[/quote]
,
,
lies on side 
and point 
lies on side 
.
and 
be points on segments 
and 
,
is parallel to 
.
be a point on line 
,
.
be the point on line 
,
.
,
.
meet at point 
.
on line 
be 
.
meet 
respectively. Let 
be the midpoint of 
lie on a circle.
such that 
satisfying 
where 
.
cannot be a perfect cube of a positive integer.
has an incenter 
Suppose 
.
.
meets 
again at point 
.
and 
.
lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet 
and 
.
.
and 
with another positive integer 
,
is a perfect square. For example, if the number on the blackboard is 
,
,
,
,
.
,
and orthocenter 
,
be an arbitrary point on the circumcircle of triangle 
and that line 
is not parallel to line 
is perpendicular to 
.
is a diameter, and 
and 
are parallel. Let 
.
and 
are perpendicular.
and 
are perpendicular.
be positive integers and 
be an odd prime such that:![\[
a^p \equiv 1 \pmod{p^n}.
\]](http://latex.artofproblemsolving.com/3/4/8/34815b8211ded2f37b0653525dc4b436356e6eee.png)
Prove that:![\[
a \equiv 1 \pmod{p^{n-1}}.
\]](http://latex.artofproblemsolving.com/0/7/a/07a4ed2ab3c94f41028080f7789cfb6e2973955c.png)
and 
and 
)
Prove that if lines 
are similar.
satisfying![\[\begin{aligned}
\begin{cases}
a+b+c+d=0,\\
a^2+b^2+c^2+d^2=12,\\
abcd=-3.\\
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/8/b/1/8b1923e0303c0aa79332ee4540ee1885a9cf3f08.png)
crosses are placed. A set of 2025 cells we will call balanced if no two of them are in the same row or column. It is known that any balanced set has at least 
crosses.
,
onto 
past 
.
be on the perpendicular bisector of 
such that 
are on the same side of ![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of 
and 
and 
such that 
.
and 
intersect the circumcircles of 
and 
at 
and 
,
and 
intersect 
and 
at 
.
Y by Adventure10, Mango247
Let 
be the points of tangency of the excribed circles of triangle 
with the sides of 
Let 
be the circumradius of triangle 
Show that ![\[
R'=\frac{1}{2r}\sqrt{2R\left(2R-h_{a}\right)\left(2R-h_{b}\right)\left(2R-h_{c}\right)}\]](//latex.artofproblemsolving.com/0/e/f/0ef9900a7f960e707c47b06d2904aea999218447.png)
where as usual,
is the circumradius of 
r is the inradius of and 
are the lengths of altitudes of
be the points of tangency of the excribed circles of triangle 
with the sides of 
Let 
be the circumradius of triangle 
Show that ![\[
R'=\frac{1}{2r}\sqrt{2R\left(2R-h_{a}\right)\left(2R-h_{b}\right)\left(2R-h_{c}\right)}\]](http://latex.artofproblemsolving.com/0/e/f/0ef9900a7f960e707c47b06d2904aea999218447.png)
where as usual,
is the circumradius of 
r is the inradius of and 
are the lengths of altitudes of 
and 
Perpendicular lines to 
through 
concur at the reflection of the incenter 
If 
touches 
then by Euler theorem we have![$\frac{[ \triangle A'B'C']}{[ \triangle ABC]}=\frac{[ \triangle XYZ]}{[ \triangle ABC]}= \frac{p(I,(O))}{4R^2}=\frac{2Rr}{4R^2}$](http://latex.artofproblemsolving.com/3/e/9/3e9d92a2285a9c08aa97cc65ad5874f8b0e1b94c.png)
![$\Longrightarrow [ \triangle A'B'C']=\frac{r}{2R} \cdot [ \triangle ABC]$](http://latex.artofproblemsolving.com/8/6/4/864bffbed67efa58144406b6bef35f7c19fc7def.png)
![$[\triangle A'B'C']=\frac{r}{2R} \cdot [ \triangle ABC]=\frac{a'b'c'}{4R'} \Longrightarrow R'=\frac{a'b'c'}{[ \triangle ABC]} \cdot \frac{R}{2r} \ \ (1)$](http://latex.artofproblemsolving.com/0/0/b/00b1c7692a96787848847dad91b2fe58ba0b0670.png)
we have
![$a'b'c'=[\triangle ABC] \sqrt{ \frac{2(2R-h_a)(2R-h_b)(2R-h_c)}{R}} \ \ (2) $](http://latex.artofproblemsolving.com/2/4/d/24d99937cdc33cbda80ce254def2df6778626515.png)
and 
we get![$R'=\frac{[\triangle ABC]}{[\triangle ABC]}\sqrt{ \frac{2(2R-h_a)(2R-h_b)(2R-h_c)}{R}} \cdot \frac{R}{2r}$](http://latex.artofproblemsolving.com/1/8/9/18991e06a28a1c7ef820d680b4727de8247c57af.png)
