyclic. Since a) implies that $\\angle ISD = \\angle IDC = \\angle... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations... | null | proof only | null | |
06hy | Let $ABC$ be an equilateral triangle. $P$ is a point inside the triangle such that $\angle CBP = 12^\circ$ and $\angle ACP = 54^\circ$. Find $\angle CAP$. | [
"We have $\\angle CAP = 42^\\circ$.\nConstruct a point $D$ such that $\\angle DAC = \\angle DCA = 72^\\circ$, and $B$ lies inside $\\triangle DCA$. Since\n$$\n\\angle BCD = \\angle ACD - \\angle ACB = 72^\\circ - 60^\\circ = 12^\\circ = \\angle CBP,\n$$\nwe have $DC // BP$. Also, note that $BD$ is the perpendicular... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 42° | |
00an | Agustin and Lucas take turns in marking one cell at a time in a $101 \times 101$ grid. Agustin starts the game. A cell cannot be marked if there are already two marked cells in its row or in its column. The one who cannot move loses. Decide which player has a winning strategy. | [
"We describe a winning strategy for the second player Lucas. It applies to any grid square $n \\times n$ with odd $n \\ge 3$. Call a row or column empty, incomplete or full at a certain moment of the game if it contain respectively 0, 1 or 2 marked cells. The strategy of Lucas has two stages.\n\n**Stage 1:** The fi... | Argentina | Argentine National Olympiad 2016 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Lucas (the second player) has a winning strategy. | |
0hal | Andriy read a big book for a month. He was reading the book according to a schedule: from 1 until 20 April he read in average 20 pages per day, from 6 until 25 April he read in average of 30 pages per day, and from 11 until 30 April he read in average of 40 pages per day. What are maximum and minimum amounts of pages, ... | [
"Let's split April into 5 segments. Let from 1 until 5 April he read in average $x$ pages, from 6 until 10 April – in average $a$ pages, from 11 until 20 April – in average $c$ pages, from 21 until 25 April – in average $b$ pages, from 26 until 30 – in average $y$ pages.\nThen equalities are true:\n$$\n\\begin{gath... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | maximum 1200, minimum 800 | |
08og | Problem:
Let $ABC$ be a triangle with $\measuredangle B = \measuredangle C = 40^{\circ}$. The bisector of the $\measuredangle B$ meets $AC$ at the point $D$. Prove that $\overline{BD} + \overline{DA} = \overline{BC}$. | [
"Solution:\nSince $\\measuredangle BAC = 100^{\\circ}$ and $\\measuredangle BDC = 120^{\\circ}$ we have $\\overline{BD} < \\overline{BC}$. Let $E$ be the point on $\\overline{BC}$ such that $\\overline{BD} = \\overline{BE}$. Then $\\measuredangle DEC = 100^{\\circ}$ and $\\measuredangle EDC = 40^{\\circ}$, hence $\... | JBMO | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
09bz | $AC = BC$ байх адил хажуут $ABC$ гурвалжинд багтсан тойрог $BC$, $CA$, $AB$ талуудыг харгалзан $A'$, $B'$, $C'$ цэгүүдэд шүргэнэ. $AA_1$ хэрчим багтсан тойргийг $D$ цэгт огтлох ба $B'D$ шулуун $AB$ талтай $E$ цэгт огтлолцоно. Тэгвэл $AE = EC'$ гэж батал. | [
"\n\n$AB'$ нь багтсан тойргийн шүргэгч учраас $\\angle AB'E = \\varphi$ гэвэл $\\angle B'A'A = \\varphi$ болно. $\\angle A = \\angle B$ учраас $\\triangle AB'E \\sim \\triangle BAA'$ болно.\n\nЭндээс\n$$\n\\frac{AB'}{AB} = \\frac{AE}{A'B}\n$$\n$AB' = AC' = C'B = \\frac{AB}{2}$ гэдгийг тооцв... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | Mongolian | proof only | null | |
049s | Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = abc$. Prove
$$
a^{5} (bc - 1) + b^{5} (ca - 1) + c^{5} (ab - 1) \geq 54\sqrt{3}.
$$ | [] | Croatia | CroatianCompetitions2011 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
090l | Determine all polynomials $f(x)$ with integer coefficients such that, for any integer $n \ge 2$, the following conditions hold:
$\bullet\ f(n) > 0.$
$\bullet\ f(n)$ divides $n^{f(n)} - 1$. | [
"**Solution:** We will show that $f(x) = (x-1)^m$ provides an answer. For $n \\ge 2$ we have $f(n) = (n-1)^m > 0$, thus the first condition is satisfied.\n\n**Lemma 1.** Let $d, t, x$ be positive integers. If $x-1$ is a multiple of $d^t$, then $x^d-1$ is a multiple of $d^{t+1}$.\n\n**Proof.** Since $x \\equiv 1 \\p... | Japan | The 35th Japanese Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | f(x) = (x - 1)^m for some nonnegative integer m | |
02x2 | Problem:
a) Quantos números de quatro algarismos têm soma de seus algarismos par?
b) Um número com dois dígitos distintos e não nulos é chamado de bonito se o dígito das dezenas é maior do que o dígito das unidades. Quantos números bonitos existem?
c) Quantos números pares de quatro dígitos podemos formar utilizando os... | [
"Solution:\na) Para que a soma seja par, devemos utilizar quatro, dois ou nenhum algarismo par. Eles podem ocupar as seguintes ordens: unidades de milhar, centenas, dezenas ou unidades.\n\ni) No primeiro caso, o zero não pode ocupar a maior ordem e as outras podem ser ocupadas por quaisquer dos 5 números pares, res... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | a) 4500, b) 45, c) 156, d) 53332.8 | |
0eb4 | A deltoid is inscribed into a circle with radius $r$. One of the sides of the deltoid is twice as long as the other. Find the ratio of the area of the deltoid to the area of the circle. | [
"Denote the lengths of the sides of the inscribed deltoid by $a$ and $2a$. Due to symmetry, the longer of the two diagonals passes through the centre of the circle. By Thales' theorem there is a right angle between the sides of length $a$ and $2a$. By Pythagoras' theorem the length of the longer diagonal is equal t... | Slovenia | National Math Olympiad in Slovenia | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 8/(5π) | |
0j6v | Problem:
Alberto, Bernardo, and Carlos are collectively listening to three different songs. Each is simultaneously listening to exactly two songs, and each song is being listened to by exactly two people. In how many ways can this occur? | [
"Solution:\n\nAnswer: $6$\n\nWe have $\\binom{3}{2} = 3$ choices for the songs that Alberto is listening to. Then, Bernardo and Carlos must both be listening to the third song. Thus, there are $2$ choices for the song that Bernardo shares with Alberto. From here, we see that the songs that everyone is listening to ... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 6 | |
0fsa | Problem:
Sei $ABCD$ ein gleichschenkliges Trapez mit $AD > BC$. Sei $X$ der Schnittpunkt der Winkelhalbierenden von $\angle BAC$ und $BC$. Sei $E$ der Schnittpunkt von $DB$ mit der Parallelen zu der Winkelhalbierenden von $\angle CBD$ durch $X$ und sei $F$ der Schnittpunkt von $DC$ mit der Parallelen zu der Winkelhalb... | [
"Solution:\n\nDa $ABCD$ ein Sehnenviereck ist, hat man $\\angle ABE = \\angle ACF$.\n\nWenn man beweist, dass die Dreiecke $\\triangle ABE$ und $\\triangle ACF$ ähnlich sind, dann hat man, dass $\\angle AED = \\angle AEB = \\angle AFC = \\angle AFD$, also ist $AEFD$ ein Sehnenviereck. Man muss also nur noch zeigen,... | Switzerland | null | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jcs | Problem:
Triangle $ABC$ satisfies $\angle B > \angle C$. Let $M$ be the midpoint of $BC$, and let the perpendicular bisector of $BC$ meet the circumcircle of $\triangle ABC$ at a point $D$ such that points $A, D, C$, and $B$ appear on the circle in that order. Given that $\angle ADM = 68^{\circ}$ and $\angle DAC = 64^... | [
"Solution:\n\nAnswer: $86^{\\circ}$\n\nExtend $DM$ to hit the circumcircle at $E$. Then, note that since $ADEB$ is a cyclic quadrilateral, $\\angle ABE = 180^{\\circ} - \\angle ADE = 180^{\\circ} - \\angle ADM = 180^{\\circ} - 68^{\\circ} = 112^{\\circ}$.\n\nWe also have that $\\angle MEC = \\angle DEC = \\angle DA... | United States | HMMT November | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 86° | |
0atx | Problem:
Let $P$ be a point in the interior of $\triangle ABC$. Extend $AP$, $BP$, and $CP$ to meet $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively. If $\triangle APF$, $\triangle BPD$, and $\triangle CPE$ have equal areas, prove that $P$ is the centroid of $\triangle ABC$. | [
"Solution:\n\nDenote by $(XYZ)$ the area of $\\triangle XYZ$. Let $w = (APF) = (BPD) = (CPE)$, $x = (BPF)$, $y = (CPD)$, and $z = (APE)$.\n\nHaving the same altitude, we get\n$$\n\\frac{BD}{DC} = \\frac{(BAD)}{(CAD)} = \\frac{2w + x}{w + y + z}\n$$\nand\n$$\n\\frac{BD}{DC} = \\frac{(BPD)}{(CPD)} = \\frac{w}{y}\n$$\... | Philippines | 15th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
03g5 | Let $n \ge 4$ be an integer number and $S_n = \{1, 2, 3, \dots, 2^n\}$. Two sets $A, B$ are given, $A \subset S_n, B \subset S_n \setminus S_{n-1}$, such that $|A| = n+1, |B| = 2$. Is it possible $ab-1$ be a perfect cube for any $a \in A, b \in B$?
