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0ce1 | Let $(a_n)_{n \ge 1}$ be a sequence of non-negative real numbers satisfying
$$
a_{n+1}^2 + a_n a_{n+2} \le a_n + a_{n+2},
$$
for all $n \ge 1$. Prove that the sequence $(a_n)_{n \ge 1}$ is bounded. | [
"To prove boundedness, it is sufficient to show that $a_n \\le 1$ for all $n \\ge 3$. Rewrite the condition in the statement in the equivalent form\n$$\na_{n+1}^2 - 1 \\le (1-a_n)(a_{n+2}-1) = (a_n-1)(1-a_{n+2}) \\quad \\text{for all } n \\ge 1.\n$$\nWe first show that $\\min(a_n, a_{n+1}) \\le 1$ for all $n \\ge 2... | Romania | THE 73rd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS - FIRST SELECTION TEST | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0ekb | Problem:
Največ koliko izmed vsot $x+y$, $x+z$, $x+w$, $y+z$, $y+w$ in $z+w$ je lahko lihih, če so $x$, $y$, $z$ in $w$ naravna števila?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6 | [
"Solution:\n\nOpazimo, da so dane vsote ravno vse možne vsote po 2 izmed števil $x$, $y$, $z$ in $w$. Če nobeno od števil $x$, $y$, $z$ in $w$ ni liho, tudi nobena vsota ni liha. Če je liho natanko 1 izmed števil, so lihe natanko 3 vsote, če sta lihi 2 števili, so lihe 4 vsote, če so liha 3 števila, so lihe 3 vsote... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | C | |
0ezu | Problem:
Circles, each with radius less than $R$, are drawn inside a square side $1000R$. There are no points on different circles a distance $R$ apart. Show that the total area covered by the circles does not exceed $340,000 R^2$. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0130 | Problem:
Every integer is coloured with exactly one of the colours BLUE, GREEN, RED, YELLOW. Can this be done in such a way that if $a, b, c, d$ are not all $0$ and have the same colour, then $3a - 2b \neq 2c - 3d$? | [
"Solution:\n\nA colouring with the required property can be defined as follows. For a non-zero integer $k$ let $k^{*}$ be the integer uniquely defined by $k = 5^{m} \\cdot k^{*}$, where $m$ is a nonnegative integer and $5 \\nmid k^{*}$. We also define $0^{*} = 0$. Two non-zero integers $k_{1}, k_{2}$ receive the sa... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
06ie | Find the largest integer $n$ satisfying the following conditions:
(i) $n^2$ can be expressed as the difference of two consecutive cubes;
(ii) $2n + 79$ is a perfect square. | [
"The answer is $181$.\nLet $n^2 = (m+1)^3 - m^3 = 3m^2 + 3m + 1$. This implies\n$$\n(2n - 1)(2n + 1) = 4n^2 - 1 = 12m^2 + 12m + 3 = 3(2m + 1)^2.\n$$\nAs $(2n - 1, 2n + 1) = (2n - 1, 2) = 1$, one of $2n - 1$ and $2n + 1$ is a square and the other is $3$ times a square.\n\n* If $2n + 1$ is a square, then $3 \\mid 2n ... | Hong Kong | IMO HK TST | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 181 | |
0jdj | Problem:
Let $\omega = \cos \frac{2\pi}{727} + i \sin \frac{2\pi}{727}$. The imaginary part of the complex number
$$
\prod_{k=8}^{13}\left(1+\omega^{3^{k-1}}+\omega^{2 \cdot 3^{k-1}}\right)
$$
is equal to $\sin \alpha$ for some angle $\alpha$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, inclusive. Find $\alpha$. | [
"Solution:\n$727 = 3^6 - 2$. Our product telescopes to\n$$\n\\frac{1-\\omega^{3^{13}}}{1-\\omega^{3^{7}}} = \\frac{1-\\omega^{12}}{1-\\omega^{6}} = 1 + \\omega^{6},\n$$\nwhich has imaginary part $\\sin \\frac{12\\pi}{727}$, giving $\\alpha = \\frac{12\\pi}{727}$."
] | United States | HMMT November | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 12π/727 | |
0fu5 | Problem:
Seien $a$ und $b$ feste positive Zahlen. Finde in Abhängigkeit von $a$ und $b$ den kleinstmöglichen Wert der Summe
$$
\frac{x^{2}}{(a y+b z)(a z+b y)}+\frac{y^{2}}{(a z+b x)(a x+b z)}+\frac{z^{2}}{(a x+b y)(a y+b x)}
$$
wobei $x, y, z$ positive reelle Zahlen sind. | [
"Solution:\n\nSei im Folgenden\n$$\nA=\\frac{x^{2}}{(a y+b z)(a z+b y)}+\\frac{y^{2}}{(a z+b x)(a x+b z)}+\\frac{z^{2}}{(a x+b y)(a y+b x)}\n$$\nDie Vermutung, dass das Minimum von $A$ für $x=y=z$ angenommen wird, ist recht naheliegend. Einsetzen von $x=y=z$ liefert den Wert $\\frac{3}{(a+b)^{2}}$. Wir geben nun dr... | Switzerland | SMO Finalrunde | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 3/(a+b)^2 | |
0bev | Let $ABCD$ be a square. Point $E$ lies in the interior of the angle $\angle CAB$ such that angle $\angle BAE$ is $15^\circ$, and lines $BE$ and $BD$ are perpendicular. Show that $AE = BD$. | [
"Lines $AC$ and $BE$ are parallel, because both are perpendicular to $DB$. Let $F$ be the foot of the perpendicular from $E$ onto $AC$. Notice that $EF = BO = \\frac{BD}{2}$, where $O$ is the centre of the square. The triangle $FAE$ has a right angle at $F$ and has $\\angle FAE$ of $30^\\circ$, hence $AE = 2EF = BD... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0f0r | Problem:
At time $1$, $n$ unit squares of an infinite sheet of paper ruled in squares are painted black, the rest remain white. At time $k + 1$, the color of each square is changed to the color held at time $k$ by a majority of the following three squares: the square itself, its northern neighbour and its eastern neig... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
078w | Let $H$ be the orthocenter of triangle $ABC$. Let $E, F$ be the feet of the $B, C$-altitudes. Let $D, M, N$ be the midpoints of segments $AH, BD, CD$ respectively, and $T$ be the intersection of line $FM$ and $EN$. Suppose $D, E, T$, and $F$ are concyclic. Prove that $DT$ passes through the circumcentre of $ABC$. | [
"Let $O$ be the circumcenter of $(ABC)$ and $J$ be the midpoint of $DO$. Now it is sufficient to prove that $TD$ and $TJ$ coincide. We first prove that $M, N, T, J$ are concyclic.\n$$\n\\angle MJN = \\angle BOC = 2\\angle BAC = \\angle EDF = 180^\\circ - \\angle FTE = \\angle 180^\\circ - \\angle NJM\n$$\nThis prov... | India | EGMO TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g94 | 令 $a$, $b$, $c$, $d$ 為任意實數且滿足 $a + b + c + d = 0$。試證：
$$
1296(a^7 + b^7 + c^7 + d^7)^2 \le 637(a^2 + b^2 + c^2 + d^2)^7.
$$ | [
"由題設的對稱性，不妨設 $a$ 最大，$d$ 最小。依題設知 $a \\ge 0$，$d \\le 0$ 且 $d = -(a+b+c) \\le 0$，即 $a+b+c \\ge 0$。\n令 $S_k = a^k + b^k + c^k + d^k$, 其中 $k$ 為正整數, 則 $S_7 = a^7 + b^7 + c^7 - (a+b+c)^7$.\n因為當 $a = -b$ 或 $b = -c$ 或 $c = -a$ 時, $S_7 = 0$. 故可假設\n$$\nS_7 = (a+b)(b+c)(c+a)[x(a^4 + b^4 + c^4) + y(a^3b + ab^3 + b^3c + bc^3 + c... | Taiwan | 二〇一五數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0e00 | Problem:
Vsota prvih petih členov aritmetičnega zaporedja je enaka $50$, razlika med petim in drugim členom pa je $9$. Izračunaj, kateri člen je sedemkrat tolikšen kot prvi. | [
"Solution:\n\nUporabimo obrazec za vsoto prvih petih členov aritmetičnega zaporedja $s_{5} = \\frac{5}{2} (2 a_{1} + 4 d)$. Upoštevamo, da je $a_{5} - a_{2} = 9$ oziroma $a_{1} + 4d - a_{1} - d = 9$. Izračunamo diferenco $d = 3$. Diferenco vstavimo v obrazec za vsoto $\\frac{5}{2} (2 a_{1} + 4 d) = 50$ in izračunam... | Slovenia | Državno tekmovanje | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 9 | |
081i | Problem:
Un pannello contiene 100 lampadine, disposte in modo da formare un quadrato di 10 righe e 10 colonne. Alcune di esse sono accese, le altre sono spente.
