and 
Prove 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 
.
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)
Y by Abdollahpour, Adventure10, Mango247, and 2 other users
Let 
be a point on the longest side 
of a triangle 
. The perpendicular bisectors of 
and 
intersect 
and 
respectively in 
and 
. Prove that the circumcenter 
of 
lies on the circumcircle of triangle 
.
be a point on the longest side 
of a triangle 
. The perpendicular bisectors of 
and 
intersect 
and 
respectively in 
and 
. Prove that the circumcenter 
of 
lies on the circumcircle of triangle 
.
) so angle TMK =angle KNC=angle C & angle TKM=angle TNM=angle A, so triangle KTM and triangle ABC are similar.
is cyclic so 
.
and 
so 
.
or 
I
so 
is cyclic.
\Rightarrow \angle KOM=\angle KBM=90^0$](http://latex.artofproblemsolving.com/9/b/9/9b920b4f4b68f9baf5a96ad6b06d3e3745b918c3.png)
so 
and 
be the midpoints of 
and 
intersect at 
,
and 
intersect at 
.
.
and 
.
,
)
.
and that 
is a parallelogram, so 
.
,
,
,
is cyclic, as desired.
and 
be the midpoints of 
is cyclic, so it is enough to show that 
,
.
.
are in this order on side 
are in this order on side 
are in this order on side 
are in this order on side 
,![\[ \frac{KT}{MS} = \frac{c-\frac{x}{\cos A}}{\frac{y}{\cos C} - a} = \frac{\cos C}{\cos A} = \frac{OT}{OS} \]](http://latex.artofproblemsolving.com/e/d/1/ed15c59066e884236f91e168af69c39fb5666cd6.png)
because multiplying yields that this is equivalent to 
,
,
vary uniformly on 
respectively, so the circumcircle of 
passes through a fixed point other than 
.
it is clear that 
so we are done!
point 
with 
or anything equivalent. Now perpendicular bisectors of 
are essentially a pencil of lines that move uniformly too (it's easy to work out what that means), so their intersections with fixed lines also move in the same fashion.
passes through a fixed point in this case. So, we're good by checking few cases.
in the same ratio as 
;
.
.
.
intersect 
And, 
.
Hence, 
labbeling.
)
and 
we have 
and 
so 
is true.
