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 an interior point on the side 
of an acute-angled triangle 
. Let the circumcircle of triangle 
intersect 
again at 
and the circumcircle of triangle 
intersect 
again at 
. Let 
, 
, and 
intersect the circumcircle of triangle again at 
, 
and 
, respectively. Let 
and 
be the incentres of triangles 
and 
, respectively. Prove that 
are concyclic.
be a median in an acute-angled triangle . A point 
is chosen on the line through 
tangent to the circumcircle of 
so that 
. The segments 
and meet at 
. Prove that the circumcenter of 
lies on the line .
be an integer. The sequence 
is defined via 
, 
and![\[ a_{n+2} = ka_{n+1} + a_n \]](http://latex.artofproblemsolving.com/b/5/e/b5e3e6f9f0b0709113498266e41b372299a2986f.png)
for any positive integer 
. Prove that there are finitely many pairs 
of positive integers such that 
.
be a triangle with 
. Let be the intersection point of the internal bisector of angle 
and the circumcircle of . Let 
be the intersection point of the perpendicular bisector of with the external bisector of angle 
. Prove that the midpoint of the segment lies on the circumcircle of triangle 
.
![$a,b,c\in[0,1]$](http://latex.artofproblemsolving.com/8/b/d/8bd8f441b63556fdf318ad140bbd3e8e358e9e1d.png)
, assume 
, compute the range of![\[f(a,b,c)=(1-a^2)^2+(1-b^2)^2+(1-c^2)^2.\]](http://latex.artofproblemsolving.com/7/c/1/7c1611d9b80f5f694afed1894acd57a93aa5b31c.png)
with orthocenter , circumcenter and incenter 
, circum-radius 
, in-radius 
.
.
satisfying
for all 
.
with![\[2x^3zf(z)+yf(y) \ge 3yz^2f(x)\]](http://latex.artofproblemsolving.com/a/b/f/abf10d7e679f6f04db556424a0a0eda26f371fd4.png)
for all 
.
, such that,
for all 
.
are primes, which satisfy:
is an integer
.
be circumscribed about circle 
, and let 
be the orthocenter of . The circle touches line at . The tangent to the circle 
at meets at 
. Let be the midpoint of 
, and let the line 
meet again at 
. The tangent to parallel to meets the line 
at 
. Prove that 
is tangent to .
be the incircle of triangle and touches at . meets again at 
. Let be 
-excenter, and 
be the midpoint of 
, respectively. Prove that 
are concyclic.
be the incenter of 
(i.e. is the center of 
) .
and 
be the antipode of 
in 
.
that maps 
we get 
are collinear ,
passes through the reflection 
of 
in the perpendicular bisector of 
.
and 
,
,
are concyclic 
are concyclic (
is an isosceles trapezoid) .
be the projection of onto and let the tangent to at cut at 
Denote 
and let 
be the midpoint of 
be the homothety that sends the to the -excircle and let 
be the homothety with center and ratio 
Since 
and 
, we obtain 
Therefore, 
Because 
is isoceles, we have 
Therefore, 
are inscribed in the circle of diameter 
, and from Power of a Point we find 
are the points of tangency of with 
, respectively, then quadrilateral 
is harmonic because 
are concurrent. Therefore, 
, and because the cevians 
are concurrent, it is well-known that 
is harmonic. Since 
is the midpoint of 
, it is well-known that 
Therefore, 
are concyclic. Finally, because is harmonic, it follows that 
, implying that 
are concyclic. 
be tangent to and at 
. Let 
meet at 
. Note that 
is tangent to , since 
is harmonic. Let 
be the point such that 
. Note that 
and 
are isosceles, and furthermore that 
(with both orthogonal to 
). Additionally, 
, and 
, so 
, the -symmedian of , is parallel to 
, the -median of 
. Therefore 
are collinear. Let be the midpoint of 
; then 
, so lies on 
. Therefore![\[
MP\cdot MN=2\cdot MQ\cdot \frac{1}{2}\cdot MK = MQ\cdot MK=MB\cdot MC
\]](http://latex.artofproblemsolving.com/4/c/a/4cae5c011b2da1858a18c583d05d74f55ff1b5b1.png)
is cyclic.
is harmonic, so by a well-known property of harmonic divisions,![\[
GL\cdot GP=GD\cdot GM = GB\cdot GC
\]](http://latex.artofproblemsolving.com/8/d/a/8daed92d4edb538d73d8749a6a4b550e93ec959a.png)
is cyclic as well, implying 
is cyclic as desired.
be the projection of onto , and consider the negative inversion at fixing 
. It is not hard to see by projecting onto 
that the image of the incircle is the perpendicular bisector of 
, so the image of , 
, is the intersection of with this perpendicular bisector. Since it is well-known that 
, dilation at with ratio 
yields that 
is an isosceles trapezoid, which implies the conclusion.
and 
be real numbers . Prove that
be reals such that 
Prove that
.
We also have that 
so 
.
.
.
.
.
.
which can be easily checked that is true by Stewart's relation: 
.
.
and 
,
and 
are symmetric wrt the perpendicular bisector of 
.