L'impianto elettrico è tale che quando si preme il pulsante corrispondente ad una qualunque delle lampadine, cambiano di stato (cioè, si accendono o si spengo... | [
"Solution:\n\na) È facile verificare che se si preme il pulsante di tutte le lampadine di una riga e di una colonna tutte le lampadine del pannello cambiano di stato un numero pari di volte, con l'unica eccezione proprio della lampadina che si trova all'incrocio della riga e della colonna considerata, che cambia st... | Italy | Gara Nazionale di Matematica | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) From any initial configuration it is possible to make all lamps lit.
b) It is possible if and only if the number of lit lamps in every row and in every column has the same parity across all rows and columns; that is, either all rows and all columns have an even number of lit lamps, or all rows and all columns have a... | |
017p | Find all positive integers $n$ such that the decimal representation of $n^2$ consists of odd digits only. | [
"The only such numbers are $n = 1$ and $n = 3$.\nIf $n$ is even, then so is the last digit of $n^2$. If $n$ is odd and divisible by $5$, then $n = 10k + 5$ for some integer $k \\ge 0$ and the second-to-last digit of $n^2 = (10k + 5)^2 = 100k^2 + 100k + 25$ equals $2$.\nThus we may restrict ourselves to numbers of t... | Baltic Way | BALTIC WAY | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 1 and 3 | |
0jnf | Problem:
Find all ordered 4-tuples of integers $(a, b, c, d)$ (not necessarily distinct) satisfying the following system of equations:
$$
\begin{aligned}
a^{2}-b^{2}-c^{2}-d^{2} & = c-b-2 \\
2 a b & = a-d-32 \\
2 a c & = 28-a-d \\
2 a d & = b+c+31
\end{aligned}
$$ | [
"Solution:\nSolution 1. Subtract the second equation from the third to get $a(c-b+1)=30$. Add the second and third to get $2 a(b+c)=-4-2 d$. Substitute into the fourth to get\n$$\n2 a(2 a d-31)=-4-2 d \\Longleftrightarrow a(31-2 a d)=2+d \\Longleftrightarrow d=\\frac{31 a-2}{2 a^{2}+1}\n$$\nwhich in particular give... | United States | HMMT February 2015 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Abstract Algebra > Other"
] | null | proof and answer | (5,-3,2,3) | |
0elc | Problem:
Določi najmanjšo možno vrednost izraza $\left|25^{m}-36^{n}\right|$, če sta $m$ in $n$ naravni števili. | [
"Solution:\n\nUpoštevanjem $25=5^{2}$ in $36=6^{2}$ lahko izraz razstavimo kot\n$$\n\\left|25^{m}-36^{n}\\right|=\\left|5^{2 m}-6^{2 n}\\right|=\\left|\\left(5^{m}\\right)^{2}-\\left(6^{n}\\right)^{2}\\right|=\\left|\\left(5^{m}+6^{n}\\right)\\left(5^{m}-6^{n}\\right)\\right|=\\left(5^{m}+6^{n}\\right)\\left|\\left... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 11 | |
0eqa | The values of $x$ satisfying the equation $(x - 7)(x + 12) = -48$ are
(A) 7 or 12
(B) -7 or 12
(C) 4 or -9
(D) -4 or 9
(E) -12 or 7 | [
"Multiplying out the equation gives $x^2 + 5x - 84 = -48$, so $x^2 + 5x - 36 = 0$. This factorizes to $(x + 9)(x - 4) = 0$, so $x = -9$ or $x = +4$."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | MCQ | C | |
0iu2 | Problem:
Let $f$ be a differentiable real-valued function defined on the positive real numbers. The tangent lines to the graph of $f$ always meet the $y$-axis 1 unit lower than where they meet the function. If $f(1)=0$, what is $f(2)$? | [
"Solution:\n\nThe tangent line to $f$ at $x$ meets the $y$-axis at $f(x)-1$ for any $x$, so the slope of the tangent line is $f'(x) = \\frac{1}{x}$, and so $f(x) = \\ln(x) + C$ for some $C$. Since $f(1) = 0$, we have $C = 0$, and so $f(x) = \\ln(x)$. Thus $f(2) = \\ln(2)$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof and answer | ln(2) | |
0394 | Consider the function $f(x) = a(|\sin x| + |\cos x|) - 3\sin 2x - 7$, where $a$ is a real parameter.
a) Prove that $f(x) = f(\frac{\pi}{2} - x) = f(\pi + x) = f(\frac{3\pi}{2} - x)$ for every $x$.
b) Find all pairs $(a, n)$, where $n$ is a positive integer, for which the equation $f(x) = 0$ has $2007$ roots in the in... | [
"a)\nWe have\n$$\n\\begin{align*}\nf(x + \\pi) &= a(|\\sin(x + \\pi)| + |\\cos(x + \\pi)|) - 3\\sin(2x + 2\\pi) - 7 \\\\\n&= a(|-\\sin x| + |-\\cos x|) - 3\\sin(2x) - 7 = f(x), \\\\\n\\\\\nf\\left(\\frac{\\pi}{2} - x\\right) &= a\\left(|\\sin\\left(\\frac{\\pi}{2} - x\\right)| + |\\cos\\left(\\frac{\\pi}{2} - x\\ri... | Bulgaria | Spring Mathematical Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | Pairs (a, n): (7, 502), (5√2, 2007), (2√2, 2007) | |
06rq | We say that a function $f: \mathbb{R}^k \rightarrow \mathbb{R}$ is a metapolynomial if, for some positive integers $m$ and $n$, it can be represented in the form
$$
f\left(x_1, \ldots, x_k\right)=\max_{i=1, \ldots, m} \min_{j=1, \ldots, n} P_{i, j}\left(x_1, \ldots, x_k\right)
$$
where $P_{i, j}$ are multivariate polyn... | [
"We use the notation $f(x)=f\\left(x_1, \\ldots, x_k\\right)$ for $x=\\left(x_1, \\ldots, x_k\\right)$ and $[m]=\\{1,2, \\ldots, m\\}$. Observe that if a metapolynomial $f(x)$ admits a representation like the one in the statement for certain positive integers $m$ and $n$, then they can be replaced by any $m' \\geq ... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
01lr | Let $S(n)$ be the sum of all digits in the decimal representation of $n \in \mathbb{N}$. How many numbers $n$, $1 \le n \le 999$, satisfy the following condition $S(n) = \text{GCD}(n, n+6)$? | [] | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 25 | |
0f65 | Problem:
Take a large number of unit squares, each with one edge red, one edge blue, one edge green, and one edge yellow. For which $m$, $n$ can we combine $mn$ squares by placing similarly colored edges together to get an $m \times n$ rectangle with one side entirely red, another entirely blue, another entirely green... | [] | Soviet Union | 18th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Exactly those rectangles where both side lengths have the same parity (both even or both odd). | |
0alp | Problem:
If $36-4 \sqrt{2}-6 \sqrt{3}+12 \sqrt{6}=(a \sqrt{2}+b \sqrt{3}+c)^{2}$, find the value of $a^{2}+b^{2}+c^{2}$.
(a) 12
(c) 14
(b) 5
(d) 6 | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | c | |
0g36 | Problem:
Sei $k$ ein Kreis mit Mittelpunkt $O$. Seien $A, B, C$ und $D$ vier unterschiedliche Punkte auf dem Kreis $k$ in dieser Reihenfolge, sodass $A B$ ein Durchmesser von $k$ ist. Der Umkreis des Dreiecks $C O D$ schneide $A C$ zum zweiten Mal in $P$. Zeige, dass $O P$ und $B D$ parallel sind. | [
"Solution:\n\nLösung 1 (Louis), Winkeljagd: Wir zeigen, dass $\\angle O D B=\\angle P O D$. Dies beweist, dass die Linien parallel sind.\nDa $O$ der Mittelpunkt des Kreises ist, haben wir $O D=O B$ und somit ist das Dreieck $O D B$ gleichschenklig an $O$. Somit ist $\\angle O D B=\\angle O B D$. Weil $O$ auf der St... | Switzerland | Vorrunde | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic... | null | proof only | null | |
0dcr | In triangle $ABC$ $\angle B = 60^{\circ}$, $O$ is the circumcenter, and $L$ is the foot of an angle bisector of angle $B$. The circumcircle of triangle $BOL$ meets the circumcircle of $ABC$ at point $D \neq B$. Prove that $BD \perp AC$. | [] | Saudi Arabia | SAUDI ARABIAN 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 > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Spiral simila... | English | proof only | null | |
02qk | Problem:
Juliana quer dar a cada uma das 26 letras $A, B, C, D, \ldots, W, X, Y, Z$ do alfabeto um valor numérico diferente de zero, de tal modo que $A \times C = B$, $B \times D = C$, $C \times E = D$, e assim por diante, até $X \times Z = Y$.
