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.![\[
C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p}
\]](http://latex.artofproblemsolving.com/4/0/7/4072235afbaa1f2d8d4c9ca044615a113a4306ce.png)
for 
, where 
is the binomial coefficient 
choose 
.
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 
).
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)
, is the intersection point of diagonals and is the midpoint of arc . Prove that line 
passes through the midpoint of segment , if and only if lines 
are parallel.
, is the intersection point of diagonals and is the midpoint of arc . Prove that line passes through the midpoint of segment , if and only if lines are parallel.
If part: If 
, then is perpendicular bisector of , so it's perpendicular bisector of ,too. So done. Only if part: Assume 
and 
. Let 
and 
. So Sine Law in 
and 
gives that 
. And also Sine Law in 
and 
gives that 
. So from here we can easily get that 
. So .Done!
then is perp. bisector of , then it is also perp. bisector of , as desired. passes through midpoint of , 
and 
intersect at point 
and, 
and 
intersect at point 
, then from cheva theorem in triangle 
we have 
and we know that 
, then 
is isosceles trapezoid, then 
, implies that is isosceles trapezoid too, which completes the proof.
such that 
where 
and 
Prove that
