with no parallel sides is inscribed in a circle 
.
are inscribed in triangles 
,
and 
,
,
,
.
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
such that 
for all 
be the 
-
and 
be the intersection points of 
and 
be the projections of 
intersects the tangent line at 
at a point 
.
.
and 
: Prove that :
board and made 
moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called 
if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
is lovely if there exist a positive integer 
and (not necessarily distinct) positive integers 
,
,
,
such that 
and 
for 
.
,
be an odd prime. Prove that:![\[\displaystyle\sum_{k=1}^{p-1}k^{2p-1} \equiv \frac{p(p+1)}{2} \pmod{p^2}\]](http://latex.artofproblemsolving.com/a/5/5/a55e75d4d9769ceec3a3ce359aac676c7c7875fd.png)
,
,
,
'![\[n=a_1+a_2+\dots+a_m,\]](http://latex.artofproblemsolving.com/6/c/2/6c2a979a5ff6c40bf2d7152c8c32088fc36f848b.png)
where 
,
,
,
.
be a function such that
for all 
.
.
Y by NO_SQUARES, Rounak_iitr
The point 
is the middle of the arc 
of the circumcircle of the triangle 
. The point 
is the center of its inscribed circle 
. The line 
intersects the circumcircle of the triangle at 
for the second time. Prove that the circle passing through the midpoints of the segments 
and 
is tangent to the circle which is symmetric to with respect to 
.
is the middle of the arc 
of the circumcircle of the triangle 
. The point 
is the center of its inscribed circle 
. The line 
intersects the circumcircle of the triangle at 
for the second time. Prove that the circle passing through the midpoints of the segments 
and 
is tangent to the circle which is symmetric to with respect to 
.
which finishes by feurbach tangency, and so it suffices to show that 
call it 
.
are collinear and 
at 
then well known 
,
over 
lies on nagellian but follows as 
is refl of nagellian in 
'
.
and hence 
.
and thus we are done as this clearly implies that 
.