a) Se Juliana der a $A$ e $B$ os valores 5 e 7, respectivamente, quais se... | [
"Solution:\n\na) Substituindo $A = 5$ e $B = 7$ em $A \\times C = B$, temos $5 \\times C = 7$ e segue que $C = \\frac{7}{5}$. Podemos agora achar $D$ substituindo os valores de $B$ e $D$ em $B \\times D = C$; obtemos $7 \\times D = \\frac{7}{5}$ e então $D = \\frac{1}{5}$. Finalmente, de $C \\times E = D$ temos $\\... | Brazil | Nível 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) C = 7/5, D = 1/5, E = 1/7. b) G = A. c) The total product equals 2010. | |
083y | Problem:
Antonio e Bernardo giocano al seguente gioco: sono date due pile di gettoni, una con $m$ gettoni e l'altra con $n$ gettoni. Ogni giocatore sceglie a turno una delle seguenti mosse:
- prendere un gettone da una delle pile;
- prendere un gettone da ciascuna delle pile;
- spostare un gettone da una pila ad un'al... | [
"Solution:\n\nSe almeno uno tra $m$ ed $n$ è dispari dispari vince Antonio, se sono entrambi pari vince Bernardo.\n\nEvidentemente l'unica configurazione in cui non sono più possibili mosse è quella in cui entrambe le pile sono vuote. In particolare, entrambe le pile avranno in quel momento un numero pari di getton... | Italy | Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Antonio wins if at least one of m or n is odd; Bernardo wins if both m and n are even. | |
0e67 | Problem:
Zapiši vsa naravna števila $x$, za katera je vrednost izraza $6(x-1)-2(3 x+2(x-1))$ večja od vrednosti izraza $5(x+1)-3 x-57$. | [
"Solution:\n\nZapišemo neenačbo $6(x-1)-2(3 x+2(x-1))>5(x+1)-3 x-57$.\n\nOdpravimo oklepaje na levi in desni strani neenačbe. Dobimo $6 x-6-10 x+4>2 x-52$.\n\nIzračunamo $-6 x>-50$ oziroma $x<\\frac{25}{3}$.\n\nZa naravna števila $1,2,3,4,5,6,7$ in $8$ je vrednost prvega izraza večja od vrednosti drugega izraza."
] | Slovenia | Državno tekmovanje | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1, 2, 3, 4, 5, 6, 7, 8 | |
09wu | Problem:
Gegeven is een driehoek $A B C$ met zijn omgeschreven cirkel en met $|A C| < |A B|$. Op de korte boog $A C$ ligt een variabel punt $D$ ongelijk aan $A$. Zij $E$ de spiegeling van $A$ in de binnenbissectrice van $\angle B D C$. Bewijs dat de lijn $D E$ door een vast punt gaat, onafhankelijk van de plek van $D$... | [
"Solution:\n\nZij $M$ het snijpunt van de binnenbissectrice van $\\angle B D C$ met de omgeschreven cirkel van $\\triangle A B C$. Omdat $D$ op de korte boog $A C$ ligt, ligt $M$ op de boog $B C$ waar $A$ niet op ligt. Er geldt $\\angle B D M = \\angle M D C$ omdat $D M$ de binnenbissectrice van $\\angle B D C$ is,... | Netherlands | IMO-selectietoets III | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0k78 | Problem:
Determine the number of subsets $S$ of $\{1,2, \ldots, 1000\}$ that satisfy the following conditions:
- $S$ has 19 elements, and
- the sum of the elements in any non-empty subset of $S$ is not divisible by 20. | [
"Solution:\n\nFirst we prove that each subset must consist of elements that have the same residue mod 20. Let a subset consist of elements $a_{1}, \\ldots, a_{19}$, and consider two lists of partial sums\n$$\n\\begin{aligned}\n& a_{1}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \\ldots, a_{1}+a_{2}+\\cdots+a_{19} \\\\\n& a_{2... | United States | HMMT February 2019 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 8 * C(50, 19) | |
0anb | Problem:
Let $v(X)$ be the sum of elements of a nonempty finite set $X$, where $X$ is a set of numbers. Calculate the sum of all numbers $v(X)$ where $X$ ranges over all nonempty subsets of the set $\{1,2,3, \ldots, 16\}$. | [
"Solution:\nThe answer is $2^{15} \\cdot 8 \\cdot 17$\n\nWe note that each $k \\in \\{1,2,3, \\ldots, 16\\}$ belongs to $2^{15}$ subsets of $\\{1,2,3, \\ldots, 16\\}$. We reason as follows: we can assign 0 or 1 to $k$ according to whether it is not or in a subset of $\\{1,2,3, \\ldots, 16\\}$. As there are 2 choice... | Philippines | AREA STAGE | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 4456448 | |
04y2 | Let $P$ be a non-degenerate polygon with $n$ sides, where $n > 4$. Prove that there exist three distinct vertices $A, B, C$ of $P$ with the following property: If $l_1, l_2, l_3$ are the lengths of the three polygonal chains into which $A, B, C$ break the perimeter of $P$, then there is a triangle with side lengths $l_... | [
"By scaling, we can assume w.l.o.g. that the perimeter of $P$ has length $2$. Let $X_1, \\dots, X_n$ be the vertices of $P$, and let $x_i = |X_i X_{i+1}|$, where $X_{n+1} = X_1$; then $\\sum_{i=1}^n x_i = 2$. Since $P$ is a non-degenerate polygon, we have that $x_i < 1$ for all $i = 1, 2, \\dots, n$. To prove the c... | Czech-Polish-Slovak Mathematical Match | Czech-Polish-Slovak Match | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0anh | Problem:
Find all possible values of $\frac{2 \cdot 3^{-x}-1}{3^{-x}-2}$, as $x$ runs through all real numbers.
(a) $(-\infty, 1 / 2) \cup (2,+\infty)$
(b) $(1 / 2,2)$
(c) $[2,+\infty]$
(d) $(0,+\infty)$ | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | (a) | |
03cx | Let $m$ and $n$ be positive integers and $m < n$. Find the number of all injective functions $f: \{1, 2, \dots, m\} \to \{1, 2, \dots, n\}$ such that for any nonempty subset $A \subseteq \{1, 2, \dots, m\}$, the set of values $f(A)$ is distinct from $A$, i.e. $f(A) \neq A$. (A function $f$ is injective if $f(x) \neq f(... | [
"Answer. $\\frac{(n-1)!}{(n-m-1)!}$. Set $B_i = \\{1, 2, \\dots, i\\}$. By induction on $m$ we prove that for any $n > m$ the number of injective functions $f: B_m \\to B_n$, that satisfy the condition of the problem equals $(n-1)(n-2)\\dots(n-m)$.\n\nFor $m=1$ we have $f(1) \\neq 1$, i.e. there are $n-1$ possible ... | Bulgaria | Bulgaria 2022 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | (n-1)!/(n-m-1)! | |
073h | Problem:
Let $ABC$ be a triangle, $I$ its in-centre; $A_{1}, B_{1}, C_{1}$ be the reflections of $I$ in $BC, CA, AB$ respectively. Suppose the circum-circle of triangle $A_{1}B_{1}C_{1}$ passes through $A$. Prove that $B_{1}, C_{1}, I, I_{1}$ are concyclic, where $I_{1}$ is the in-centre of triangle $A_{1}B_{1}C_{1}$. | [
"\nNote that $IA_{1} = IB_{1} = IC_{1} = 2r$, where $r$ is the in-radius of the triangle $ABC$. Hence $I$ is the circum-centre of the triangle $A_{1}B_{1}C_{1}$.\n\nLet $K$ be the point of intersection of $IB_{1}$ and $AC$. Then $IK = r$, $IA = 2r$ and $\\angle IKA = 90^{\\circ}$. It follow... | India | INMO | [
"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",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
0j73 | Problem:
Let $A=\{1,2,3, \ldots, 9\}$. Find the number of bijective functions $f: A \rightarrow A$ for which there exists at least one $i \in A$ such that
$$
\left|f(i)-f^{-1}(i)\right|>1
$$ | [
"Solution:\nAnswer: 359108\n\nWe count the complement - the number of functions $f$ such that for all $i \\in A$, $\\left|f(i)-f^{-1}(i)\\right| \\leq 1$.\n\nThe condition is equivalent to $|f(f(i))-i| \\leq 1$ for all $i \\in A$. If $f(j)=j$, the inequality is automatically satisfied for $i=j$. Otherwise, if $f(f(... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 359108 | |
0kl6 | Problem:
For positive integers $a$, $b$, $c$, $x$, $y$, $z$ such that $a x y = b y z = c z x$, can $a + b + c + x + y + z$ be prime? | [
"Solution:\n\nDividing by $x y z$, we get $a / x = b / y = c / z$. Let this fraction in lowest terms be $m / n$. We then have\n\n$$\nm + n \\mid a + x,\\ b + y,\\ c + z,\n$$\n\nso $m + n$ is a nontrivial factor of $a + b + c + x + y + z$, so $a + b + c + x + y + z$ is not prime."
] | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
096q | Problem:
Să se demonstreze, că oricare numere reale $a$ şi $b$ satisfac inegalitatea
$$
\sqrt{(a-3)^{2}+b^{2}}+\sqrt{a^{2}+(b-4)^{2}} \geqslant 5
$$
Când are loc egalitatea? | [
"Solution:\nMetoda 1. Inegalitatea din enunţ se scrie astfel: $\\sqrt{(a-3)^{2}+b^{2}} \\geqslant 5-\\sqrt{a^{2}+(b-4)^{2}}$.\nFie $a$ şi $b$ două numere reale arbitrare. Dacă partea dreaptă a inegalităţii este negativă, atunci inegalitatea este adevărată. Dacă ea este nenegativă, se ridică ambele părţi la pătrat:\... | Moldova | A 63-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Equality holds precisely for all real pairs satisfying 4a + 3b = 12 with 0 ≤ a ≤ 3 (equivalently, the point lies on the segment joining the two fixed points). | |
08z8 | There is a convex quadrilateral $ABCD$ and a point $P$ inside such that the lines $AP$ and $AD$ are orthogonal and the lines $BP$ and $CD$ are orthogonal. When $AB = 7$, $AP = 3$, $BP = 6$, $AD = 5$, $CD = 10$, find the area of triangle $ABC$. | [
"We can assume $A$, $B$, $C$, $D$ are in counter-clockwise order. Since the lines $AP$ and $AD$, the lines $BP$ and $CD$ are orthogonal respectively, if we rotate the triangle $APB$ counter-clockwise by $90^\\circ$ then $AP$ and $AD$ are parallel and $BP$ and $CD$ are parallel. The points $P$, $A$, $B$ and $D$, $A$... | Japan | Japan 2022 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 245/6 | |
0atq | Problem:
The sequence $2, 3, 5, 6, 7, 8, 10, 11, \ldots$ is an enumeration of the positive integers which are not perfect squares. What is the 150th term of this sequence? | [] | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 162 | |
08le | Problem:
If the real numbers $x_{1}, x_{2}, \ldots, x_{2008}$ are such that $0 < x_{i} < 1$, for any $i$, show that
$$
1 + \sum_{1 \leq i < j \leq 2008} x_{i} x_{j} > \sum_{i=1}^{2008} x_{i}
$$ | [] | JBMO | 2008 Shortlist JBMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0az7 | Problem:
Given that $\tan x + \cot x = 8$, find the value of $\sqrt{\sec^{2} x + \csc^{2} x - \frac{1}{2} \sec x \csc x}$. | [
"Solution:\nNote that\n$$\n\\tan x + \\cot x = \\frac{\\sin x}{\\cos x} + \\frac{\\cos x}{\\sin x} = \\frac{\\sin^{2} x + \\cos^{2} x}{\\sin x \\cos x} = \\frac{1}{\\sin x \\cos x}\n$$\nThis means that $\\sin x \\cos x = \\frac{1}{8}$.\nNow,\n$$\n\\begin{aligned}\n\\sqrt{\\sec^{2} x + \\csc^{2} x - \\frac{1}{2} \\s... | Philippines | 20th Philippine Mathematical Olympiad | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | 2√15 | |
00uf | Let $\mathbb{R}_{>0}$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that for all positive real numbers $x$ and $y$,
$$
f(x^{2023} + f(x)f(y)) = x^{2023} + yf(x).
$$ | [
"Answer: $f(x) = x$ for all $x \\in \\mathbb{R}_{>0}$. We first show that $f$ is bijective.\n* $f$ is injective since $f(y_1) = f(y_2)$ implies\n$$f(x^{2023} + f(x)f(y_1)) = f(x^{2023} + f(x)f(y_2)) \\implies x^{2023} + y_1f(x) = x^{2023} + y_2f(x),$$ \nhence $y_1 = y_2$.\n* $f$ is surjective since for any positive... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = x | |
02oa | Problem:
Fração radical - Se $\frac{\sqrt{x}}{\sqrt{y}}=5$, quanto é $\frac{x+y}{2 y}$ ?
(a) $\frac{5}{2}$
(b) $3 \sqrt{2}$
(c) $13 y$
(d) $\frac{25 y}{2}$
(e) 13 | [
"Solution:\n\nElevando ao quadrado ambos os membros de $\\frac{\\sqrt{x}}{\\sqrt{y}}=5$, obtemos $\\frac{x}{y}=25$. Assim,\n$$\n\\frac{x+y}{2 y}=\\frac{1}{2} \\times \\frac{x+y}{y}=\\frac{1}{2} \\times\\left(\\frac{x}{y}+\\frac{y}{y}\\right)=\\frac{1}{2} \\times\\left(\\frac{x}{y}+1\\right)=\\frac{1}{2} \\times(25+... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | e | |
0d53 | Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}$ such that
$$
f\left(\frac{x}{y}\right)=f(x)+f(y)-f(x) f(y)
$$
for all $x, y \in \mathbb{R}_{>0}$. Here, $\mathbb{R}_{>0}$ denotes the set of all positive real numbers. | [
"Define $g: \\mathbb{R}_{>0} \\rightarrow \\mathbb{R}$ by $g(x)=1-f(x)$ for all $x>0$. The functional equation can be rewritten $g(x / y)=g(x) g(y)$ for all $x, y>0$.\n\nPutting $y=x=1$, we get $g(1)=g(1)^2$. This means that $g(1)=0$ or $g(1)=1$.\n\n1. Assume $g(1)=0$. For all $x>0$, we have $g(x)^2=g(x / x)=g(1)=0... | Saudi Arabia | SAMC 2015 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English, Arabic | proof and answer | Either f(x) = 0 for all x > 0, or f(x) = 1 for all x > 0. | |
0k2f | Problem:
Let $ABCD$ be a convex quadrilateral so that all of its sides and diagonals have integer lengths. Given that $\angle ABC = \angle ADC = 90^{\circ}$, $AB = BD$, and $CD = 41$, find the length of $BC$.
Proposed by: Anders Olsen | [
"Solution:\n\nLet the midpoint of $AC$ be $O$ which is the center of the circumcircle of $ABCD$. $ADC$ is a right triangle with a leg of length $41$, and $41^{2} = AC^{2} - AD^{2} = (AC - AD)(AC + AD)$. As $AC, AD$ are integers and $41$ is prime, we must have $AC = 840$, $AD = 841$. Let $M$ be the midpoint of $AD$.... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Number Theory > Divisibility / Factorizati... | null | proof and answer | 580 | |
0dem | Let $ABC$ be a non-isosceles triangle with incenter $I$, circumcenter $O$ and a point $D$ on segment $BC$ such that $(BID)$ cuts segment $AB$ at $E \neq B$ and $(CID)$ cuts segment $AC$ at $F \neq C$. Circle $(DEF)$ cuts segments $AB$, $AC$ again at $M$, $N$. Let $P = IB \cap DE$ and $Q = IC \cap DF$. Prove that $EN$, ... | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homo... | null | proof only | null | |
073d | Let $a, b, c$ be non-negative real numbers such that $a+b \le c+1$, $b+c \le a+1$ and $c+a \le b+1$. Prove that
$$
a^2 + b^2 + c^2 \le 2abc + 1.
