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.
with orthocenter 
. Extend 
to meet 
at 
. The angle bisector of 
meets the midpoint of 
. If 
, compute the area of 
.
intersect its circumcircle 
at 
. The tangets to from 
determine a triangle 
, where 
is relative to 
and so on. Prove that lines 
are concurrent at a point 
and that belongs to the euler line of .
be a cyclic quadrilateral whose opposite sides are not parallel. Suppose points 
lie in the interiors of segments 
respectively, such that 
Let 
intersect 
at 
, and 
intersect 
at 
. Prove that lines 
and 
are either parallel or coincide.
. Prove that:
be reals such that 
Prove that 
Let be reals such that 
Prove that 
be a fixed integer. There are 
beads on a circular necklace. You wish to paint the beads using 
colors, such that among any 
consecutive beads every color appears at least once. Find the largest value of 
for which this task is 
possible.
This easy via the binomial theorem for the quantity is just 
, but how do we arrive at this using the I-E-P?
with side length of 
. The point is selected on the diagonal 
such that 
. Denote the circumcenters of 
as 
. Find the length of 
be an equilateral triangle. Let and 
be points on 
and , respectively, and let 
and 
be points on such that 
is a rectangle. If 
and the area of is 
, what is the length of 
?
is 
, so, 
is equilateral with side length . Now, in right triangle 
, 
by trigonometrical ratios (
).
from at . Its length is 
. Thus, in right triangle 
, the length of is 
.
and 
.
and 
.
Where 
such that :
units, also 
,
u
+
=
, now 
is 
, so