(Dragomir Grozev) | [
"Answer: NO. Let us argue by contradiction. Arrange the numbers in $A$ and $B$ in increasing order $1 \\le a_1 < a_2 < \\dots < a_{n+1} \\le 2^n$ and $2^{n-1} < b_1 < b_2 \\le 2^n$. Apparently, there exists an index $i \\le n$ such that $a_i < a_{i+1} \\le 2a_i$. We denote:\n\n$$\na_i b_1 - 1 = q_1^3, \\quad a_{i+1... | Bulgaria | TST for BMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | NO | |
0hz3 | Problem:
How many ways are there to cover a $3 \times 8$ rectangle with 12 identical dominoes? | [
"Solution:\n\nTrivially there is 1 way to tile a $3 \\times 0$ rectangle, and it is not hard to see there are 3 ways to tile a $3 \\times 2$. Let $T_{n}$ be the number of tilings of a $3 \\times n$ rectangle, where $n$ is even. From the diagram below we see the recursion $T_{n}=3 T_{n-2}+2\\left(T_{n-4}+T_{n-6}+\\l... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 153 | |
032d | Problem:
Let $f(x) = x^{2} - a x + a^{2} - 4$, where $a$ is a real number. Find all $a$, for which:
a) the equation $f(x) = 0$ has two real roots $x_{1}$ and $x_{2}$ such that $|x_{1}^{3} - x_{2}^{3}| \leq 4$;
b) the inequality $f(x) \geq 0$ holds for all integers $x$. | [
"Solution:\n\na.\nUsing Vieta's formulas we get $x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} = (x_{1} + x_{2})^{2} - x_{1} x_{2} = 4$. Hence $|x_{1}^{3} - x_{2}^{3}| \\leq 4 \\Longleftrightarrow |x_{1} - x_{2}| \\leq 1$. Let $D = 16 - 3 a^{2}$ be the discriminant of $f(x)$. Then $D \\geq 0$ and $|x_{1} - x_{2}| = \\sqrt{D}... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) a ∈ [−4√3/3, −√5] ∪ [√5, 4√3/3].
b) a ∈ (−∞, (−1 − √13)/2] ∪ [(1 + √13)/2, ∞). | |
07mk | Find the least $k$ for which the number $2010$ can be expressed as the sum of the squares of $k$ integers. | [
"The number $2010$ is a multiple of $3$ but not of $9$. The sum of two squares can be a multiple of $3$ only if the two squares involved are both multiples of $3$ – which means they must be multiples of $9$ as well. So $2010$ is not the sum of two squares.\n\nOne can write $2010$ in various ways as the sum of three... | Ireland | Irish Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | proof and answer | 3 | |
0boo | Find all positive integers that have exactly $8$ positive divisors, among which three are primes of the form $a$, $\overline{bc}$ and $\overline{cb}$, given that $a + \overline{bc} + \overline{cb}$ is a perfect square and $a, b, c$ are digits, with $b < c$. | [
"Suppose $x$ is an integer with the given properties. Then $x$ is a multiple of $y = a \\cdot \\overline{bc} \\cdot \\overline{cb}$. Since $y$ has eight divisors, it follows that $x = y$.\n\nThe numbers $\\overline{bc}$ and $\\overline{cb}$ are distinct primes, so $\\overline{bc} \\in \\{13, 17, 37, 79\\}$. The con... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 2015 | |
05f3 | Problem:
Soit $ABC$ un triangle, $\Gamma$ son cercle circonscrit et $\Omega$ un autre cercle passant par les points $A$ et $B$. La droite $(AC)$ coupe le cercle $\Omega$ en un point $D$ et la tangente au cercle $\Gamma$ en $B$ coupe $\Omega$ en un point $E$.
Montrer que les droites $(BC)$ et $(DE)$ sont parallèles.
... | [
"Solution:\n\nLa droite $(BE)$ est tangente au cercle $\\Omega$ en $B$, donc par le théorème de l'angle tangent, $\\widehat{EBA} = \\widehat{BCA}$.\n\nLes points $D$, $A$, $B$ et $E$ sont cocycliques donc $\\widehat{EDA} = 180^\\circ - \\widehat{EBA}$.\n\nOn déduit que $\\widehat{EDA} = 180^\\circ - \\widehat{BCA}$... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0f3r | Problem:
Are there any solutions in positive integers to $a^4 = b^3 + c^2$? | [
"Solution:\n\nWe have $b^3 = (a^2 - c)(a^2 + c)$, so one possibility is that $a^2 \\pm c$ are both cubes. So we want two cubes whose sum is twice a square. Looking at the small cubes, we soon find $8 + 64 = 2 \\cdot 36$ giving $6^4 = 28^2 + 8^3$. Multiplying through by $k^{12}$ gives an infinite family of solutions... | Soviet Union | ASU | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0hu3 | Problem:
Prove that, for positive integers $n$ and $m$,
$$
\operatorname{gcd}\left(2^{m}-1,2^{n}-1\right)=2^{\operatorname{gcd}(m, n)}-1
$$ | [
"Solution:\nOur main method of attacking the gcd is the following fact:\n$$\n\\operatorname{gcd}(m, n)=\\operatorname{gcd}(m+k n, n)\n$$\nfor positive integers $n$ and $k$. (Proof: If $d \\mid m$ and $d \\mid n$, then $d \\mid m+k n$; if $d \\mid m+k n$ and $d \\mid n$, then $d \\mid m$. Thus the pairs on both side... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
00dw | In the quadrilateral $ABCD$, whose sides are $AB$, $BC$, $CD$ and $DA$, $\angle ABC = \angle BCD = 150^\circ$, $AB = 18\ \text{cm}$ and $BC = 24\ \text{cm}$. Outside the quadrilateral $ABCD$ we draw the equilateral triangles $APB$, $BQC$ and $CRD$. Then we draw the segments $PQ$ and $QR$ and thus a pentagon $APQRD$ is ... | [
"First, let us notice that since $APB$ and $CDR$ are equilateral, we have $AB = AP$ and $CD = DR$. Hence, the difference between the perimeters of $APQRD$ and $ABCD$ equals\n$$\n(AD + DR + QR + PQ + AP) - (AD + CD + BC + AB) = PQ + QR - BC.\n$$\nBy hypothesis we know that $BC = 24$ and $\\text{per}(APQRD) - \\text{... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 10 | |
012e | Problem:
A spider and a fly are sitting on a cube. The fly wants to maximize the shortest path to the spider along the surface of the cube. Is it necessarily best for the fly to be at the point opposite to the spider? ("Opposite" means "symmetric with respect to the center of the cube".) | [
"Solution:\n\nSuppose that the side of the cube is $1$ and the spider sits at the middle of one of the edges. Then the shortest path to the middle of the opposite edge has length $2$. However, if the fly goes to a point on this edge at distance $s$ from the middle, then the length of the shortest path is\n$$\n\\min... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | No | |
09gw | a) Prove that there are no odd numbers $x, y, z$ such that $xy+1, yz+1, zx+1$ are all perfect squares.
