AoPS Academy
the point 
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
at 
, and to the lines 
and 
at 
and 
, respectively. The lines 
and 
meet at 
, and the lines 
and 
meet at 
Let 
be the point of intersection of the lines 
and , and let 
be the point of intersection of the lines 
and 
Prove that is the midpoint of 
opposite the vertex 
is the circle that is tangent to the line segment , to the ray beyond 
, and to the ray beyond 
.)
be complex numbers with 
. Find the greatest possible value of the expression 
digit numbers are there
? and 
?
and 
then is this uniformly convergence on 
comment on uniformly convergence on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
where in general it is should be uniformly convergence.
and trying to solve
integer matrices having the following properties:
the entries of are (positive) prime numbers,
there exists a integer matrix such that 
and the determinant of is the square of a prime number.
such that the equation in the unknown 
:
admits 
roots: 
s.t. 
.
be a triangle, and let 
and 
be the orthogonal projections of onto the internal bisectors from and . Prove that 
is parallel to .
be the incenter. Then 
from 
concyclic. Then 
so 
, so 
.
hits at 
, then is the reflection of on the -bisector and -bisector, resp., and consequently is the image of 
after a homothety from with ratio 
, implying 
.
be feet of on angle bisectors of and respectively. We have ![\[\angle IXY = \angle IAY = \angle CAY - \angle CAI = 90^{\circ} - \frac{C}{2} - \frac{A}{2} = \frac{B}{2}.\]](http://latex.artofproblemsolving.com/0/3/4/034c95103a410189f239bf1a9969a66fdfff6cc6.png)
= 
.
+ 
= 
.
= 
= 
- 
. = 
- 
and = - 
- 
. (from the sum of the angles of the triangle FAC) + 
+
= (*) = 2
= 2
,...(**)
=
, so we're done.
![\[\sum_{m=1}^\infty\sum_{n=1}^\infty\dfrac{m^2n}{3^m(n3^m+m3^n)}.\]](http://latex.artofproblemsolving.com/7/d/6/7d6247fbcfd6eac156ce774f727d7e95c470de52.png)
![$\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{lcm(1,2,3,...,n)}=e$](http://latex.artofproblemsolving.com/a/1/3/a139e5cb990f3386de109c701f8edc71b4054712.png)
