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.
and 
. Prove that
Equality holds when 
or 
and . Prove that
Equality holds when or 
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?
satisfies 
. If 
, find the maximum possible value of 
.
and a constant 
, it is given that 
and 
are also friendly polynomials. Prove that 
.
is incenter of 
and 
inresects 
at 
.In circumcircle of , 
is midpoint of arc 
and 
intersect this circle at 
. If 
intersects 
at 
, prove that: 
.
be the point at infinity on , 
the point at infinity on . We first show that 
lie on a hyperbola. Let 
be the midpoint of arc , we claim 
is the isogonal conjugate of the desired hyperbola. It only suffices to show 
lie on this hyperbola as it is well known it passes through 
. However, 
by Fact 5, hence after reflecting about 
, we see that is the isogonal conjugate of by alternate angles implying parallelism. Similarly, is the isogonal conjugate of . Finally, the isogonal of is and because is on 
, is sent to the point at infinity along the isogonal conjugate of our hyperbola. Hence we simply want 
. But well known 
(somewhere in EGMO), and also 
(all of this is easy angle chase). Finally Pascal on 
finishes.
. Now use ratio lemma on triangle 
.![\[\frac{DN}{MN}=\frac{DA}{AM}\times \frac{\sin \angle DAT}{\sin\angle MAT}\]](http://latex.artofproblemsolving.com/1/a/8/1a8eba0efad6ff5719563d3b13a7ac8ea8b4009d.png)
We want this to be equal to 
. Well known 
if is midpoint of arc . Now, 
. Thus we just want to prove that![\[\frac{CI}{AM}\times\frac{\sin\angle DAT}{\sin \angle A}=1\]](http://latex.artofproblemsolving.com/1/5/2/152f8ecf3b811a05b58c62d62d38d2c44a5f47de.png)
or as 
by Fact 5,![\[\frac{CI}{AM}=\frac{\sin\angle A}{\sin\angle DAT}=\frac{\sin\angle IMC}{\sin\angle CIM}\]](http://latex.artofproblemsolving.com/2/b/3/2b3ea8e26e112907e716e5c4d5bb07c0998443d4.png)
because of bowtie, and also because 
and also because 
.
and 
.
and 
.
Where 
such that :
and let 
be the point at infinity of 
.
finishes.