b) Prove that there are infinitely many even numbers $x, y, z$ such that $xy+1, yz+1, zx+1$ are all perfect squares. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
017e | The point $L$ is the internal point of the side $AC$ of the isosceles triangle $ABC$ ($AB = BC$). The circle $\omega$ goes through $B$ and is tangent to $AC$ at $L$. It intersects the line $AB$ at points $B$ and $D$ and the line $BC$ at points $B$ and $E$. Let $M$ be the midpoint of the segment $DE$ and let $N \neq L$ ... | [
"If both $D$ and $E$ lie on the sides of the triangle, or if both of them lie on the extensions of those sides, then the point $N$ lies on the segment $AC$. However, on the segment $AC$ there is only one point $N$ for which $\\frac{AN}{CN} = \\frac{AL}{CL}$. As $L$ is such a point we have a contradiction to $N \\ne... | Baltic Way | BALTIC WAY | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
060s | Problem:
Soient $a, b, c$ trois réels strictement positifs. Montrer que
$$
\frac{a^{4}+1}{b^{3}+b^{2}+b}+\frac{b^{4}+1}{c^{3}+c^{2}+c}+\frac{c^{4}+1}{a^{3}+a^{2}+a} \geqslant 2
$$ | [
"Solution:\n\nOn commence par appliquer l'inégalité arithmético-géométrique aux trois termes de l'énoncé :\n$$\n\\frac{a^{4}+1}{b^{3}+b^{2}+b}+\\frac{b^{4}+1}{c^{3}+c^{2}+c}+\\frac{c^{4}+1}{a^{3}+a^{2}+a} \\geqslant 3 \\sqrt[3]{\\frac{\\left(a^{4}+1\\right)\\left(b^{4}+1\\right)\\left(c^{4}+1\\right)}{\\left(b^{3}+... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof only | null | |
0aa6 | Problem:
Let $a, b, \alpha, \beta$ be real numbers such that $0 \leq a, b \leq 1$, and $0 \leq \alpha, \beta \leq \frac{\pi}{2}$. Show that if
$$
a b \cos (\alpha-\beta) \leq \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}
$$
then
$$
a \cos \alpha+b \sin \beta \leq 1+a b \sin (\beta-\alpha)
$$ | [
"Solution:\nThe condition can be rewritten as\n$$\na b \\cos (\\alpha-\\beta)=a b \\cos \\alpha \\cos \\beta+a b \\sin \\alpha \\sin \\beta \\leq \\sqrt{\\left(1-a^{2}\\right)\\left(1-b^{2}\\right)}\n$$\nSet $x=a \\cos \\alpha$, $y=b \\sin \\beta$, $z=b \\cos \\beta$, $t=a \\sin \\alpha$. We can now rewrite the con... | Nordic Mathematical Olympiad | The 31st Nordic Mathematical Contest | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0l0f | Problem:
Compute the sum of all positive integers $x$ such that $(x-17) \sqrt{x-1} + (x-1) \sqrt{x+15}$ is an integer. | [
"Solution:\nFirst, we prove the following claim.\n\nClaim 1. If integers $a, b, c, d, n$ satisfy $a$ and $c$ are nonzero, $b$ and $d$ are nonnegative, and $a \\sqrt{b} + c \\sqrt{d} = n$, then either $n = 0$ or both $b$ and $d$ are perfect squares.\n\nProof. We know $a \\sqrt{b} = n - c \\sqrt{d}$. Squaring both si... | United States | HMMT November | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 11 | |
0fq4 | Problem:
Determina el máximo valor posible de la expresión
$$
27 a b c + a \sqrt{a^{2} + 2 b c} + b \sqrt{b^{2} + 2 c a} + c \sqrt{c^{2} + 2 a b}
$$
siendo $a, b, c$, números reales positivos tales que $a + b + c = \frac{1}{\sqrt{3}}$. | [
"Solution:\n\nEn primer lugar se observa que cuando $a = b = c = \\frac{1}{3 \\sqrt{3}}$ el valor que toma la expresión es $\\frac{2}{3 \\sqrt{3}}$, lo cual sugiere conjeturar que\n$$\n27 a b c + a \\sqrt{a^{2} + 2 b c} + b \\sqrt{b^{2} + 2 c a} + c \\sqrt{c^{2} + 2 a b} \\leq \\frac{2}{3 \\sqrt{3}}\n$$\nPara proba... | Spain | LIII Olimpiada matemática Española | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 2/(3*sqrt(3)) | |
09ht | What is the maximum number of pawns that can be placed on a $10 \times 10$ chess board in such a way that a knight placed anywhere on the board attacks at most one pawn? (We allow the knight to be placed in a cell with a pawn.) | [] | Mongolia | Round 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 28 | |
03h8 | Problem:
If $A$ and $B$ are fixed points on a given circle not collinear with centre $O$ of the circle, and if $XY$ is a variable diameter, find the locus of $P$ (the intersection of the line through $A$ and $X$ and the line through $B$ and $Y$). | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | The locus is the circle through A and B whose center is the intersection point of the tangents to the given circle at A and at B. | |
02df | A graph has 100 points. Given any four points, there is one joined to the other three. Show that one point must be joined to all 99 other points. What is the smallest number possible of such points (that are joined to all the others)? | [
"Suppose that no point is joined to all the others. Then given any point $X$ we can find $Y$ not joined to $X$. So take arbitrary $A$ and $C$. Then take $B$ not joined to $A$ and $D$ not joined to $C$. Then the four points $A, B, C, D$ do not meet the required condition. Contradiction.\n\nSo find $X_1$ joined to al... | Brazil | III OBM | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 97 | |
0c2r | Let $ABC$ be a triangle, and let $M$ be a point on the side $AC$, different from both $A$ and $C$. The line through $M$ and parallel to $BC$ crosses the side $AB$ at $N$. The segments $BM$ and $CN$ cross at $P$, and the circles $BNP$ and $CMP$ meet again at $Q$. Show that the angles $BAP$ and $CAQ$ are equal. | [
"\n\nInvert from $A$ with radius $\\sqrt{AB \\cdot AM}$, then reflect in the internal bisectrix of the angle $BAC$, to obtain an involution $\\varphi$.\nSince the lines $BC$ and $MN$ are parallel, $AB \\cdot AM = AC \\cdot AN$, so $\\varphi B = M$ and $\\varphi C = N$. The circles $ABM$ and... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | null | proof only | null | |
0eb6 | Let $ABC$ be a right triangle with the right angle at $C$, such that $|BC| = a$ and $|AC| = b$. Let $D$ be a point on the opposite side of the line $AC$ from $B$, such that the triangle $ACD$ is similar to the triangle $ABC$. Let $E$ be a point on the line $CD$, such that $\angle EBC$ is a right angle. Find the area of... | [
"The triangles $ACD$ and $ABC$ are similar, so $\\angle CAD = \\angle BAC$. We notice that the point $E$ does not lie on the same side of the line $BC$ as $A$, so $\\angle BCE = 180^\\circ - \\angle ACB - \\angle DCA = 90^\\circ - \\angle DCA = \\angle CAD = \\angle BAC$. We conclude that the triangles $ABC$ and $C... | Slovenia | National Math Olympiad in Slovenia | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (2ab^4 + 2a^3b^2 + a^5) / (2b(a^2 + b^2)) | |
04vd | Prove the claim: If we choose any four factors of $720$, then one of them divides the product of the other three. | [
"Given the decomposition $720 = 2^4 \\cdot 3^2 \\cdot 5$, the number $720$ has exactly three prime factors: $2$, $3$ and $5$. So, each of its factors is of the form $2^\\alpha \\cdot 3^\\beta \\cdot 5^\\gamma$, where $\\alpha, \\beta, \\gamma$ are non-negative integers (satisfying the inequalities $\\alpha \\le 4$,... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
04cr | Let $a$, $b$ and $c$ be the lengths of sides of a triangle with area $P$. Prove that $P < \frac{1}{6}(a^2 + b^2 + c^2)$. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | English | proof only | null | |
0du2 | Alice starts with some coins in a box. On the $n^{\text{th}}$ minute (for $n = 1, 2, 3, \dots$), she can choose to put an additional coin into the box or do nothing, and then she writes the number $2c^2 - n^2$ on the board where $c$ is the number of coins in the box. Show that for any positive integer $m$, there will e... | [
"Let $c_n$ be the number of coins on the $n^{\\text{th}}$ minute. Consider the ordered pair $(x_n, y_n) = (c_n, n - c_n)$, noting that each minute either $x_n$ or $y_n$ increases by 1. The number written on the $n^{\\text{th}}$ minute is\n$$\n2c_n^2 - n^2 = 2x_n^2 - (x_n + y_n)^2 = x_n^2 - 2x_ny_n - y_n^2.\n$$\nIt ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof only | null | |
0jp2 | Problem:
Are there integers $a, b, c, d$ which satisfy $a^{4}+b^{4}+c^{4}+2016=10 d$? | [
"Solution:\n\nThe answer is no. Look at the equation in base $5$. Observe that $0^{4}=0$, $1^{4}=1=1_{5}$, $2^{4}=16=31_{5}$, $3^{4}=81=311_{5}$, $4^{4}=256=2011_{5}$, so each of $a^{4}, b^{4}, c^{4}$ must end in $0$ or $1$ in base $5$. On the other hand $10 d - 2016$ ends with $4$ in base $5$. This is impossible."... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | No | |
0hfj | Nonzero real numbers $x_1, x_2, \dots, x_n$ satisfy the following conditions:
$$
x_1 - \frac{1}{x_2} = x_2 - \frac{1}{x_3} = \dots = x_{n-1} - \frac{1}{x_n} = x_n - \frac{1}{x_1}.
