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and 
with equal radii intersect at P and Q. Points B and C are located on the circles and so that they are inside the circles and , respectively. Also, points X and Y distinct from P are located on and , respectively, so that:
The intersection point of the circumcircles of triangles XPC and YPB is called S. Prove that BC, XY and QS are concurrent.
with 
and 
, let 
be the midpoint of 
. Choose point 
on the extension of 
past 
and point 
on segment 
such that lies on 
. Let 
be on the opposite side of 
from 
such that 
and 
. Let 
intersect the circumcircle of 
again at 
, and let 
intersect the circumcircle of 
again at 
. Prove that , 
, and 
are collinear.
which satisfy ![\[f(x)\cdot f(y)\mid (1+2x)\cdot f(y)+(1+2y)\cdot f(x)\]](http://latex.artofproblemsolving.com/6/4/f/64f2720634f57b4d88db823b9931379386ac1c27.png)
for all non-negative integers 
and 
.
, and square 
. Point and 
are on the circle, and 
is tangent to the circle at point 
. Let represent the midpoint of 
and 
represent the intersection between and circle. Prove that 
.
![$\color[rgb]{0.35,0.35,0.35}\displaystyle\sum_{n=0}^\infty\frac{\sin (nx)}{3^n}$](http://latex.artofproblemsolving.com/a/f/3/af33799c6924b61378db323645ea3e1b9c4e8c9d.png)
if ![$\color[rgb]{0.35,0.35,0.35}\sin x=1/3$](http://latex.artofproblemsolving.com/c/e/a/cea5c46445caa512d94d655350170fb19518a2c8.png)
and ![$\color[rgb]{0.35,0.35,0.35} 0\le x\le \pi/2$](http://latex.artofproblemsolving.com/9/8/b/98b7d75c944002b8b9c11dbe8cca0d402e6fbb67.png)
.![$\color[rgb]{0.35,0.35,0.35}v_1=2$](http://latex.artofproblemsolving.com/2/0/4/204b13717fb1dec258e4917232ca9c96344031d2.png)
, ![$\color[rgb]{0.35,0.35,0.35}v_2=4$](http://latex.artofproblemsolving.com/b/c/2/bc24e9589aaf75cb848e9fddeef43fb7a6fea2c3.png)
and ![$\color[rgb]{0.35,0.35,0.35} v_{n+1}=3v_n-v_{n-1}$](http://latex.artofproblemsolving.com/8/7/3/8734d09c683e906771abf049d89b273196175585.png)
, prove that ![$\color[rgb]{0.35,0.35,0.35}v_n=2F_{2n-1}$](http://latex.artofproblemsolving.com/b/1/e/b1efbac9eef93c5649dab70be54ae2286ca5b2f0.png)
, where the terms ![$\color[rgb]{0.35,0.35,0.35}F_n$](http://latex.artofproblemsolving.com/0/2/5/025c95f57f5317da3d9739c1620448d8f7cf4e47.png)
are the Fibonacci numbers.
be a prime number. Define the set![\[
M_q = \left\{ x \in \mathbb{Z}^* \,\middle|\, \sqrt{x^2 + 2q^{2025} x} \in \mathbb{Q} \right\}.
\]](http://latex.artofproblemsolving.com/9/8/b/98b266a869cf5686cea86be51152ef234486993e.png)
.
be the midpoint of 
, then 
. Suppose, without loss of generality, that 
, we know that 
is a perpendicular bisector of 
, if 
then 
. through 
intersects and 
at 
and , respectively, so 
. We see that 
so 
(*). Since 
and 
we have 
, that is, is the reflection of wrt 
then 
is isosceles with 
, then by (*) we have that is the centroid of so 
and since 
we have that 
. Let's apply the bisector theorem in 
be a positive root of the equation 
.
be real numbers so that 
and 
Show that
.
![\[ \frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}. \]](http://latex.artofproblemsolving.com/c/f/0/cf007b7ce24aa95493617d1f4011e451e7c49126.png)
