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.
be the circumcircle of acute triangle 
. Two tangents of from 
and 
intersect at 
, 
and 
intersect at 
. Point 
, 
are on 
and 
such that 
and 
.
are concyclic.
the centre of the circle passing through . 
, 
are difined similarly. Prove that 
, 
, 
are concurrent.
> 
that satisfy 
, for all 
€ .
be a fixed positive integer. Given a set 
of 
points in the plane such that no three are collinear and no four concyclic, let 
be the number of circles 
that contain 
in their interior, and let ![\[m(S)=a_1+a_2+\cdots + a_n.\]](http://latex.artofproblemsolving.com/4/9/b/49b928b1f20e2d1799a1234c7f25a2bc3d62d0dd.png)
Prove that there exists a positive integer 
depending only on 
such that the points of 
are the vertices of a convex polygon if and only if 
such that 
.
, then prove that the following inequality holds:![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
such that has exactly 2000 prime divisors and divides 
?
and 
is a subset of 
such that if 
then 
Find the smallest of 
that satisfy the following property: if 
, 
, 
, 
are all the positive divisors of 
with 
, then 
divides 
for every 
.
, in which 
: 
, but the proof is important.
and 
Prove that
satisfying ![$$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$$](http://latex.artofproblemsolving.com/f/6/e/f6ed10f7fff1cc94edd8f451e75718a0916a8bfa.png)
holds for all 
be non-negative real numbers such that 
Prove that
and 
are chosen on the sides and of the triangle 
. The segments 
and 
intersect at the point 
. Given that 
and 
, prove that the points , , , and lie on the same circle.
and are chosen on the sides and of the triangle . The segments and intersect at the point . Given that and , prove that the points , , , and lie on the same circle.
suchthat 
isoscles 
isoscles trapizum 
isoscles
foom congrunty of 
Cyclic eqilateral
be the perpendicular bisector of . Let 
be the symmetric of wrt . Since 
is the symmetric of triangles 
and 
are congruent, so 
and 
. Also, since 
and are collinear then 
and are also collinear. Hence we have 
and therefore 
and we are done.
in the plane such that 
is a parallelogram.
doing simple angle chase using isoceles property of triangle and paralllel lines IS midpoint of arc
in thecircumcircle of 
; 
is cyclic forcing 
also to be..
and 
.