$$
For which $n$ do the numbers $x_1, x_2, \dots, x_n$ all have to be equal? | [
"Suppose that $x_i = x_{i+1}$ for some $i$ (from now on we denote $x_{n+k} = x_k$.) Then from the equality $x_i - \\frac{1}{x_{i+1}} = x_{i+1} - \\frac{1}{x_{i+2}}$ we get $x_{i+1} = x_{i+2}$, then $x_{i+2} = x_{i+3}$ and so on. So we will get that all numbers are equal.\n\nSuppose now that we don't have equal adja... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | All odd positive integers n | |
0boy | Given a positive real number $t$, determine the sets $A$ of real numbers containing $t$, for which there exists a set $B$ of real numbers depending on $A$, $|B| \ge 4$, such that the elements of the set $AB = \{ab: a \in A, b \in B\}$ form a finite arithmetic progression. | [
"The required sets are $\\{t\\}$, $\\{-t, t\\}$, $\\{0, t\\}$ and $\\{-t, 0, t\\}$. It is readily checked that the elements of the Minkowski product of each of these sets and the set $\\{-1, 0, 1, 2\\}$ form a finite arithmetic progression.\n\nNow, let $A$ and $B$ be sets of real numbers satisfying the conditions i... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | {t}, {-t, t}, {0, t}, {-t, 0, t} | |
02ll | A positive integer is *dapper* if at least one of its multiples begins with $2008$. For example, $7$ is dapper because $200858$ is a multiple of $7$ and begins with $2008$. Observe that $200858 = 28694 \times 7$.
Prove that every positive integer is dapper. | [
"Let $n$ be any positive integer. Choose $k$ to be an integer greater than the number of digits of $n$. So the interval $[2008 \\cdot 10^k, 2009 \\cdot 10^k[$, which contains $10^k > n$ integer consecutive numbers, has a multiple of $n$. So every positive integer $n$ is dapper."
] | Brazil | XXX OBM | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | English | proof only | null | |
0jvw | Problem:
A cylinder with radius $15$ and height $16$ is inscribed in a sphere. Three congruent smaller spheres of radius $x$ are externally tangent to the base of the cylinder, externally tangent to each other, and internally tangent to the large sphere. What is the value of $x$? | [
"Solution:\n\nLet $O$ be the center of the large sphere, and let $O_{1}, O_{2}, O_{3}$ be the centers of the small spheres. Consider $G$, the center of equilateral $\\triangle O_{1} O_{2} O_{3}$. Then if the radii of the small spheres are $r$, we have that $OG = 8 + r$ and $O_{1}O_{2} = O_{2}O_{3} = O_{3}O_{1} = 2r... | United States | HMMT November 2016 | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | (15√37 − 75)/4 | |
0fxa | Problem:
Finde alle natürlichen Zahlen $n>1$, sodass $(n-1)!$ durch $n$ teilbar ist. | [
"Solution:\nWir nennen $n$ gut, falls $n$ ein Teiler von $(n-1)!$ ist. Sei $p$ der kleinste Primteiler von $n$. Dann ist die natürliche Zahl $\\frac{n}{p}$ entweder gleich $1$ oder gleich $p$ oder grösser als $p$. Wir unterscheiden nun diese drei Fälle.\n\n- Ist $n=p$ prim, dann ist jede der Zahlen $1,2, \\ldots,(n... | Switzerland | Vorrundenprüfung | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All n > 1 except primes and n = 4 (equivalently, all composite numbers other than 4). | |
02wz | Problem:
a) Considere um primo $p$ que divide $10^{n}+1$ para algum $n$ inteiro positivo. Por exemplo, $p=7$ divide $10^{3}+1$. Analisando o período principal da representação decimal de $\frac{1}{p}$, verifique que o número de vezes que o dígito $i$ aparece é igual ao número de vezes que o dígito $9-i$ aparece para ca... | [
"Solution:\na) Podemos escrever $10^{n}+1=p \\cdot a$ onde $a$ é um número com não mais que $n$ dígitos na base 10, digamos $a=a_{1} a_{2} \\ldots a_{n}$. Queremos dizer com isso que cada número $a_{i}$ é um dos dígitos de $a$. Mesmo que ele possua estritamente menos que $n$ dígitos, podemos colocar alguns $a_{i}$'... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2 | |
0i5n | Problem:
We call a set of professors and committees on which they serve a university if
(1) given two distinct professors there is one and only one committee on which they both serve,
(2) given any committee, $C$, and any professor, $P$, not on that committee, there is exactly one committee on which $P$ serves and no ... | [
"Solution:\n\nLet $C$ be any committee. Then there exists a professor $P$ not on $C$ (or else there would be no other committees). By axiom 2, $P$ serves on a committee $D$ having no common members with archetype $C$. Each of these committees has at least two members, and for each $Q \\in C, R \\in D$, there exists... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6 | |
0cw4 | 100 distinct positive integers are arranged in a circle. Vasya divided each number by its clockwise neighbor with remainder, and he obtained only two distinct remainders. Petya divided each number by its counter-clockwise neighbor with remainder. Prove that Petya obtained 100 distinct remainders.
(N. Agakhanov, S. Berl... | [
"Пронумеруем числа по часовой стрелке $a_1, a_2, \\dots, a_{100}$ так, чтобы число $a_{100}$ было наименьшим. Тогда остаток от деления $a_{100}$ на $a_1$ будет равен $a = a_{100}$ (ибо $a_1 > a_{100}$), а остаток $b$ от деления $a_{99}$ на $a_{100}$ будет меньше, чем $a_{100}$. Значит, $a > b$ — единственные остатк... | Russia | Regional round | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English; Russian | proof only | null | |
0dzt | Problem:
Izračunaj $\sin 2x$, če je $\sin x - \cos x = \frac{1}{2}$. | [
"Solution:\n\nZapišemo zvezo za $\\sin 2x = 2 \\sin x \\cos x$ in $(\\sin x - \\cos x)^2 = \\sin^2 x - 2 \\sin x \\cos x + \\cos^2 x$,\nkamor vstavimo $\\sin x - \\cos x = \\frac{1}{2}$. Dobimo enačbo $\\left(\\frac{1}{2}\\right)^2 = 1 - 2 \\sin x \\cos x$. Izračunamo\n$2 \\sin x \\cos x = \\frac{3}{4} = \\sin 2x$.... | Slovenia | Državno tekmovanje | [
"Precalculus > Trigonometric functions"
] | null | final answer only | 3/4 | |
0csd | В четырёхугольнике $ABCD$ стороны $AD$ и $BC$ параллельны. Докажите, что если биссектрисы углов $DAC$, $DBC$, $ACB$ и $ADB$ образовали ромб, то $AB = CD$. | [
"Пусть $O$ — точка пересечения диагоналей $AC$ и $BD$ (см. рис. 1). Биссектрисы углов $ADB$ и $DAC$ пересекаются в центре $O_1$ окружности, вписанной в треугольник $AOD$, а биссектрисы углов $ACB$ и $DBC$ — в центре $O_2$ окружности, вписанной в треугольник $BOC$. Значит, точки $O_1$ и $O_2$ лежат на биссектрисах в... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordin... | null | proof only | null | |
03ey | Let $p > q$ be primes, such that $240 \nmid p^4 - q^4$. Find the maximal value of $\frac{q}{p}$.
