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)
be a convex hexagon with 
and 
. which is equidistant from sides 
and 
.
and 
are the centers of mass of 
and 
, show that 
.
be a convex hexagon with and .
which is equidistant from sides and . and are the centers of mass of and , show that .
, , to form an equilateral triangle; the only possible point is its center.
, , to form an equilateral triangle; the only possible point is its center. is the incenter.
such that 
where 
and 
Prove that
,
is the triangle formed by lines 
.
are the centroids of triangles 
.
is similar to three triangles above.
.
,
,
,
is similar to the previous three (this is probably the only statement ever where complex numbers give the cleanest proof). But spotting it requires you to think synthetically about the configuration. I'm not sure how a "blind" complex numbers approach would work here. To be fair, if it weren't for the complex numbers part, this problem absolutely wouldn't have landed here.