is inscribed into a circle 
.
be an altitude of this triangle. A ray 
meets 
.
is tangent to 
denote the set of positive integers. For any positive integer 
,
is called 
for all 
.
be real numbers such that 
Prove that
Let 
Prove that
Prove that
is written on the blackboard. Every time, Carmela can erase the number 
on the black board and replace it with a new number 
,
is a perfect square. Prove or disprove that all positive integers 
can be written exactly once on the blackboard.
be a triangle inscribed in a circle, where the tangents to the circle at points 
and 
intersect at the point 
.
and 
.
intersect at point 
be the reflection of 
and 
intersect at point 
,
.
.
such that for all positive integer 
,
Where 
denotes the number of divisors of 
.
,
such that for all positive integer 
.
then![\[\sum |\pm z_1\pm z_2\pm z_3|^2=2^3\sum_{i=1}^3|z_k|^2.\]](http://latex.artofproblemsolving.com/6/d/1/6d1cd28a47704aa0cac2a6c6b572816fc1db790a.png)
Generalize the problem.
and 
be their respective pedal triangles. Let![\[ X_1=P_2Q_3\cap P_3Q_2,\quad X_2=P_1Q_3\cap P_3Q_1,\quad X_3=P_1Q_2\cap P_2Q_1 \]](http://latex.artofproblemsolving.com/7/f/e/7fe3ec4f69ec074635b2202daad9371e3bb8c72d.png)
Prove that the points 
,
and 
lie on the line 
.
be the intersections of 
with the sides of 
are the midpoints of 
,
,
and 
.
![$ a,b\in[\frac{1}{2},1] $](http://latex.artofproblemsolving.com/0/d/2/0d24f742e1f7db553b3a8795c061c796f07a897f.png)
.
Let ![$ a,b\in[\frac{1}{2},2] $](http://latex.artofproblemsolving.com/a/8/d/a8d4c8205bac0e34129f582a8f255355eda12e0b.png)
.
Y by Davi-8191, tenplusten, HsuAn, Adventure10
Let 
be a fixed acute-angled triangle. Consider some points 
and 
lying on the sides 
and 
, respectively, and let 
be the midpoint of 
. Let the perpendicular bisector of intersect the line 
at 
, and let the perpendicular bisector of 
intersect the lines and at 
and 
, respectively. If the quadrilateral 
is cycle, prove that 
.
be a fixed acute-angled triangle. Consider some points 
and 
lying on the sides 
and 
, respectively, and let 
be the midpoint of 
. Let the perpendicular bisector of intersect the line 
at 
, and let the perpendicular bisector of 
intersect the lines and at 
and 
, respectively. If the quadrilateral 
is cycle, prove that 
.
then 
at its midpoint 
Since 
is the reflection of 
on 
then it follows by symmetry that 
cuts 
again at 
forming the isosceles trapezoid 
Thus again, by symmetry, if 
cuts 
then 
and 
are isogonals WRT 
are tangents of 
as 
,
specified in the problem statement). Now, let 
be the orthocenter of 
denote 
,
are reflections of 
in 
,
,
are collinear. From 
we obtain 
From 
we find 
It follows that 
Now, because the reflection 
in 
,
(well-known). Meanwhile, since 
By Power of a Point, we obtain 
and 
Therefore, ![\[\frac{AX}{MX} \cdot \frac{MB}{AU} = \frac{AH}{MB} \cdot \frac{MV}{AV} = \frac{AH}{2MB} = \frac{OM}{MB},\]](http://latex.artofproblemsolving.com/f/1/b/f1b6a4f57e9f08c9f20ad8d2904c47b750824a98.png)
where 
denotes the center of 
Then by side-angle-side similarity, we deduce that 
,
and passes through 
on 
respectively. Let 
intersect 
at ![\[
\angle FXE=\angle GXH=\angle TKS=180-\angle FAE
\]](http://latex.artofproblemsolving.com/7/f/1/7f1594c4446e957f0151b4e1140b1ebd033906da.png)
is cyclic. By Brokard's Theorem, 
,
to 
,
.
,
.
,
m
be two such points on the arc 
are concyclic, 
are also concyclic which is not possible as they lie on the same side of 
so that 
)
.
.
.
goes through midpoint of chord 
,
- 
is tangent to 
and result follows.
outside 
is cyclic
the circle with radius zero centered at 
we have that 
goes from the modpoints of 
so 
then 
is tangent t
.
so 
. It's 
is 