(Angel Gushev) | [
"$\\square$"
] | Bulgaria | Bulgarian Winter Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof and answer | 5/7 | |
0b6b | Let $ABC$ be a triangle, with $AB \le BC$. Let $M$ be the midpoint of the side $BC$, $N$ be the foot of the angle bisector of $\angle B$, $O$ the meeting point of the straight lines $AM$ and $AB$, and $P$ the meeting point of the straight lines $CO$ and $AB$. Show that $4AP \le AB + BC$. | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
01vd | Let $A_1H_1$, $A_2H_2$, $A_3H_3$ be altitudes and $A_1L_1$, $A_2L_2$, $A_3L_3$ be bisectors of an acute angled triangle $A_1A_2A_3$.
Prove the inequality $S(L_1L_2L_3) \ge S(H_1H_2H_3)$ where $S$ stands for the area of a triangle. | [
"One can prove the following\n**Lemma.** If $\\alpha$, $\\beta$, $\\gamma$ are angles of a triangle then\n$$\n\\cos \\alpha + \\cos \\beta + \\cos \\gamma + \\cos 2\\alpha + \\cos 2\\beta + \\cos 2\\gamma \\ge 0.\n$$\nNow, if $a$, $b$, $c$ are the side lengths of the triangle, $\\alpha$, $\\beta$, $\\gamma$ are cor... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Ge... | English | proof only | null | |
0h9n | In isosceles triangle $ABC$ with the vertex in $B$ there are altitudes $BH$ and $CL$. Point $D$ is such that $BDCH$ is a rectangle. Find the angle $DLH$.
(Bogdan Rublyov) | [
"Let us denote by $O$ the intersection of diagonals of rectangle $HBDC$ (Fig. 3). Since $\\triangle CBL$ has right angle, $BO = LO = DO$, therefore, $HO = LO = DO$. Hence, $\\triangle DHL$ is the right triangle with the hypotenuse $DH$, which yields $\\angle DLH = 90^\\circ$."
] | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 90° | |
0adj | Four boys Andrej, Bojan, Vasko and Goce are collecting post stamps. Andrej has as many post stamps as Bojan and Vasko have together. Goce has five times less post stamps than Andrej, and Bojan has four times more post stamps than Vasko. If they together have 5016 post stamps, how many post stamps each of the four boys ... | [
"Let $A$, $B$, $V$ and $G$ be the numbers of post stamps that Andrej, Bojan, Vasko and Goce have correspondingly. According to the first condition of the problem we obtain $A = B + V$, $A = 5G$, $B = 4V$. Then we have that $A = B + V = 4V + V = 5V$. But $A = 5G$ so we get that $V = G$. Then according to the second ... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Andrej 2280, Bojan 1824, Vasko 456, Goce 456 | |
021l | Problem:
Construímos dois triângulos equiláteros: $ABE$ interno e $BFC$ externo ao quadrado $ABCD$. Prove que os pontos $D$, $E$ e $F$ se localizam na mesma reta. | [] | Brazil | null | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof only | null | |
0een | Problem:
Točke $A(2,-1)$, $B(-1,-5)$ in $C(-3,-13)$ so oglišča trikotnika $ABC$. Izračunaj dolžino težničnice na stranico $BC$. Rezultat naj bo natančen in delno korenjen. Točki $A$ in $B$ prezrcali čez abscisno os. Dobljeni točki označi z $A'$ in $B'$. Nariši štirikotnik $AA'B'B$ in izračunaj njegovo ploščino.
![](a... | [
"Solution:\n\nZapis koordinat razpolovišča stranice $BC$: $R(-2,-9)$\n\nUgotovitev, da je dolžina težiščnice enaka razdalji med točkama $A$ in $R$\n\nIzračun dolžine težiščnice:\n$$\nt_{BC} = d(A, R) = \\sqrt{(-2-2)^2 + (-9+1)^2}\n$$\n\nZapis rešitve:\n$$\nt_{BC} = 4\\sqrt{5}\n$$\n\nPrezrcaljeni točki $A$ in $B$ te... | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | median length = 4*sqrt(5); area = 18 | |
0eye | Problem:
The medians divide a triangle into 6 smaller triangles. 4 of the circles inscribed in the smaller triangles have equal radii. Prove that the original triangle is equilateral. | [
"Solution:\n\nDenote the side lengths by $a$, $b$, $c$ and the corresponding median lengths by $m_a$, $m_b$, $m_c$. The six small triangles all have equal area. [Let the areas be $t_1, \\ldots, t_6$. It is obvious that the adjacent pairs have equal height and equal base, so we have $t_1 = t_2$, $t_3 = t_4$, $t_5 = ... | Soviet Union | 2nd ASU | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Mi... | null | proof only | null | |
0d1j | For a positive integer $n$, find the first decimal of the number:
$$
a_n = \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n}.
$$ | [
"The numbers $a_1 = \\frac{1}{2}$ and $a_2 = \\frac{7}{12}$ have the first decimal 5. The number\n$$\na_3 = \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} = \\frac{37}{60} > 0.6\n$$\nhas the first decimal 6.\n\nWe will prove that for every $n \\ge 3$, we have $0.6 < a_n < 0.7$. Notice that\n$$\na_{n+1} - a_n = \\frac{1... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | The first decimal digit is 5 for n = 1, 2, and 6 for all n ≥ 3. | |
02vj | Problem:
A figura a seguir representa um triângulo $A B C$, dois quadrados $A B D E$ e $A C F G$, ambos construídos sobre os lados do triângulo $A B C$, e outro triângulo $A E G$.
Dizemos que os triângulos $A B C$ e $A E G$, posicionados dessa forma em relação a dois quadrados, são triângulos ... | [
"Solution:\n\na. Como a soma de todos os ângulos ao redor de um ponto é $360^{\\circ}$, temos\n$$\n\\begin{aligned}\n\\angle B A C+\\angle C A G+\\angle G A E+\\angle E A B & =360^{\\circ} \\\\\n\\angle B A C+90^{\\circ}+\\angle G A E+90^{\\circ} & =360^{\\circ} \\\\\n\\angle B A C+\\angle G A E & =180^{\\circ}\n\\... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hpc | Problem:
If $x \geq 4$ is a real number prove that
$$
\sqrt{x}-\sqrt{x-1} \geq \frac{1}{x}
$$ | [
"Solution:\nNotice that\n$$\n\\sqrt{x}-\\sqrt{x-1}=\\frac{(\\sqrt{x}-\\sqrt{x-1}) \\cdot (\\sqrt{x}+\\sqrt{x-1})}{\\sqrt{x}+\\sqrt{x-1}}=\\frac{1}{\\sqrt{x}+\\sqrt{x-1}}\n$$\nNow the required inequality is equivalent to $x \\geq \\sqrt{x}+\\sqrt{x-1}$ or after dividing both sides by $\\sqrt{x}$:\n$$\n\\sqrt{x} \\ge... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0ejz | Problem:
Izračunaj vrednost izraza $0 . \overline{27} \cdot\left(x^{2}+2 y x+y^{2}\right):\left(3 y^{2}-3 x\right)$, če velja $x-y=5$ in $\frac{7 x-7 y}{10}-\left(\frac{2}{2 x+y}\right)^{-1}=5^{0}$. | [
"Solution:\nŠtevilo $0 . \\overline{27}$ zapišemo v obliki okrajšanega ulomka $\\frac{3}{11}$. Enačbo $\\frac{7 x-7 y}{10}-\\left(\\frac{2}{2 x+y}\\right)^{-1}=5^{0}$ preoblikujemo v $\\frac{7 \\cdot(x-y)}{10}-\\frac{2 x+y}{2}=1$. Upoštevamo 1. enačbo $x-y=5$ in dobimo $\\frac{7 \\cdot 5}{10}-\\frac{2 x+y}{2}=1$. T... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških i strokovnih šol Odbirno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | -5/11 | |
0dr7 | Let $0 < a_1 < a_2 < \dots < a_n$ be real numbers. Prove that
$$
\left(\frac{1}{1+a_1} + \frac{1}{1+a_2} + \dots + \frac{1}{1+a_n}\right)^2 \le \frac{1}{a_1} + \frac{1}{a_2-a_1} + \frac{1}{a_3-a_2} + \dots + \frac{1}{a_n-a_{n-1}}.
