Found problems: 85335
Let $x,y,z $ be positive reals such that $\sqrt{a}=x(y-z)^2$, $\sqrt{b}=y(z-x)^2$ and $\sqrt{c}=z(x-y)^2$. Prove that
\[a^2+b^2+c^2 \geq 2(ab+bc+ca)\]
In square $ABCD$ with $AB = 10$, point $P, Q$ are chosen on side $CD$ and $AD$ respectively such that $BQ \perp AP,$ and $R$ lies on $CD$ such that $RQ \parallel PA.$ $BC$ and $AP$ intersect at $X,$ and $XQ$ intersects the circumcircle of $PQD$ at $Y$. Given that $\angle PYR = 105^{\circ},$ $AQ$ can be expressed in simplest radical form as $b\sqrt{c}-a$ where $a, b, c$ are positive integers. Find $a+b+c.$
Let $a,b,c>1$. Solve in $\mathbb{R}$ the equation $\log_{a+b}(a^x+b)=\log_b((b+c)^x-c)$.
[i]Mihai Opincariu[/i]
If $yz:zx:xy=1:2:3$, then $\dfrac{x}{yz}:\dfrac{y}{zx}$ is equal to
$\textbf{(A) }3:2\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:4\qquad\textbf{(D) }2:1\qquad \textbf{(E) }4:1$
Two people write a $2k$-digit number, using only the numbers $1, 2, 3, 4$ and $5$. The first number on the left is written by the first of them, the second - the second, the third - the first, etc. Can the second one achieve this so that the resulting number is divisible by $9$, if the first seeks to interfere with it? Consider the cases $k = 10$ and $k = 15$.
A plane geometric figure of $n$ sides with the vertices $A_1,A_2,A_3,\dots, A_n$ ($A_i$ is adjacent to $A_{i+1}$ for every $i$ integer where $1\leq i\leq n-1$ and $A_n$ is adjacent to $A_1$) is called [i]brazilian[/i] if:
I - The segment $A_jA_{j+1}$ is equal to $(\sqrt{2})^{j-1}$, for every $j$ with $1\leq j\leq n-1$.
II- The angles $\angle A_kA_{k+1}A_{k+2}=135^{\circ}$, for every $k$ with $1\leq k\leq n-2$.
[b]Note 1:[/b] The figure can be convex or not convex, and your sides can be crossed.
[b]Note 2:[/b] The angles are in counterclockwise.
a) Find the length of the segment $A_nA_1$ for a brazilian figure with $n=5$.
b) Find the length of the segment $A_nA_1$ for a brazilian figure with $n\equiv 1$ (mod $4$).
Find all positive integers $n=p_1p_2 \cdots p_k$ which divide $(p_1+1)(p_2+1)\cdots (p_k+1)$ where $p_1 p_2 \cdots p_k$ is the factorization of $n$ into prime factors (not necessarily all distinct).
Determine all pairs of integer numbers $(x, y)$ such that:
$$\frac{(x - y)^2}{x + y} = x - y + 6$$
Let $ABC$ be a triangle in which $CA>BC>AB$. Let $H$ be its orthocentre and $O$ its circumcentre. Let $D$ and $E$ be respectively the midpoints of the arc $AB$ not containing $C$ and arc $AC$ not containing $B$. Let $D'$ and $E'$ be respectively the reflections of $D$ in $AB$ and $E$ in $AC$. Prove that $O, H, D', E'$ lie on a circle if and only if $A, D', E'$ are collinear.
Let $N$ be the number of polynomials $P(x_1, x_2, \dots, x_{2016})$ of degree at most $2015$ with coefficients in the set $\{0, 1, 2 \}$ such that $P(a_1,a_2,\cdots ,a_{2016}) \equiv 1 \pmod{3}$ for all $(a_1,a_2,\cdots ,a_{2016}) \in \{0, 1\}^{2016}.$
Compute the remainder when $v_3(N)$ is divided by $2011$, where $v_3(N)$ denotes the largest integer $k$ such that $3^k | N.$
[i]Proposed by Yang Liu[/i]
In the diagram of rectangles below, with lengths as labeled, let $A$ be the area of the rectangle labeled $A$, and so on. Find $36A+6B+C+6D$.
[asy]
size(3cm);
real[] A = {0,8,13};
real[] B = {0,7,12};
for (int i = 0; i < 3; ++i) {
draw((A[i],0)--(A[i],-B[2]));
draw((0,-B[i])--(A[2],-B[i]));
}
label("8", (4,0), N);
label("5", (10.5,0),N);
label("7", (0,-3.5),W);
label("5", (0,-9.5),W);
label("$A$", (4,-3.5));
label("$B$", (10.5,-3.5));
label("$C$", (10.5,- 9.5));
label("$D$", (4, -9.5));
[/asy]
[i]2018 CCA Math Bonanza Team Round #1[/i]
Prove that the numbers $ A $, $ B $, $ C $ defined by the formulas
$$
A = tg \beta tg \gamma + 5,\\
B = tg \gamma tg \alpha + 5,\\
C = tg \alpha tg \beta + 5,$$
where $ \alpha>0 $, $ \beta > 0 $, $ \gamma > 0 $ and $ \alpha + \beta + \gamma = 90^\circ $, satisfy the inequality
$$ \sqrt{A} + \sqrt{B} + \sqrt{C} < 4 \sqrt{3}.$$
In a country, mathematicians chose an $\alpha> 2$ and issued coins in denominations of 1 ruble, as well as $\alpha ^k$ rubles for each positive integer k. $\alpha$ was chosen so that the value of each coins, except the smallest, was irrational. Is it possible that any natural number of rubles can be formed with at most 6 of each denomination of coins?