$$ | [
"Adding the first two, we get $2b \\le 2$ so that $b \\le 1$. Similarly, we get $c \\le 1$ and $a \\le 1$. Put $\\alpha = 1 - a$, $\\beta = 1 - b$ and $\\gamma = 1 - c$. Then $0 \\le \\alpha, \\beta, \\gamma \\le 1$ and\n$$\n\\alpha + \\beta\\gamma, \\quad \\beta + \\gamma \\ge \\alpha, \\quad \\gamma + \\alpha \\g... | India | Indija TS 2007 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0ble | Find four consecutive numbers, knowing that they are obtained by adding a prime number (not necessary the same) to $91$, $109$, $124$ and $148$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | 150, 151, 152, 153 | |
0d57 | What is the maximum number of bishops that can be placed on an $8 \times 8$ chessboard such that at most three bishops lie on any diagonal? | [
"If the chessboard is colored black and white as usual, then any diagonal is a solid color. So we may consider bishops on black and white squares separately.\n\nIn one direction, the lengths of the black diagonals are $2, 4, 6, 8, 6, 4$, and $2$. Each of these can have at most three bishops, except the first and la... | Saudi Arabia | SAMC 2015 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English, Arabic | proof and answer | 38 | |
068o | Find all integer triads $a, b, c$, with $a > 0 > b > c$, and sum $a+b+c=0$ such that the number $N = 2017 - a^3b - b^3c - c^3a$ is a perfect square. | [
"2. First we observe that for every selection of six points, one per each sector, an hexagon (convex or non-convex) is created containing A.\nFrom these six points we can create totally $\\binom{6}{3} = 20$ triangles. We will count how many of them contain A. For any two points lying in two opposite sectors the thi... | Greece | 34th Hellenic Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | a=36, b=-12, c=-24 | |
0erd | A sequence of $2016$ terms is constructed as follows: The first two terms of the sequence are both equal to $3$. Starting from the third term, each subsequent term is the sum of the preceding two terms. Each of the terms of this sequence $3$, $3$, $6$, $9$, $...$ is now divided by $2$ and the remainders are added. What... | [
"The first six terms of the sequence are $3$, $3$, $6$, $9$, $15$, $24$, $...$, which are Odd, Odd, Even, Odd, Odd, Even, $...$, since Odd $+$ Odd $=$ Even and Odd $+$ Even $=$ Odd. The pattern continues in the same way in cycles of length $3$, with two odd numbers and one even number in each cycle. After division ... | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | C | |
0ffu | Problem:
Un círculo tiene el centro sobre el lado $AB$ del cuadrilátero inscriptible $ABCD$. Los otros tres lados son tangentes al círculo. Demostrar que $AD + BC = AB$. | [] | Spain | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cce | Let $n \in \mathbb{N}$, $n \ge 2$. Determine all $z \in \mathbb{C}$ such that:
$$
|z^{n+1} - z^n| \ge |z^{n+1} - 1| + |z^{n+1} - z|
$$ | [
"We notice that $z = 1$ is a solution and $z = 0$ is not a solution, so we consider $z \\in \\mathbb{C} \\setminus \\{0, 1\\}$. Considering $w = \\frac{1}{z}$ and multiplying the given inequality by $\\frac{1}{|z|^{n+1}}$, we have:\n$$\n|1 - w| \\ge |1 - w^{n+1}| + |1 - w^n| \\ge |(1 - w^{n+1}) - (1 - w^n)| = |w|^n... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof and answer | All complex numbers that are nth roots of unity or (n+1)th roots of unity, i.e., U_n ∪ U_{n+1}. | |
0co9 | Let $\omega$ be the circumcircle of a triangle $ABC$. A point $K$ is chosen on the bisector line of angle $BAC$ so that $K$ lies inside triangle $ABC$. Line $CK$ intersects $\omega$ at points $C$ and $M$. A circle $\Omega$ touches line $CM$ at $K$, and intersects segment $AB$ at points $A$ and $P$. Circles $\Omega$ and... | [
"Поскольку $CK$ касается $\\Omega$, имеем $\\angle APK = \\angle AKC$.\n\nПусть прямая $MP$ пересекает вторично окружность $\\omega$ в точке $Q'$. Тогда имеем $\\angle AQ'MP = \\angle AQ'M = \\angle ACM = 180^\\circ - \\angle AKC - \\angle KAC = 180^\\circ - \\angle APK - \\angle PAK = \\angle AKP$.\n\nЗначит, точк... | Russia | Final round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0d7j | For any positive integer $n$, show that there exists a positive integer $m$ such that $n$ divides $2016^{m} + m$. | [
"We generalize the problem to the following problem:\nLet $a$ be a given positive integer. For every natural $n$, there is a positive integer $m$ such that $n$ divides $a^{m} + m$.\n\nIn fact, we proceed by induction on $n$. Obviously this statement holds for $n = 1$. Now assume $n > 1$ and this statement holds for... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | English | proof only | null | |
0dud | Problem:
Naj bosta $E$ in $F$ taki notranji točki na stranici $AB$ pravokotnika $ABCD$, da je $|AE| = |EF|$. Pravokotnica na $AB$ skozi točko $E$ seka diagonalo $AC$ v točki $G$, daljici $FD$ in $BG$ pa se sekata v točki $H$. Dokaži, da imata trikotnika $FBH$ in $GHD$ enaki ploščini. | [
"Solution:\n\nNaj $p_{XYZ}$ označuje ploščino trikotnika $XYZ$. Računajmo\n$$\n\\begin{aligned}\np_{DGH} - p_{HFB} = & \\left(p_{DAF} - p_{DAG} - p_{AEG} - p_{EFG} - p_{GFH}\\right) \\\\\n& - \\left(p_{EBG} - p_{EFG} - p_{GFH}\\right) = \\\\\n= & p_{DAF} - p_{DAG} - p_{AEG} - p_{EBG}\n\\end{aligned}\n$$\nOznačimo z... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
074z | Prove that there are infinitely many positive integers $m$ for which there exist consecutive odd positive integers $p_m, q_m$ ($\equiv p_m + 2$) such that the pairs $(p_m, q_m)$ are all distinct and
$$
p_m^2 + p_m q_m + q_m^2, \quad p_m^2 + m p_m q_m + q_m^2
$$
are both perfect squares. | [
"Consider the relations $p^2 + p q + q^2 = u^2$ and $p^2 + m p q + q^2 = v^2$. Thus $(m - 1) p q = (v - u)(v + u)$. Suppose we choose $m - 1 = r^2$ where $r$ is a positive integer and $v - u = r p$ and $v + u = r q$. Then $u = (r q - r p)/2$ and $4(p^2 + p q + q^2) = (r q - r p)^2$. This leads to\n$$\n(r^2 - 4) p^2... | India | Indija TS 2010 | [
"Number Theory > Diophantine Equations > Pell's equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0c55 | A set $A \subset \{1, 2, 3, \dots, 2019\}$ will be called *prime differences free* if the difference of any two of its elements is not a prime. Find the maximum number of elements of a prime differences free set.
Vasile Pop | [] | Romania | SHORTLISTED PROBLEMS FOR THE 70th NMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 505 | |
0539 | In Wonderland, the government of each country consists of exactly $a$ men and $b$ women, where $a$ and $b$ are fixed natural numbers and $b > 1$. For improving of relationships between countries, all possible working groups consisting of exactly one government member from each country, at least $n$ among whom are women... | [
"Let $r$ be the number of countries in Wonderland. If the minimal number of women in working groups is $n = 0$ then forming a working group means just choosing one government member from each country. Thus there are $(a+b)^r$ different working groups. This number can be prime only if $r = 1$ because $a+b \\ge b > 1... | Estonia | IMO Team Selection Contest | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1 | |
0di2 | Determine whether or not it is possible to partition the set of positive integers in infinite subsets $A_1, A_2, \dots$ such that for every positive integer $k$, the sum of elements of $A_k$ is $k + 2023$.