$$ | [
"By Cauchy-Schwarz inequality,\n$$\n\\text{LHS} \\le \\left( \\frac{1}{a_1} + \\frac{1}{a_2 - a_1} + \\dots + \\frac{1}{a_n - a_{n-1}} \\right) \\left( \\frac{a_1}{(1+a_1)^2} + \\frac{a_2 - a_1}{(1+a_2)^2} + \\dots + \\frac{a_n - a_{n-1}}{(1+a_n)^2} \\right).\n$$\nNote that $\\frac{a_1}{(1+a_1)^2} \\le \\frac{a_1}{... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
05rb | Problem:
On dit qu'un ensemble $B$ d'entiers est un intervalle d'entiers s'il existe des entiers $i \leqslant j$ tels que $B=\{i, i+1, \ldots, j\}$, et on note $\mathcal{J}$ l'ensemble des intervalles d'entiers. Par ailleurs, si $A=\{a_1, \ldots, a_k\}$ est un ensemble d'entiers, avec $a_1<\ldots<a_k$, on pose
$$
f(A)... | [
"Solution:\n\nDans toute cette solution, on notera $E_n$ l'ensemble des parties de $\\{1, \\ldots, n\\}$ et $S_n$ l'ensemble des parties $A$ de $\\{1, \\ldots, n\\}$ telles que $\\{1, n\\} \\subseteq A$. Si $A$ est un ensemble et $k$ un entier, on notera également $A+k$ l'ensemble $\\{x+k \\mid x \\in A\\}$.\n\nPou... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0bn9 | Show that there are positive odd integers $m_1 < m_2 < \dots$ and positive integers $n_1 < n_2 < \dots$ such that $m_k$ and $n_k$ are relatively prime, and $m_k^4 - 2n_k^4$ is a perfect square for each index $k$. | [
"Let $m$ and $n$ be relatively prime positive integers such that $m$ is odd and $m^4 - 2n^4$ is a perfect square, e.g., $m = 3$ and $n = 2$. Write $\\ell^2 = m^4 - 2n^4$, so $\\ell^4 = (m^4 - 2n^4)^2 = (m^4 + 2n^4)^2 - 8m^4n^4$, and $\\ell^4 - 8m^4n^4 - (m^4 + 2n^4)^2 = -16m^4n^4 = -(2mn)^4$. Multiply the latter by... | Romania | 2015 Ninth STARS OF MATHEMATICS Competition | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0jk5 | Problem:
What is the smallest positive integer $n$ which cannot be written in any of the following forms?
- $n=1+2+\cdots+k$ for a positive integer $k$.
- $n=p^{k}$ for a prime number $p$ and integer $k$.
- $n=p+1$ for a prime number $p$.
- $n=p q$ for some distinct prime numbers $p$ and $q$ | [
"Solution:\nAnswer: 40 The first numbers which are neither of the form $p^{k}$ nor $p q$ are $12,18,20,24,28,30,36,40, \\ldots$. Of these $12,18,20,24,30$ are of the form $p+1$ and 28,36 are triangular. Hence the answer is 40 ."
] | United States | HMMT November 2014 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 40 | |
05rr | Problem:
On dit qu'une paire d'entiers $(a, b)$ est chypriote si $a \geqslant b \geqslant 2$, si $a$ et $b$ sont premiers entre eux, et si $a+b$ divise $a^{b}+b^{a}$.
Démontrer qu'il existe une infinité de paires chypriotes distinctes. | [
"Solution:\n\nPosons $k = a - b$. Puisque $a$ et $b$ sont premiers entre eux, on sait que $a \\neq b$, donc que $k \\geqslant 1$. Ainsi,\n$$\na^{b} + b^{a} \\equiv a^{b} + (-a)^{a} \\equiv a^{b}\\left(1 + (-1)^{a} a^{k}\\right) \\pmod{a+b}\n$$\nOr, $a$ est premier avec $b$ donc avec $a+b$ aussi.\nD'après le théorèm... | France | Préparation Olympique Française de Mathématiques | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0cfn | Determine the functions $f : \mathbb{Q} \to \mathbb{Q}$ and $g : \mathbb{Q} \to \mathbb{Q}$ for which
$$
g(f(x+y)) = f(x) + y g(y) - a x, \forall x, y \in \mathbb{Q},
$$
where $a \in \mathbb{Q}^*$ is given. | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = a x for all rational x and g(x) = 0 for all rational x | |
0i0m | Problem:
Prove that any integer greater than or equal to $7$ can be written as a sum of two relatively prime integers, both greater than $1$. (Two integers are relatively prime if they share no common positive divisor other than $1$. For example, $22$ and $15$ are relatively prime, and thus $37=22+15$ represents the n... | [
"Solution:\n\nFirst note that if $d$ divides two integers $a$ and $b$, then $d$ must also divide their difference $a-b$. Therefore, consecutive positive integers are always relatively prime (difference is $1$). Likewise, if $a$ and $b$ are both odd with a difference of $2$ or $4$, then $a$ and $b$ are relatively pr... | United States | 2nd Bay Area Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof only | null | |
06e8 | On a planet there are $3 \times 2005!$ aliens and $2005$ languages. Each pair of aliens communicates with each other in exactly one language. Show that there are $3$ aliens who communicate with each other in one common language. | [
"We use terminologies in graph theory. We shall prove by induction that in a complete graph of $3 \\cdot k!$ vertices, if we colour every edge in one of $k$ colours, then there exists a monochromatic triangle.\n\nFor the base case $k=1$, since there are $3$ vertices but only one colour, the result holds trivially.\... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0h17 | Prove that for any collection $a_1, a_2, \dots, a_{2011}$ of real numbers with $a_{2011} \neq 0$ there exists a function $f: \mathbb{R} \to \mathbb{R}$, such that for any real $x$ we have:
$$
a_1 f(x) + a_2 f(f(x)) + \dots + a_{2011} \underbrace{f(f(f\dots f(x)\dots))}_{2011} = x.
$$ | [
"We will search for the function $f$ in the form $f(x) = kx$ with $k \\neq 0$. Then\n$$\n\\underbrace{f(f(f\\dots f(x)\\dots))}_{n} = k^n x.\n$$\nSo the equality from the problem statement becomes:\n$$\na_1 kx + a_2 k^2 x + \\dots + a_{2011} k^{2011} x = x.\n$$\nWe cancel $x$ and obtain the following equation\n$$\n... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof only | null | |
0dhl | On a $9 \times 9$ board, several cells are shaded in such a way that from any shaded cell you can get to any other shaded cell, visiting only the shaded cells and moving only between cells neighboring with a side. Determine the largest possible perimeter of the shaded region. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 120 | |
04rl | Touching circles $k_1(S_1, r_1)$ and $k_2(S_2, r_2)$ lie in a right-angled triangle $ABC$ with the hypotenuse $AB$ and legs $AC = 4$ and $BC = 3$ in such way, that the sides $AB$, $AC$ are tangent to $k_1$ and the sides $AB$, $BC$ are tangent to $k_2$. Find radii $r_1$ and $r_2$, if $4r_1 = 9r_2$. (Pavel Novotný)
![](... | [
"The hypotenuse $AB$ has length $AB = 5$ with respect to Pythagoras' theorem. Then for angles in the triangle there is $\\cos \\alpha = \\frac{4}{5}$, $\\cos \\beta = \\frac{3}{5}$,\n$$\n\\cot \\frac{\\alpha}{2} = \\sqrt{\\frac{1 + \\cos \\alpha}{1 - \\cos \\alpha}} = 3,\n$$\n$$\n\\cot \\frac{\\beta}{2} = \\sqrt{\\... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | r1 = 45/47, r2 = 20/47 | |
02f3 | The sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = 8$, $a_2 = 18$, $a_{n+2} = a_{n+1} \cdot a_n$.
