[/quote]
,
satisfying 
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.
be an acute triangle with 
,
be the projection from 
.
be a point on the extension of 
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 
,
and 
such that 
.
and 
intersect the circumcircles of 
and 
at 
and 
,
and 
intersect 
and 
at 
.
.
such that 
![\[
C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p}
\]](http://latex.artofproblemsolving.com/4/0/7/4072235afbaa1f2d8d4c9ca044615a113a4306ce.png)
for 
,
is the binomial coefficient 
choose 
.
satisfy that: 
and 
.
intersect at points 
and 
and 
.
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)
be a function such that for every regular 
-
we have 
.
for all reals 
.
be positive integers, where 
.
such that
.
.
is the sum of digits of 
taken in base 
)
Y by Adventure10, Mango247
The quadrilateral 
is inscribed in a circle which has diameter 
. Points 
and 
are symmetric to 
and 
with respect to the lines and 
respectively. If the lines 
and intersect at 
, and the lines and 
intersect at 
, prove that 
is perpendicular to .
is inscribed in a circle which has diameter 
. Points 
and 
are symmetric to 
and 
with respect to the lines and 
respectively. If the lines 
and intersect at 
, and the lines and 
intersect at 
, prove that 
is perpendicular to .
.
(i.e. the circumcircle of 
we get
.
thus
.
imply that 
thus 
,
is the internal bisector of 
.
thus 
is harmonic.
is a harmonic quadruple.Thus
.
.
and 
are symmetric to 
, 
,
.
is circumscribed. Than we need to prove that 
.
.
and 
.
and 
.
.