*Remark*: a partition of a set $X$ is a collection of subsets of $X$ such that every element of $X$ is contained ... | [
"The answer is No. Suppose such partition exists. Then for every positive integer $k$, we have\n$$\nB_k = A_1 \\cup A_2 \\cup \\dots \\cup A_k \\subset \\{1, 2, \\dots, k + 2023\\},\n$$\nsince all elements of $A_i$ are at most $i + 2023$ for every $i \\in \\{1, 2, \\dots, k\\}$ and\n$$\n\\sum_{b \\in B_k} b = \\sum... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | No | |
0kuf | Problem:
Suppose $x$, $y$, and $z$ are real numbers greater than $1$ such that
$$
\begin{aligned}
& x^{\log_{y} z} = 2 \\
& y^{\log_{z} x} = 4, \text{ and } \\
& z^{\log_{x} y} = 8
\end{aligned}
$$
Compute $\log_{x} y$. | [
"Solution:\nTaking $\\log_{2}$ both sides of the first equation gives\n$$\n\\begin{aligned}\n& \\log_{2} x \\log_{y} z = 1 \\\\\n& \\frac{\\log_{2} x \\log_{2} z}{\\log_{2} y} = 1\n\\end{aligned}\n$$\nPerforming similar manipulations on the other two equations, we get\n$$\n\\begin{aligned}\n& \\frac{\\log_{2} x \\l... | United States | HMMT November 2023 | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | sqrt(3) | |
0ld1 | Given a positive integer $k$. Find the number of non-negative integers that do not exceed $10^k$ and satisfy the following conditions
i) $n$ is divisible by $3$,
ii) The digits of $n$ in decimal representation are in the set $\{2, 0, 1, 5\}$. | [
"Denote by $S = \\{2, 0, 1, 5\\}$ and\n$$\nA(n, i) = \\{\\overline{x_n x_{n-1} \\dots x_1} : x_j \\in S \\text{ and } x_1 + \\dots + x_n \\equiv i \\pmod{3}\\}.\n$$\nLet $a_n, b_n$ and $c_n$ be the cardinal number of $A(n, 0)$, $A(n, 1)$ and $A(n, 2)$ respectively. Since a natural number is divisible by $3$ if and ... | Vietnam | VMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | If k is a multiple of three: (4^k + 2) / 3; otherwise: (4^k − 1) / 3. | |
0g4s | Problem:
Let $n$ be a positive integer. Roger has a $(2 n+1) \times (2 n+1)$ square garden. He puts down fences to divide his garden into rectangular plots. He wants to end up with exactly two horizontal $k \times 1$ plots and exactly two vertical $1 \times k$ plots for each even integer $k$ between $1$ and $2 n+1$, as... | [
"Solution:\nConsider the 4 largest plots Roger will fence off. We will prove they will comprise the border of the garden.\n\nConsider a vertical $1 \\times 2k$ piece. Clearly, one of its short (horizontal) edges must touch the border, because otherwise there would be a narrow margin of width smaller than $1$ on eit... | Switzerland | Second round 2023 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^n | |
0er8 | In $\triangle RUG$, $RU = 10$ cm, $UG = 21$ cm, $RG = 17$ cm and $RY$ is perpendicular to $UG$ at point $Y$. The length of $RY$, in cm, is
(A) 8 (B) 5 (C) 7 (D) 6 (E) 8.1 | [
"Let $h = RY$ and $x = UY$, so $GY = 21 - x$. By Pythagoras' theorem, $h^2 = 10^2 - x^2$ and also $h^2 = 17^2 - (21-x)^2$, so $100 - x^2 = 289 - 441 + 42x - x^2$. This gives $42x = 100 - 289 + 441 = 252$ and therefore $x = 252 \\div 42 = 6$. Finally, $h^2 = 100 - 6^2 = 64$, so $h = 8$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | MCQ | A | |
0iz9 | Problem:
You are given two diameters $AB$ and $CD$ of circle $\Omega$ with radius $1$. A circle is drawn in one of the smaller sectors formed such that it is tangent to $AB$ at $E$, tangent to $CD$ at $F$, and tangent to $\Omega$ at $P$. Lines $PE$ and $PF$ intersect $\Omega$ again at $X$ and $Y$. What is the length o... | [
"Solution:\n\nAnswer: $\\frac{4\\sqrt{2}}{3}$.\n\nLet $O$ denote the center of circle $\\Omega$. We first prove that $OX \\perp AB$ and $OY \\perp CD$. Consider the homothety about $P$ which maps the smaller circle to $\\Omega$. This homothety takes $E$ to $X$ and also takes $AB$ to the line tangent to circle $\\Om... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 4√2/3 | |
0gde | 找出所有正整數 $x, y, z, t$ 滿足
$$
xy - zt = x + y = z + t
$$
且 $xy$ 和 $zt$ 都是完全平方數。 | [
"解：無正整數解。\n設 $xy = a^2$ 且 $zt = c^2$。\n若 $x+y=z+t$ 爲奇數，則 $xy$ 和 $zt$ 都是偶數，得 $xy-zt = x+y = z+t$\n也是偶數，矛盾。令 $s = \\frac{x+y}{2}$，由前可知 $s$ 是整數。令 $b = \\frac{|x-y|}{2}$，$d = \\frac{|z-t|}{2}$，\n則原題條件可得：\n$$\ns^2 = a^2 + b^2 = c^2 + d^2\n$$\n和\n$$\n2s = a^2 - c^2 = d^2 - b^2.\n$$\n由於上兩式中，$a, d$ 和 $b, c$ 對稱，我們只須證明上二式在 $... | Taiwan | 二〇一九數學奧林匹亞競賽第二階段選訓營,模擬競賽(二) | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | No positive integer solutions. | |
0hom | Problem:
An isolated island has the shape of a circle. Initially there are $9$ flowers on the circumference of the island: $5$ of the flowers are red and the other $4$ are yellow. During the summer $9$ new flowers grow on the circumference of the island according to the following rule: between $2$ old flowers of the s... | [
"Solution:\n\nThe answer is \"no\". Assume that we got all red flowers in the year $n$ for the first time. Then in the year $n-1$ all the flowers were yellow. We will prove that this is impossible.\n\nLet's change the weird story into the one with the flowers labeled by $1$ (instead of red) and $-1$ (instead of yel... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
03x4 | Prove that in an arithmetic progression consisting of 40 distinct positive integers, at least one of the numbers cannot be written as $2^k + 3^l$, where $k, l$ are nonnegative integers. (Posed by Chen Yonggao) | [
"Suppose on the contrary that there exist 40 distinct positive integers in arithmetic progression such that each term can be written as $2^k + 3^l$, and denote this sequence by $a, a+d, a+2d, \\ldots, a+39d$, where $a, d$ are positive integers. Let\n$$\nm = \\lfloor \\log_2 (a + 39d) \\rfloor, \\quad n = \\lfloor \... | China | China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Other"
] | English | proof only | null | |
0cv8 | Let $\Omega$ be the circumcircle of a scalene triangle $ABC$ with $\angle ACB = 60^\circ$. The points $A'$ and $B'$ are chosen on the internal angle bisectors of the angles $BAC$ and $ABC$, respectively, so that $AB' \parallel BC$ and $BA' \parallel AC$. The line $A'B'$ meets $\omega$ at points $D$ and $E$. Prove that ... | [
"Let $N$ be the midpoint of the arc $ABC$, and $T$ be the midpoint of the lesser arc $NC$. Prove that $N$ and $T$ lie on $A'B'$. For this purpose, notice that $AN = BN = AB = A'B = B'A$.\n\nFrom the parallelism of the lines $AB'$ and $BC$, we get that $\\angle AB'B = \\angle CBB' = \\angle ABB'$. Therefore, $AB' = ... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | English; Russian | proof only | null | |
047o | Find all positive integers $m$ such that there exists an infinite set $A$ of positive integers satisfying: for any $m$ distinct elements $a_1, a_2, \dots, a_m$ in $A$, both $a_1+a_2+\dots+a_m$ and $a_1a_2\cdots a_m$ are square-free.
*Note: A positive integer $n$ is called square-free if it is not divisible by the squa... | [
"**Proof:** We first prove a lemma.\n**Lemma:** For integers $m \\ge 2$, $s \\ge 1$, and a sequence $1 = x_1 < \\dots < x_s$ where every sum $\\sum_{1 \\le j \\le m} x_{i_j}$ ($1 \\le i_1 \\le \\dots \\le i_m \\le s$) is square-free, there exists an integer $x > x_s$ such that:\n* $x$ is coprime with each $x_i$ ($1... | China | 2025 International Mathematical Olympiad China National Team Selection Test | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | All square-free positive integers m. | |
0ag1 | Let $f: \mathbb{R} \to \mathbb{R}$ and let:
$$
f(f(x)+f(y))+f(f(x)-f(y)) = x^k f(x) + y^k f(y)
$$
hold, where $k$ is a given natural number. What values can $f(1)$ have? (easier case: $k=2$). | [
"For $x = y = 0$ we get $f(2a) = -a$ (where $a = f(0)$).\n\nFor $x = 0$ and $y = 2a$ we get $a - a = -a (2a)^k$, so $a = 0$.\n\nFor $x = y = 1$ we get $f(2b) = 2b$ (where $b = f(1)$).\n\nFor $x = 2b$ and $y = 0$ we get $4b = (2b)^{k+1}$, so $b = 0$ or $b = \\frac{k\\sqrt{2}}{2}$.\n\nFor $b = 0$, $f(x) = 0$ for ever... | North Macedonia | null | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | If k = 2: f(1) ∈ {0, sqrt(2)/2}. If k ≠ 2: f(1) = 0. | |
0gcw | 已知 $a, b, c, d$ 為正實數且滿足 $a + b + c + d = 4$。試證：
$$
\sum_{\text{cyc}} \frac{3a^3}{a^2 + ab + b^2} + \sum_{\text{cyc}} \frac{2ab}{a+b} \geq 8.
$$ | [
"Notice that\n$$\n\\sum_{\\text{cyc}} \\frac{a^3 - b^3}{a^2 + ab + b^2} = \\sum_{\\text{cyc}} (a-b) = 0.\n$$\nHence\n$$\n\\text{LHS} = \\sum_{\\text{cyc}} \\frac{3a^3 + 3b^3}{2(a^2 + ab + b^2)} + \\sum_{\\text{cyc}} \\frac{2ab}{a+b}.\n$$\nNow\n$$\n\\frac{2ab}{a+b} = a+b-\\frac{a^2+b^2}{a+b},\n$$\nwhich means we onl... | Taiwan | 二〇一九數學奧林匹亞競賽第二階段選訓營，獨立研究（一） | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0777 | Problem:
Let $n$ and $M$ be positive integers such that $M > n^{n-1}$. Prove that there are $n$ distinct primes $p_{1}, p_{2}, p_{3}, \ldots, p_{n}$ such that $p_{j}$ divides $M + j$ for $1 \leq j \leq n$. | [
"Solution:\n\nIf some number $M + k$, $1 \\leq k \\leq n$, has at least $n$ distinct prime factors, then we can associate a prime factor of $M + k$ with the number $M + k$ which is not associated with any of the remaining $n - 1$ numbers.\n\nSuppose $M + j$ has less than $n$ distinct prime factors. Write\n$$\nM + j... | India | Indian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
04we | Let $n > 1$ be a positive integer and $p$ its greatest prime factor. For each nonempty subset of divisors of the number $n$, write on the board the sum of its elements. Suppose that in this way we have written more than $p$ numbers from the set $\{1, 2, \dots, p+2\}$, and that we have not written any number from this s... | [] | Czech Republic | National Round | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0a5y | Problem:
Find the sum of the smallest and largest possible values for $x$ which satisfy the following equation.