Find all terms which are perfect squares. | [
"We have $a_n = 2^{b_n} 3^{c_n}$, where $b_{n+2} = b_n + b_{n+1}$, and $c_{n+2} = c_n + c_{n+1}$.\n\n$c_1$ and $c_2$ are even, so all $c_n$ are even.\n\n$b_1$ and $b_2$ are odd, so $b_n$ is even for $n$ multiple of $3$.\n\nWe need both $b_n$ and $c_n$ even for $a_n$ to be a perfect square.\n\nSo the answer is all $... | Brazil | XV OBM | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All terms with indices divisible by 3 (a3, a6, a9, ...). | |
0l5o | Problem:
Compute the unique 5-digit positive integer $abcde$ such that $a \neq 0$, $c \neq 0$, and
$$abcde = (ab + cde)^2.$$ | [
"Solution:\n\nLet $ab = X$ and $cde = Y$. The original equation is equivalent to $1000X + Y = (X + Y)^2$. Taking this modulo 999, we get $(X + Y)^2 \\equiv X + Y \\pmod{999}$. Therefore, $27 \\cdot 37 = 999$ divides $(X + Y)(X + Y - 1)$. Since $\\gcd(X + Y, X + Y - 1) = 1$, each of 27 and 37 can divide at most one ... | United States | HMMT February | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 88209 | |
06g0 | Given $2009$ points in the plane so that no three points lie on a line, show that there exist at least $1344021$ triangles, each of which is formed by any three of the $2009$ points, and each of the triangles does not contain any of the remaining $2006$ points. | [
"Fix one of the points $A$. We label the other points $B_1, B_2, \\dots, B_{2008}$ such that the rays $AB_1, AB_2, \\dots, AB_{2008}$ align in the clockwise direction. Then it is easy to see that among\n$$\n\\triangle AB_1B_2, \\triangle AB_2B_3, \\dots, \\triangle AB_{2007}B_{2008}, \\triangle AB_{2008}B_1,\n$$\na... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
05mi | Problem:
Déterminer tous les polynômes $P$ à coefficients dans $\mathbb{Z}$ tels que, pour tous $p$ premier et $u$ et $v$ dans $\mathbb{Z}$ tels que $p \mid u v-1$ on ait : $p \mid P(u) P(v)-1$. | [
"Solution:\n\nSoit $f$ un tel polynôme, et $n \\geqslant 0$ son degré. Clairement $f$ n'est pas le polynôme nul.\n\nPosons $g(X) = X^{n} f\\left(\\frac{1}{X}\\right)$. Alors $g$ est un polynôme à coefficients entiers.\n\nSoit $x$ un entier non nul, et $p$ un nombre premier avec $p > \\max \\left(|x|, |f(x) g(x) - x... | France | TEST DU GROUPE A ET DES CANDIDATES À L'ÉPREUVE EGMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | P(X) = ± X^n for some nonnegative integer n | |
02uu | Problem:
Um grupo de 10 estudantes participa de uma competição de matemática formada por equipes de 4 estudantes. Sabemos que quaisquer duas das equipes possuem exatamente um estudante em comum.
a) Qual o número máximo de equipes de que um estudante pode participar? Forneça um exemplo de distribuição de 10 alunos ond... | [
"Solution:\n\na) Considere um estudante $A$ que participa do maior número de equipes e digamos que ele esteja em uma equipe com os três estudantes $B, C$ e $D$. Qualquer outra equipe que também tenha $A$ como um de seus membros, deverá conter outros três estudantes que não estão no conjunto $\\{B, C, D\\}$. Como ex... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) 3; for example, one student appears in three teams while the other nine students are split into three disjoint groups of three. b) No. | |
0dxj | Problem:
Poenostavi izraz $ (\sin 2x - 2 \cos x) \frac{\tan x}{1 - \sin^2 x} \cdot (1 + \sin x) $. | [
"Solution:\n\nUporabimo zvezo za $\\sin 2x = 2 \\sin x \\cdot \\cos x$ in nato iz prvega faktorja izpostavimo skupni faktor $2 \\sin x (\\sin x + 1)$. Uredimo drugi faktor $\\frac{\\tan x}{1 - \\sin^2 x} = \\frac{\\frac{\\sin x}{\\cos x}}{(1 - \\sin x)(1 + \\sin x)}$. Izraz krajšamo in dobimo $-2 \\sin x$.\n\nIzraz... | Slovenia | 7. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Precalculus > Trigonometric functions"
] | null | final answer only | -2 sin x | |
055n | An acute angle with vertex $A$ and size $\alpha$ is given on a plane. Points $B_0$ and $B_1$ are chosen on different sides of the angle in such a way that $\angle AB_0B_1 = \beta$. Whenever points $B_0, B_1, \dots, B_{n-1}$ are defined, the next point $B_n$ on side $AB_{n-2}$ is allowed to be defined in such a way that... | [
"Let $B_n$ for some $n > 1$ be definable (Fig. 30). By construction, $A$, $B_{n-2}$, and $B_n$ are collinear, whereby $A$ cannot lie between $B_{n-2}$ and $B_n$. Since $B_{n-1}B_{n-2} = B_{n-1}B_n$ and $B_{n-2} \\neq B_n$, the triangle $B_{n-2}B_{n-1}B_n$ is isosceles, so that $\\angle AB_{n-2}B_{n-1} + \\angle AB_... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Largest n equals (β − 90°)/α + 1 if (β − 90°)/α is a nonnegative integer; otherwise largest n equals ⌊β/α⌋ + 1. | |
04l6 | Determine the product $(1 + \tan 1^\circ)(1 + \tan 2^\circ)\cdots(1 + \tan 45^\circ)$. | [
"Let $P = (1 + \\tan 1^\\circ)(1 + \\tan 2^\\circ)\\cdots(1 + \\tan 45^\\circ)$.\n\nRecall that $1 + \\tan x = \\frac{\\sin x + \\cos x}{\\cos x}$.\nBut we can use the tangent addition formula:\n\n$$\n\\tan(45^\\circ + x) = \\frac{\\tan 45^\\circ + \\tan x}{1 - \\tan 45^\\circ \\tan x} = \\frac{1 + \\tan x}{1 - \\t... | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 2^23 | |
0k3p | Problem:
Give an example of a strictly increasing function $f: \mathbb{R} \rightarrow [0,1]$ with the property that
$$
f(x+y) \leq f(x)+f(y)
$$
for any real numbers $x$ and $y$. | [
"Solution:\nWe claim that\n$$\nf(x) = \\frac{1}{1 + e^{-x}}\n$$\nworks fine. This function is strictly increasing by definition, so all that's left to do is check the inequality\n$$\n\\frac{1}{1 + e^{-(x+y)}} \\leq \\frac{1}{1 + e^{-x}} + \\frac{1}{1 + e^{-y}}\n$$\nLetting $a = e^{-x}$ and $b = e^{-y}$, it's the sa... | United States | Berkeley Math Circle | [
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | f(x) = 1/(1 + e^{-x}) | |
02wn | Problem:
Vinte e cinco garotos e vinte e cinco garotas estão sentados ao redor de uma mesa. Prove que é possível encontrar uma pessoa que tem garotas como vizinhas. | [
"Solution:\n\nSuponha que não haja uma pessoa que possua duas garotas como vizinhas. Marquemos com $-1$ cada garoto e com $+1$ cada garota. Cada pessoa ou possui como vizinhos $(-1,-1)$ ou $(+1,-1)$. Somando todos os 50 pares de números associados aos vizinhos de cada pessoa, contaremos a contribuição de cada um du... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
06mz | There is a set of $n$ 01-sequences of length 200. Every pair of 01-sequences differ at least at 101 positions. (For example, the two 01-sequences of length 6, 111100 and 010001 differ at four positions, 1st, 3rd, 4th and 6th positions, counting from the left.) Is it possible that $n \ge 101$? | [
"No, it is not possible. Indeed, let $n_i$ be the number of 01-sequences in the set with 1 at the $i$th entry, where $1 \\le i \\le 200$. We also let $S$ be the total number of different positions between all pairs of 01-sequences in the set. Since every pair differs at least at 101 positions, we have $S \\ge 101 \... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | No; the maximum possible size is 100. | |
0kk5 | Problem:
Find the smallest positive integer $n$ such that the divisors of $n$ can be partitioned into three sets with equal sums. | [
"Solution:\nI claim the answer is $120$. First, note that $120 = 2^{3} \\cdot 3 \\cdot 5$, so the sum of divisors is $(1+2+4+8)(1+3)(1+5) = 15 \\cdot 4 \\cdot 6 = 360$. Thus, we need to split the divisors into groups summing to $120$. But then we can just take $\\{120\\}$, $\\{20, 40, 60\\}$, $\\{1, 2, 3, 4, 5, 6, ... | United States | HMMT November 2021 | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 120 | |
0b7z | Consider $v, w$ two distinct non-zero complex numbers. Prove that
$$
|zw + \bar{w}| \leq |zv + \bar{v}|,
$$
for any $z \in \mathbb{C}$, $|z| = 1$, if and only if there exists $k \in [-1, 1]$ such that $w = kv$. | [
"If there exists $k \\in [-1, 1]$ such that $w = kv$, the inequality is obvious. Conversely, let $t > 1$ such that $w - tv \\ne 0$. Setting\n$$\nz = \\frac{t\\bar{v} - \\bar{w}}{w - tv},\n$$\nwe have $|z| = 1$ and moreover\n$$\nzw + \\bar{w} = \\frac{t(w\\bar{v} - \\bar{w}v)}{w - tv},\n$$\n$$\nzv + \\bar{v} = \\fra... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0ecs | What is the value of the expression $\sqrt{0.04^3}$?
