AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
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. for which it is always possible to color crosses in two colors such that any balanced set has crosses of both colors.
be an acute triangle with 
, and let 
be the projection from 
onto 
. Let 
be a point on the extension of 
past such that 
. Let 
be on the perpendicular bisector of 
such that and 
are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across 
and 
be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and 
, respectively. Let 
and 
intersect 
and 
at 
and 
. Let 
intersect at 
. Prove that 
.
![\[
C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p}
\]](http://latex.artofproblemsolving.com/4/0/7/4072235afbaa1f2d8d4c9ca044615a113a4306ce.png)
for 
, where 
is the binomial coefficient 
choose 
.
satisfy that: 
and 
. Find the maximum: 
be a convex quadrilateral. Suppose that the circles with diameters and 
intersect at points 
and . Let 
and 
. Prove that the points , , and are concyclic. and are not the diagnols)
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 
-gon 
we have 
. Prove that 
for all reals 
.
be positive integers, where 
. It is given that there exist a positive integer 
such that
. Prove that the above equation is true for all positive integers 
. (Here 
is the sum of digits of 
taken in base 
).
, and let 
and 
be the feet of altitudes from vertices 
and 
to the opposite sides. Denote 
and 
be the circumcenter and orthocenter of triangle 
. Given that 
, prove that 
is a parallelogram.
. Since 
is isosceles, the perpendicular bisector through 
and meets at . With the similar reasoning, we can prove that the perpendicular bisector through and will meet at . Therefore, since 
and 
, then 
is a parallelogram.
. So, 
and are isosceles. Let be the midpoint of . Since is isosceles, 
. Since is the midpoint of chord , 
. Hence,
are collinear and 
. Since 
. Similarly, 
so is a parallelogram, as desired. 
be the mid-points of sides 
, respectively,
and 
are right isosceles triangles,
Let ![\[\begin{vmatrix}
1 & 1 & 0 \\
1 & \sin 2B & \sin 2C \\
1 & 0 & \tan C
\end{vmatrix}
=
\sin 2B \tan C - [\tan C -\sin 2C] =0 \implies \cos 2x \cos x = \cos x -\sin x \sin 2x
\]](http://latex.artofproblemsolving.com/e/0/4/e04f8336b1fbac44f2fc0e9259845db24e309a4d.png)
Which is true, hence, 
and 
is right-isosceles 
, but since, 
satisfying 
such that 
.
lie on the perpendicular bisector of 
lie on the perpendicular bisector of 