The number of pairs of integers $ (m,n)$ satisfying the equation
\[ m^3 \plus{} 6m^2 \plus{} 5m \equal{} 27n^3 \plus{} 9n^2 \plus{} 9n \plus{} 1\]
is
$ \text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ \text{Infinitely many}$
Let $ABCD$ be an isosceles trapezoid with bases $AB$ and $CD$. The incircle of the triangle $BCD$ touches $CD$ at $E$. Point $F$ is chosen on the bisector of the angle $DAC$ such that the lines $EF$ and $CD$ are perpendicular. The circumcircle of the triangle $ACF$ intersects the line $CD$ again at $G$. Prove that the triangle $AFG$ is isosceles.
For every positive integer $n \geq 2$, Prove that there is no $n-$tuple of distinct complex numbers $(x_1,x_2,\dots,x_n)$ such that for each $1 \leq k \leq n$ following equality holds.
$\prod_{\underset{i \neq k}{1 \leq i \leq n}}^{ } (x_k - x_i) = \prod_{\underset{i \neq k}{1 \leq i \leq n}}^{ } (x_k + x_i) $
(20 points)
A polynomial $P(x)$ with integer coefficients satisfies the following: if $F(x)$, $G(x)$, and $Q(x)$ are polynomials with integer coefficients satisfying $P\Big(Q(x)\Big)=F(x)\cdot G(x)$, then $F(x)$ or $G(x)$ is a constant polynomial. Prove that $P(x)$ is a constant polynomial.
Let $A, B, C, D, E$ be integers, $B \neq 0$ and $F=AD^{2}-BCD+B^{2}E \neq 0$. Prove that the number $N$ of pairs of integers $(x, y)$ such that \[Ax^{2}+Bxy+Cx+Dy+E=0,\] satisfies $N \le 2 d( \vert F \vert )$, where $d(n)$ denotes the number of positive divisors of positive integer $n$.
Let $ABC$ be an acute triangle. Find the locus of the points $M$, in the interior of $\bigtriangleup ABC$, such that $AB-FG= \frac{MF.AG+MG.BF}{CM}$, where $F$ and $G$ are the feet of the perpendiculars from $M$ to the lines $BC$ and $AC$, respectively.
Suppose that for a function $f: \mathbb{R}\to \mathbb{R}$ and real numbers $a<b$ one has $f(x)=0$ for all $x\in (a,b).$ Prove that $f(x)=0$ for all $x\in \mathbb{R}$ if
\[\sum^{p-1}_{k=0}f\left(y+\frac{k}{p}\right)=0\]
for every prime number $p$ and every real number $y.$
If $$\frac{x_1}{x_1+1} = \frac{x_2}{x_2+3} = \frac{x_3}{x_3+5} = ...= \frac{x_{1006}}{x_{1006}+2011}$$ and $x_1+x_2+...+x_{1006} = 503^2$, determine the value of $x_{1006}$.
Consider a function $ f:\mathbb{R}\longrightarrow\mathbb{R} $ that admits primitives. Prove that:
$ \text{(i)} $ Every term (function) of the sequence functions $ \left( h_n\right)_{n\ge 2}:\mathbb{R}\longrightarrow\mathbb{R} $ defined, for any natural number $ n $ as $ h_n(x)=x^nf\left( x^3 \right) , $ is primitivable.
$ \text{(ii)} $ The function $ \phi :\mathbb{R}\longrightarrow\mathbb{R} $ defined as
$$ \phi (x) =\left\{ \begin{matrix} e^{-1/x^2} f(x),& \quad x\neq 0 \\ 0,& \quad x=0 \end{matrix} \right. $$
is primitivable.
[i]Cristinel Mortici[/i]
Prove that any uncountable subset of the Euclidean $ n$-space contains an countable subset with the property that the distances between different pairs of points are different (that is, for any points $ P_1 \not\equal{} P_2$ and $ Q_1\not\equal{} Q_2$ of this subset, $ \overline{P_1P_2}\equal{}\overline{Q_1Q_2}$ implies either $ P_1\equal{}Q_1$ and $ P_2\equal{}Q_2$, or $ P_1\equal{}Q_2$ and $ P_2\equal{}Q_1$). Show that a similar statement is not valid if the Euclidean $ n$-space is replaced with a (separable) Hilbert space.
Let $\mathbb{N}_0$ and $\mathbb{Z}$ be the set of all non-negative integers and the set of all integers, respectively. Let $f:\mathbb{N}_0\rightarrow\mathbb{Z}$ be a function defined as
\[f(n)=-f\left(\left\lfloor\frac{n}{3}\right\rfloor \right)-3\left\{\frac{n}{3}\right\} \]
where $\lfloor x \rfloor$ is the greatest integer smaller than or equal to $x$ and $\{ x\}=x-\lfloor x \rfloor$. Find the smallest integer $n$ such that $f(n)=2010$.
Let $a_1, a_2, \ldots, a_n$ be $n$ positive integers that are not divisible by each other, i.e. for any $i \neq j$, $a_i$ is not divisible by $a_j$. Show that
\[ a_1+a_2+\cdots+a_n \ge 1.1n^2-2n. \]
[i]Note:[/i] A proof of the inequality when $n$ is sufficient large will be awarded points depending on your results.