(A) $\frac{1}{5}$
(B) $\frac{1}{25}$
(C) $\frac{1}{125}$
(D) $0.04$
(E) $0.016$ | [
"We calculate $\\sqrt{0.04^3} = \\sqrt{\\left(\\frac{4}{100}\\right)^3} = \\left(\\sqrt{\\frac{4}{100}}\\right)^3 = \\left(\\frac{2}{10}\\right)^3 = \\left(\\frac{1}{5}\\right)^3 = \\frac{1}{125}$."
] | Slovenia | National Math Olympiad 2015 – First Round | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | C | |
063v | Problem:
Man bestimme die kleinste reelle Konstante $C$ mit folgender Eigenschaft:
Für fünf beliebige positive reelle Zahlen $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$, die nicht unbedingt verschieden sein müssen, lassen sich stets paarweise verschiedene Indizes $i, j, k, l$ finden, so dass
$$
\left|\frac{a_{i}}{a_{j}}-\frac... | [
"Solution:\n\nDer gesuchte Wert ist $C=\\frac{1}{2}$.\n\nZunächst beweisen wir, dass $C \\leq \\frac{1}{2}$ gilt. Dazu nehmen wir oBdA $a_{1} \\leq a_{2} \\leq a_{3} \\leq a_{4} \\leq a_{5}$ an und betrachten die fünf Brüche $\\frac{a_{1}}{a_{2}}, \\frac{a_{3}}{a_{4}}, \\frac{a_{1}}{a_{5}}, \\frac{a_{2}}{a_{3}}, \\... | Germany | 1. Auswahlklausur | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1/2 | |
0dgv | Let $a$, $b$, $c > 0$. Prove that
$$
\frac{a+b}{a^2+b^2} + \frac{b+c}{b^2+c^2} + \frac{c+a}{c^2+a^2} \le \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
095k | Problem:
Să se rezolve, în funcţie de valorile parametrului real $a$, sistemul de ecuaţii
$$
\left\{
\begin{array}{l}
y-4 = a(x-2), \\
\dfrac{2y}{x+|x|} = \sqrt{y}
\end{array}
\right.
$$
în $\mathbb{R} \times \mathbb{R}$. | [
"Solution:\nObservăm că pentru $x \\leq 0$, ecuaţia a doua a sistemului n-are sens.\nMai observăm că perechea ordonată $(2 ; 4)$ este soluţie a sistemului, oricare ar fi valorile reale ale parametrului $a$. Sistemul dat este echivalent cu:\n\n\n\n1. Fie $a=0$, atunci primul sistem al totali... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | For a in [0, 2]: S = {(2, 4)}. For a in (-∞, 0) ∪ {4}: S = {(2, 4), (2 − 4/a, 0)}. For a in (2, 4) ∪ (4, ∞): S = {(2, 4), (2 − 4/a, 0), (a − 2, (a − 2)^2)}. | |
0al6 | Let $Q(x) = a_{2023}x^{2023} + a_{2022}x^{2022} + \dots + a_1x + a_0 \in \mathbb{Z}[x]$. For any odd prime number $p$ define the polynomial $Q_p(x) = a_{2023}x^{p-2} + a_{2022}x^{p-2} + \dots + a_1x + a_0^{p-2}$. It is known that there are infinitely many primes $p$ for which $\frac{Q_p(x)-Q(x)}{p}$ is an integer for a... | [] | North Macedonia | Team Selection Test for IMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof and answer | 1 | |
0erb | Devious Steve either lies for the whole day or tells the truth for the whole day. Which one of the following statements can he never say?
(A) "Yesterday I told the truth"
(B) "Yesterday I lied"
(C) "Today I am telling the truth"
(D) "Today I am lying"
(E) "Tomorrow I shall lie" | [
"Statements A, B and E can be made irrespective of whether Devious Steve is lying or telling the truth. Statement C can also be made whether he is telling the truth or lying. However, if statement D is true, then Steve is a liar, which contradicts the truth of statement D. Similarly if statement D is false, then St... | South Africa | South African Mathematics Olympiad First Round | [
"Discrete Mathematics > Logic"
] | English | MCQ | D | |
0e3r | Find all pairs of positive integers $m$ and $n$, such that $\frac{m^2 n}{m^2 + n}$ is a prime number. | [] | Slovenia | Selection Examinations for the IMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (2,4), (2,12) | |
09gl | In the triangle $ABC$ the points $P$ and $Q$ are chosen on the side $BC$ such that $\angle BAP = \angle ACB$ and $\angle CAQ = \angle ABC$. Let $N$ be a point on the line $AP$ such that $AP = PN$ and $M$ be a point on the line $AQ$ such that $AQ = QM$. The lines $BN$ and $CM$ intersect at the point $L$ and let $K$ be t... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation... | English | proof only | null | |
0eps | The last digit of $2011 \times 2013 \times 2015 - 2010 \times 2012 \times 2014$ is
(A) 0
(B) 1
(C) 2
(D) 4
(E) 5 | [
"The last digit of $2011 \\times 2013 \\times 2015$ is a five, and the last digit of $2010 \\times 2012 \\times 2014$ is zero, so the last digit of the difference will be $5 - 0 = 5$."
] | South Africa | South African Mathematics Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | E | |
0gsa | In an acute triangle $ABC$, let $D$ be the midpoint of $[BC]$ and $P$ be a point on $[AD]$. The interior angle bisectors of $ABP$ and $ACP$ intersect at $Q$. Let $BQ \perp QC$. Prove that $Q \in [AP]$. | [
"By angle chasing, we have\n$$\n\\begin{aligned}\n\\angle BAC + \\angle BPC &= 2\\angle BAC + \\angle ABP + \\angle ACP \\\\\n&= 2(\\angle BAC + \\angle ABQ + \\angle ACQ) = 2\\angle BQC = 180^{\\circ}.\n\\end{aligned}\n$$\nLet $R$ be a point on $[PD]$ such that $PD = DR$. Since $BD = DC$ and $PD = DR$, we conclude... | Turkey | Team Selection Test for EGMO 2019 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03qi | The rule of an "obstacle course" specifies that at the $n$th obstacle a person has to toss a die $n$ times. If the sum of points in these $n$ tosses is bigger than $2^n$, the person is said to have crossed the obstacle.
(1) At most how many obstacles can a person cross?
(2) What is the probability that a person cross... | [
"Since the die is fair, the probability of any of the six numbers appearing is the same.\n\n(1) Since the highest point of a die is $6$, and $6 \\times 4 > 2^4$, $6 \\times 5 < 2^5$, it is impossible that the sum of points appearing in $n$ tosses is bigger than $2^n$ if $n \\ge 5$. This means it is an impossible ev... | China | China Mathematical Competition (Hainan) | [
"Statistics > Probability > Counting Methods > Combinations"
] | English | proof and answer | (1) 4; (2) 100/243 | |
0268 | Problem:
2. Número premiado - Um número de 6 algarismos é "premiado" se a soma de seus primeiros 3 algarismos é igual à soma de seus 3 últimos algarismos. Por exemplo $342531$ é premiado pois $3+4+2=5+3+1$.
a. Qual é o maior e o menor número premiado, com 6 algarismos diferentes?
b. Mostre que a soma de todos os núm... | [
"Solution:\n\n(a) O maior número premiado tem de começar com $98$. Assim o número procurado é da forma: $98abcde$. Por hipótese temos: $9+8+a=b+c+d$. Para que $a$ seja máximo precisamos que $b+c+d$ seja máximo, e isto acontece quando $b=7$, $c=6$ e $d=5$. Neste caso, $a=1$ e consequentemente, o maior número premiad... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Largest: 981765; Smallest: 108234; moreover, the sum of all such six-digit numbers with distinct digits is divisible by 13. | |
0hkg | Problem:
Is the number $\left|2^{3000}-3^{2006}\right|$ bigger or smaller than $\frac{1}{2}$? | [
"Solution:\n\nNotice that $\\left|2^{3000}-3^{2006}\\right|$ is a non-negative integer, so it is either $0$ or bigger than $1/2$. However, this number is not $0$ since $2^{3000} \\neq 3^{2006}$, hence $\\left|2^{3000}-3^{2006}\\right| > \\frac{1}{2}$."
] | United States | Berkeley Math Circle Monthly Contest 3 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | greater than 1/2 |
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