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 intersection of 
and If the tangent from to circle (not the other) is' not parallel to 
by Brianchon theorem 
, and tangent from to circle are concurrent. is parallel to contradiction. So is constant.
. A very nice solution.
to line 
which maps 
is a perspectivity. In order to do this we must show that it's a homographic (preserving the cross-ratio) transformation and it maps 
.
(from line to line ; it's wel-known that this sort of transformation is homographic (even if 
is a conic, not only a circle)) and 
(from line to line ).. In this case 
, and we're done.
and a point (which doesn't belong to 
or 
). Take a variable line 
through which cuts 
in 
. The function 
is called perspectivity and it'sa a particular case of function from to which preserves the cross-ratio of four points.
s.t. 
preserves the cross-ratio, it can be shown (I won't show it here; try searching the Net for stuff like "homographic transformation", "perspectivity" etc.) that this is a perspectivity iff 
, where 
. This is exactly what I did here.
is a perspectivity from line to line (i.e. passes through a fixed point), we must show that by this transformation the intersection point of the two lines and turns into itself. This means we want to prove that when coincides with 
, by constructing 
, we also find it to coincide with . That's why the last line begins with 
and ends with 
(what we want to show is the implication 
).
is intersection of and If the tangent from to circle (not the other) is' not parallel to by Brianchon theorem , and tangent from to circle are concurrent. is parallel to contradiction. So is constant. , and tangent from to circle are concurrent by Brianchon theorem
such that 
where 
and 
Prove that
