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.
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 
.)
such that if 
and 
are distinct rational numbers satisfying 
or 
, then 
? Justify your answer.
be a positive integer. Consider the following equation: ![\[ \{x\}+\{2x\}+ \dots + \{nx\} = \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2nx \rfloor\]](http://latex.artofproblemsolving.com/8/2/9/8299b3ceb1650418c0a373536cb1d2931900efd9.png)
a) For 
, solve the given equation in 
., the equation has at most 
real solutions.
and 
intersect at two points 
and 
. Let 
be the line tangent to these circles at 
and 
, respectively, so that lies closer to than . Let 
be the line parallel to and passing through the point , with 
on and 
on . Lines 
and 
meet at 
; lines 
and meet at 
; lines 
and meet at 
. Show that 
.
is such that 
for any positive integers 
. Assume there exists a positive integer 
such that 
for all positive integers 
. Determine all possible values of 
.
satisfy that 
. Prove that: 
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of are either all odd or all even.
be real numbers . Prove that : There exist positive integer 
such that 
Where 
be an isosceles triangle (
) with incenter 
. Circle 
passes through and and is tangent to 
. intersects and circumcircle of at 
and 
, respectively. Let be the midpoint of and 
be the midpoint of 
. Prove that 
, 
and are concurrent.
and 
of triangle intersect in 
. A perpendicular 
from to 
is drawn. Circumcircles and 
intersect at and 
.
.
intersect at . Denote 
the antipodal points of in 
and 
, respectively. From we draw a line parallel to side intersecting again at . The main claim is that 
are collinear (Which proves the problem statement). That being said, we're going to point out that 
.
, 
should be collinear. A final observation would be that as and are diametrically opposite in 
, we should have 
, implying 
. Now, we will do some angle chasing:![\[ \angle QXL = \angle ZXL = \angle ZBL = \angle ZBC - \angle LBC = ... = 90^{\circ} - \angle ABC \]](http://latex.artofproblemsolving.com/5/2/4/524c0d16d3d921309924d55fcf2716f8c87b4072.png)
![\[ \angle QXT = \angle QHT = \angle THF - \angle QHF = \angle BHD - \angle ACF = ... = 90^{\circ} - \angle ABC \]](http://latex.artofproblemsolving.com/b/9/e/b9ecbafcc9960de22572b6d83f17238af90ac937.png)
And we should be done! 
isn't something you see everyday 
and 
.
