and intouch triangle 
.
at 
and have center 
.
be the midpoint of arc 
on the circumcircle. Suppose quadrilateral 
is cyclic such that 
lies on 
.
of complex numbers satisfying 
and 
.
of a variate 
is defined as follows:![\[
F(x) =
\begin{cases}
A, & -\infty < x < -1, \\
B, & -1 \leq x < 0, \\
C, & 0 \leq x < 2, \\
D, & 2 \leq x < \infty.
\end{cases}
\]](http://latex.artofproblemsolving.com/0/9/0/0903946b6a44e9bfc3a1adf8dc0ad5ac013e92c1.png)
are constants. Determine the values of 
and 
of the equation 
which follows 
(b) Solve for real 
where 
![$ f:[0,1]->R$](http://latex.artofproblemsolving.com/e/c/f/ecf649bc1479dd69a79ed8389297b15557c29200.png)
bounded
be defined on 
.
for any 
is 
.
Y by HWenslawski
Let 
be an isosceles triangle with 
and 
. Its circumcircle 
has center 
is the midpoint of 
and 
is the foot of the perpendicular from 
to 
. With center and radius 
we draw a circle which internally intersects 
at the point 
and the circle at the points 
and 
, such that lies on the small arc and on the small arc . Prove that the lines 
, 
, 
are concurrent.
Brazitikos Silouanos, Greece
be an isosceles triangle with 
and 
. Its circumcircle 
has center 
is the midpoint of 
and 
is the foot of the perpendicular from 
to 
. With center and radius 
we draw a circle which internally intersects 
at the point 
and the circle at the points 
and 
, such that lies on the small arc and on the small arc . Prove that the lines 
, 
, 
are concurrent.Brazitikos Silouanos, Greece
meet 
is isosceles trapezoid. It is sufficient to prove that 
is parallelogram. Let 
.
is cyclic, and by power bash of 
and 
+ Pythagorean theorem for 
,
,
is perpendicular bisector of 
,
and that 
;
and 
are parallel because 
is isosceles)
and 
.
on a plane, define a function ![\[ f(Y) = \operatorname{Pow}_{\omega_1}(Y) - \operatorname{Pow}_{\omega_2}(Y).\]](http://latex.artofproblemsolving.com/6/0/3/6032041509c44ac83ffad583010ae54c12f91a01.png)
Let 
.
.
.![\[ f(M) = \frac{f(B)+f(C)}{2} = \frac{-BD^2+CD^2}{2} = \frac{CD^2(1 - \tan^2 \frac{\pi}{8})}{2} = CD^2(\sqrt{2}-1).\]](http://latex.artofproblemsolving.com/a/5/7/a579c6e9218b788aa8e534e0ee93a753185fbc06.png)
.
,![\[ f(F) = -AF \cdot CF = -(AC-CD)CD = -CD^2(\sqrt{2}-1). \]](http://latex.artofproblemsolving.com/9/0/e/90ee2b9fb5227b3f028b0a6d7a2f6f2f48c652a1.png)
is equivalent to showing that 
.![\[ \frac{MX}{XF} = \frac{MC\sin{\frac{\pi}{4}}}{CF\sin{\frac{\pi}{8}}} = \frac{BC\sin{\frac{\pi}{4}}}{2CD\sin{\frac{\pi}{8}}} = \frac{\sin{\frac{\pi}{4}}}{2\sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}}} = 1. \ \ \square\]](http://latex.artofproblemsolving.com/b/0/9/b0965978cbb7bfb8b2e94ae54a691ccf3fb44e62.png)