$$9^{x + 1} + 2187 = 3^{6x - x^{2}}$$ | [
"Solution:\nFirst we prove that the given equation has at least one root. To do this we consider the following function.\n\n$$f(x) = 9^{x + 1} + 2187 - 3^{6x - x^{2}}$$\n\nNote that this function is continuous. Note also that $f(0) = 9 + 2187 - 1 > 0$ and $f(3) = 9^{4} + 2187 - 3^{9} < 0$. Therefore by the intermed... | New Zealand | NZMO Round One | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | 5 | |
015o | A king has 10 fools. Each fool amuses the king by weeks that start and end at Sunday midnight, whereby for every $n = 1, \dots, 10$, there exists a fool whose every two consecutive working weeks are separated by exactly $n$ free weeks. When no fool is present, the king feels bored. How many consecutive days at most is ... | [
"Let $G$ be a work schedule of the fools that enables the maximum number of consecutive joyful weeks. If 2 and 4 have a common working week then every working week of 4 is also a working week of 2. Consecutively, in the schedule obtained from $G$ by shifting the working weeks of 4 by one week, all weeks that are jo... | Baltic Way | Baltic Way SHL | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2009 | |
0eru | Which term must be removed from the sequence $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{6}$, $\frac{1}{10}$ and $\frac{1}{12}$ so that the sum of the remaining terms is equal to 1?
(A) $\frac{1}{2}$ (B) $\frac{1}{4}$ (C) $\frac{1}{6}$ (D) $\frac{1}{10}$ (E) $\frac{1}{12}$ | [
"If all terms are brought to $60$, the least common multiple (LCM), then all the numerators except $6$ have a factor $5$. Since the required numerator is $60$, which also has a factor $5$, the term that must be dropped is $\\frac{1}{10}$.\n\nAlternatively,\n$$\n\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{6} + \\frac{1}... | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | MCQ | D | |
03bi | Find all functions $f : \mathbb{R} \to \mathbb{R}$, bounded in the interval $(0, 1)$ and such that
$$
x^2 f(x) - y^2 f(y) = (x^2 - y^2) f(x + y) - xy f(x - y)
$$
for all $x, y \in \mathbb{R}$. | [
"Note that when $x > y + 1/2$ run through the interval $(0, n)$ then $x + y$ runs through the interval $(0, 2n - 1/2)$. Straightforward induction shows that $f$ is bounded in all intervals $(0, 2k + 1/2)$. When $0 < x < y$ it follows that $f$ is also bounded in the intervals $(-2k, 0)$. Thus, $f$ is bounded in ever... | Bulgaria | Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = a x for some real constant a | |
0kh6 | Which of the following conditions is sufficient to guarantee that integers $x$, $y$, and $z$ satisfy the equation
$$
x(x - y) + y(y - z) + z(z - x) = 1?
$$
(A) $x > y$ and $y = z$
(B) $x = y - 1$ and $y = z - 1$
(C) $x = z + 1$ and $y = x + 1$
(D) $x = z$ and $y - 1 = x$
(E) $x + y + z = 1$ | [] | United States | 2021 AMC 10 B Fall | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
0et7 | A positive integer $k$ is said to be *visionary* if there are integers $a > 0$ and $b \ge 0$ such that $a \cdot k + b \cdot (k+1) = 2020$. How many visionary integers are there? | [
"All lower case variables in this solution denote integers. Let $X$ denote the set of visionary integers. We show that $X = \\{\\lfloor \\frac{2020}{n} \\rfloor : 1 \\le n \\le 2020\\}$.\n\nIf $k$ is a visionary integer, then there exist $a > 0$ and $b \\ge 0$ such that $a k + b(k + 1) = 2020$, i.e., $2020 = k(a + ... | South Africa | The South African Mathematical Olympiad Third Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 88 | |
07m3 | Suppose $x, y, z$ are positive numbers such that $x + y + z = 1$. Prove that
$$
(a) \ xy + yz + zx \geq 9xyz;
$$
$$
(b) \ xy + yz + zx < \frac{1}{4} + 3xyz.
$$ | [
"(a) Through two uses of the arithmetic mean-geometric mean inequality,\n$$\n\\begin{aligned}\nxyz &= \\sqrt[3]{(xy)(yz)(zx)} \\cdot \\sqrt[3]{xyz} \\\\\n&\\le \\frac{xy + yz + zx}{3} \\cdot \\frac{x+y+z}{3} = \\frac{xy + yz + zx}{9}\n\\end{aligned}\n$$\nwith equality iff $x = y = z = 1/3$. Hence,\n$$\nxy + yz + zx... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
09e7 | a_1, a_2, ..., a_{49} is a permutation of the set $\{2, 3, ..., 50\}$. Denote $S_1 = a_1$, $S_2 = a_1 + a_2$, ..., $S_{49} = a_1 + a_2 + ... + a_{49}$. Find the number of different sequences $S_1, S_2, ..., S_{49}$ in which no one of $S_i$ is divisible by $3$. | [
"Let $A_0, A_1, A_2$ be subsets of $\\{2, 3, ..., 49\\}$ whose elements give remainder $0$, $1$, $2$ after dividing by $3$. Note that $A_0 \\cup A_1 \\cup A_2 = \\{2, 3, ..., 49\\}$. Define a mapping $f : \\{A_0, A_1, A_2\\} \\to \\{0, 1, 2\\}$ as follows: if $x \\in A_i$ then we put $x = i$. Therefore, to form a s... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | English | proof and answer | 48!/32!*16!*17! | |
0736 | Problem:
Let $m$ and $n$ be positive integers such that the equation $x^{2}-m x+n=0$ has real roots $\alpha$ and $\beta$. Prove that $\alpha$ and $\beta$ are integers if and only if $[m \alpha]+[m \beta]$ is the square of an integer. (Here $[x]$ denotes the largest integer not exceeding $x$.) | [
"Solution:\nIf $\\alpha$ and $\\beta$ are both integers, then\n$$\n[m \\alpha]+[m \\beta]=m \\alpha+m \\beta=m(\\alpha+\\beta)=m^{2}\n$$\nThis proves one implication.\n\nObserve that $\\alpha+\\beta=m$ and $\\alpha \\beta=n$. We use the property of integer function: $x-1<[x] \\leq x$ for any real number $x$. Thus\n... | India | INMO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
03nw | Problem:
William is thinking of an integer between 1 and 50, inclusive. Victor can choose a positive integer $m$ and ask William: "does $m$ divide your number?", to which William must answer truthfully. Victor continues asking these questions until he determines William's number. What is the minimum number of question... | [
"Solution:\n\nThe minimum number is 15 questions.\n\nFirst, we show that 14 or fewer questions is not enough to guarantee success. Suppose Victor asks at most 14 questions, and William responds with \"no\" to each question unless $m=1$. Note that these responses are consistent with the secret number being 1. But si... | Canada | CMO 2023 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 15 | |
06c2 | Prove that $S = \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2n + 1}$ is not an integer if $n \ge 1$. | [
"Let $l = \\text{lcm}(3, 5, \\dots, 2n+1)$, and let $k = v_3(l)$. Note that $k \\ge 1$ since $n \\ge 1$. Since $2 \\cdot 3^k$ is even, and $3 \\cdot 3^k > 2n+1$ (or otherwise $3^{k+1} \\mid l$), only the term $3^k$ among $3, 5, \\dots, 2n+1$ is divisible by $3^k$. Therefore, $\\frac{l}{3}S = \\frac{l}{3^{k+1}} + m$... | Hong Kong | HKG TST | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
08eo | Problem:
Quanto vale la somma delle cifre del numero $20^{21} + \left(10^{2021} + 21\right)^2$?
