.
,
and 
,
at 
.
is tangent to 
.
and 
be the midpoints of segments 
,
intersects 
and 
,
.
such that for any 
:
,
units horizontally (left or right); or
and 
,
to 
using at most 
moves.
such that 
for all positive integers 
.
and 
,
,
,
and 
respectively, and the points 
are collinear.
where 
is the inverse gamma function, and 
is the inverse of 
is a polynomial with integer coefficients, with 
odd. Suppose that 
for all 
Prove that 
is nonzero for all 
,
Prove that 
.
converges then so does 
for integers,
= 
= 
, as 
the 
which gets us the 
form therefore L'hopital came to my mind , which might be a completely wrong intuition, anyway what should i do to find that limit
![\[I=\int\limits_{-\infty}^{\infty}e^{x-10\cosh (2x)}\mathrm dx.\]](http://latex.artofproblemsolving.com/5/b/6/5b61a79360c164fcd935eb613e12ab056847a93c.png)
Y by Adventure10
The inscribed circle 
of the non-isosceles acute-angled triangle 
touches the side 
at the point 
. Suppose that 
and 
are the centres of inscribed circle and circumcircle of triangle respectively. The circumcircle of triangle 
intersects 
at the points 
and 
. Prove that 
is equal to the radius 
of .
of the non-isosceles acute-angled triangle 
touches the side 
at the point 
. Suppose that 
and 
are the centres of inscribed circle and circumcircle of triangle respectively. The circumcircle of triangle 
intersects 
at the points 
and 
. Prove that 
is equal to the radius 
of .
is the reflection of 
on the perpendicular bisector of 
,
.
is cyclic. So 
.
in 
.
we will be done since this implies 
being cyclic and therefore 
.
the intersection of 
.
.
.
.
which is just simple angle chase.
So I moved your post here and deleted the old thread.
.
are symmetric in 
,
.
we get 
so 
as required.
.
.
.
.
is cyclic so 
which implies 
.
which gives 
as 
be the orthocenter. Then since 
hence 
is an isosceles trapezoid, so 
.
