AoPS Academy
have incenter 
and intouch triangle 
. Let the circumcircle of intersect 
at 
and have center 
. Let 
be the midpoint of arc 
on the circumcircle. Suppose quadrilateral 
is cyclic such that 
lies on 
. Show that 
.
be a semicircle with diameter 
and midpoint 
. Let 
be a point on different from 
and 
.
touches in a point 
, the segment 
in a point 
, and additionally the segment 
. The circle 
touches in a point 
and additionally the segments 
and .
and 
are perpendicular.
and 
. Let 
be non-negative real numbers, such that
holds for all 
and 
. Denote
Prove that
?
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 it being given that 
and 
of the equation 
which follows 
(b) Solve for real 
![$ f:[0,1]->R$](http://latex.artofproblemsolving.com/e/c/f/ecf649bc1479dd69a79ed8389297b15557c29200.png)
continously function with 
bounded
be defined on 
. Show that the dimension of the kernel of 
for any 
is 
.
intersect at . Since 
, 
is cyclic, so 
(well-known). Thus, 
is cyclic, so 
, so since the circumcenter lies on the perpendicular bisector of , we must have 
.
meet at 
. Then 
are concyclic, therefore , the Simson-line of w.r.t. the triangle 
, passes through the midpoint of .
, and 
. Then 
, 
., , collinear, 
for 
and since 
, ![\[ \frac{1}{k-a}+ \frac{1}{k-\frac{1}{a}}=0 \implies p=\frac{1}{2} (a+\frac{1}{a}) \]](http://latex.artofproblemsolving.com/4/9/c/49ccf30703a265f7ea09ce00056f7ef4592fc89f.png)
Consider the transformation 
. Then it suffices to show 
, 
, 
are collinear. ![\[ \frac{\frac{b}{a}-1}{b-\frac{1}{a}}=\frac{\frac{a}{b}-1}{a-\frac{1}{b}} \implies \frac{b-a}{ab-1}=\frac{a-b}{1-ab} \]](http://latex.artofproblemsolving.com/2/3/1/2317f9cbc20ffda5fba86a202ddb5a1ebf6252f1.png)
which is true.
be unit circle then 
, 
, 
and 
, so we need 
colinear but by the composed shifting 
and colinearity lemma all we need is 
.
thus bisects as desired, and done 
.
where 
and 
,
is right-angled at 
.
all lie on the circle with diameter 
,![\[ \angle BHP = \angle CAP = \angle B = \angle HBA \quad \quad \text{(1)} \]](http://latex.artofproblemsolving.com/8/3/0/830a1ba5da967f7e79eb2354bdebd3d7b0a4ced6.png)
Now let 
,
shows that 
.
also implies that ![\[ \angle HAB = \frac{\pi}{2} - \angle HBA = \frac{\pi}{2} - \angle BHP = \angle PHA = \angle MHA \]](http://latex.artofproblemsolving.com/c/d/4/cd45bb734fcf01545cf54cfaeca16ce43d4e1368.png)
This implies that 
.
as required.
is a cyclic quad. Angle chasing yields 
.
Now, we proceed with complex numbers. Let the circumcircle of 
be the unit circle and let the complex numbers 
and 
be complex numbers representing 
and 
Then, by the Complex Foot formula, we get 
Also, note that the midpoint of 
and 
We wish to show that 
and 
The latter can be verified by expanding algebraically, so we are done. 
.
which is equivalent to
It is obviously true.