(A) 21
(B) 30
(C) 37
(D) 42
(E) Un numero maggiore di 100. | [
"Solution:\n\nLa risposta è $(\\mathbf{D})$. Possiamo espandere il quadrato\n$$\nN = 20^{21} + \\left(10^{2021} + 21\\right)^2 = 10^{4042} + 42 \\cdot 10^{2021} + 2^{21} \\cdot 10^{21} + 21^2\n$$\nOsserviamo che non ci sono riporti tra gli addendi scritti, dunque il risultato è la somma della somma delle cifre di $... | Italy | Italian Mathematical Olympiad, February Round | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
0f4i | Problem:
The circle $C$ has center $O$ and radius $r$ and contains the points $A$ and $B$. The circle $C'$ touches the rays $OA$ and $OB$ and has center $O'$ and radius $r'$. Find the area of the quadrilateral $OAOB$. | [] | Soviet Union | 16th ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | r r' | |
0h5s | $$
\operatorname{ctg}[x] \cdot \operatorname{ctg}\{x\} = 1.
$$
Where $[a]$ is the integer part of $a$, and $\{a\} = a - [a]$. | [
"Transform the equation to the form:\n$$\n\\cos[x] \\cdot \\cos\\{x\\} = \\sin[x] \\cdot \\sin\\{x\\},\n$$\nif $\\sin[x] \\cdot \\sin\\{x\\} \\ne 0$. Then we have\n$$\n\\cos[x] \\cdot \\cos\\{x\\} - \\sin[x] \\cdot \\sin\\{x\\} = \\cos([x] + \\{x\\}) = \\cos x = 0.\n$$\nHence $x = \\frac{1}{2}\\pi + \\pi k, k \\in ... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | x = pi/2 + pi*k, where k is any integer | |
03hk | Problem:
A sequence of numbers $a_{1}, a_{2}, a_{3}, \ldots$ satisfies
(i) $a_{1}=\frac{1}{2}$,
(ii) $a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n}$ \quad $(n \geq 1)$.
Determine the value of $a_{n}$ \quad $(n \geq 1)$. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1/(n(n+1)) | |
05dv | Problem:
Let $A B C D E F$ be a convex hexagon such that $\angle A=\angle C=\angle E$ and $\angle B=\angle D=\angle F$ and the (interior) angle bisectors of $\angle A, \angle C$, and $\angle E$ are concurrent.
Prove that the (interior) angle bisectors of $\angle B, \angle D$, and $\angle F$ must also be concurrent.
N... | [
"Solution:\n\nDenote the angle bisector of $A$ by $a$ and similarly for the other bisectors. Thus, given that $a, c, e$ have a common point $M$, we need to prove that $b, d, f$ are concurrent. We write $\\angle(x, y)$ for the value of the directed angle between the lines $x$ and $y$, i.e. the angle of the countercl... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2020 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate M... | null | proof only | null | |
0fn9 | Sean $A$ y $B$ dos conjuntos tales que:
i) $A \cup B$ es el conjunto de los enteros positivos.
ii) $A \cap B$ es el vacío.
iii) Si dos enteros positivos tienen como diferencia a un primo mayor que $2013$, entonces uno de ellos está en $A$ y el otro en $B$.
Hallar todas las posibilidades para los conjuntos $A$ y $B$. | [
"La única partición con esta propiedad es $N = \\{1, 3, 5, \\dots\\} \\cup \\{2, 4, 6, \\dots\\}$, de modo que $B$ siempre es el conjunto de los números positivos pares.\n\nSea $A \\cup B = N$ una partición admisible. La idea es aplicar la condición a un par fijo de primos gemelos $p$ y $q = p + 2$ de $[2013, 3013]... | Spain | Olimpiada Iberoamericana de Matemáticas | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | Exactly two: either A is all odd positive integers and B is all even positive integers, or A is all even positive integers and B is all odd positive integers. | |
0hws | Problem:
Let $n$ and $k$ be positive integers with $n < \sqrt{(k-1) 2^{k}}$. Prove that it is possible to color each element of the set $\{1,2, \ldots, n\}$ red or green such that no $k$-term arithmetic progression is monochromatic. | [
"Solution:\n\nLet $A$ be the number of $k$-term arithmetic progressions in $\\{1,2, \\ldots, n\\}$. For any common difference $d$, an arithmetic progression of difference $d$ fits in $\\{1,2, \\ldots, n\\}$ iff its initial term $a$ satisfies\n$$\n1 \\leq a < a + (k-1)d \\leq n\n$$\nwhich is equivalent to\n$$\n1 \\l... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0ikt | Problem:
Let
$$
f(x) = x^{4} - 6x^{3} + 26x^{2} - 46x + 65.
$$
Let the roots of $f(x)$ be $a_{k} + i b_{k}$ for $k = 1, 2, 3, 4$. Given that the $a_{k}, b_{k}$ are all integers, find $|b_{1}| + |b_{2}| + |b_{3}| + |b_{4}|$. | [
"Solution:\nThe roots of $f(x)$ must come in complex-conjugate pairs. We can then say that $a_{1} = a_{2}$ and $b_{1} = -b_{2}$; $a_{3} = a_{4}$ and $b_{3} = -b_{4}$. The constant term of $f(x)$ is the product of these, so $5 \\cdot 13 = (a_{1}^{2} + b_{1}^{2})(a_{3}^{2} + b_{3}^{2})$. Since $a_{k}$ and $b_{k}$ are... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 10 | |
0l3q | Problem:
A weird checkerboard is a coloring of an $8 \times 8$ grid constructed by making some (possibly none or all) of the following 14 cuts:
- the 7 vertical cuts along a gridline through the entire height of the board,
- and the 7 horizontal cuts along a gridline through the entire width of the board.
The divided ... | [
"Solution:\nWe can focus on only the black cells of the grid, which we need $32$ of. Moreover, the number of black squares in the bottom row and leftmost column uniquely determine the total number of black squares. Suppose that there are $x$ black cells in the bottom row and $y$ black cells in the leftmost column. ... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 7735 | |
0jso | Problem:
An $8$ by $8$ grid of numbers obeys the following pattern:
1) The first row and first column consist of all $1$s.
2) The entry in the $i$th row and $j$th column equals the sum of the numbers in the $(i-1)$ by $(j-1)$ sub-grid with row less than $i$ and column less than $j$.
What is the number in the $8$th ... | [
"Solution:\n\nAnswer: $2508$\n\nLet $x_{i, j}$ be the number in the $i$th row and the $j$th column. Then if $i, j \\geq 2$, $x_{i+1, j+1} - x_{i+1, j} - x_{i, j+1} + x_{i, j}$ only counts the term $x_{i, j}$ since every other term is added and subtracted the same number of times. Thus $x_{i+1, j+1} = x_{i+1, j} + x... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2508 | |
0adz | Тревна патека обиколува правоаголно асфалтно игралиште со димензии $36$ дм и $12$ м (како на цртежот). Ширината $d$ на патеката е еднаква на четвртината од полупериметарот на игралиштето. Колку метри жица е потребно да се огради тревната патека од игралиштето, ако тоа се заобиколи со $5$ реда на жица? Колку столбови се... | [
"$12$ м $= 120$ дм. Ширината на патеката е $d = (120 + 36) : 4$, $d = 39$ дм.\n\nЗначи едната страна на тревникот е долга $2 \\times 39 + 120 = 198$ дм, а другата е долга $2 \\times 39 + 36 = 114$ дм.\n\nЗа оградување на еден ред ограда на тревникот е потребно: $2(198 + 114) + 2(120 + 36) = 624 + 312 = 936$ дм.\n\n... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | Macedonian, English | final answer only | 468 meters of wire; 156 posts | |
00q9 | The sequence $a_0, a_1, \dots$ is defined by the initial conditions $a_0 = 1$, $a_1 = 6$ and the recursion $a_{n+1} = 4a_n - a_{n-1} + 2$ for $n > 1$. Prove that $a_{2^k-1}$ has at least three prime factors for every positive integer $k > 3$. | [
"Consider the sequence $b_0, b_1, \\dots$, defined by the initial conditions $b_0 = 1$, $b_1 = 2$, and the recursion $b_{n+1} = 4b_n - b_{n-1}$ for $n \\ge 1$. We have $a_n = b_{n+1} - 1$, for all $n \\ge 0$. In particular, $a_{2^k-1} = b_{2^k} - 1$, $k \\ge 0$. It is not hard to see that the general formula for $b... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Pell's equations"
] | English | proof only | null | |
06db | There are $212$ points inside or on a circle with radius $1$. Prove that there are at least $2001$ pairs of these points having distances at most $1$. | [
"Let $A, B, C, D, E, F$ be $6$ distinct points on the circle such that\n$$\n\\overline{AB} = \\overline{BC} = \\overline{CD} = \\overline{DE} = \\overline{EF} = \\overline{FA}.\n$$\nLet $O$ be the centre of the circle. Note that any two points in the same sector among $AOB, BOC, COD, DOE, EOF, FOA$ have distance at... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
04qj | Find all positive integers $k$ such that
$$
\frac{m+n}{m^2-kmn+n^2}
$$
is not a composite number for any positive integers $m$ and $n$. (Borna Vukorepa) | [
"Considering a fixed positive integer $k$ with the given property, by plugging $(m, n) \\leftarrow (k, 1)$ and $(m, n) \\leftarrow (k^2 + k - 1, k + 1)$ we find that both\n$$\n\\frac{k+1}{k^2 - k^2 + 1} = k + 1\n$$\nand\n$$\n\\frac{(k^2 + k - 1) + (k + 1)}{(k^2 + k - 1)^2 - k(k^2 + k - 1)(k + 1) + (k + 1)^2} = \\fr... | Croatia | Croatian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | k = 1 |
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