AoPS Academy
, inscribed in the circle 
, satisfies 
and 
, and 
is the intersection point of the diagonals 
and 
. The circle with center 
and radius 
intersects in two points 
and 
. Prove that the line 
is tangent to the circles with diameters 
and 
.
satisfying the equation![\[
(f(x+y))^2= f(x^2) + f(2xf(y) + y^2), \quad \forall x, y \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/2/5/e/25eed911f303ed7cc4e03765a2f943ad4641c451.png)
be a triangle inscribed in a circle, where the tangents to the circle at points 
and 
intersect at the point 
. Let 
be a point on the arc (not containing ) such that 
and 
. Let the lines 
and 
intersect at point 
. Let 
be the reflection of with respect to the line . The lines 
and 
intersect at point 
, and intersects the circumcircle of again at point 
. lies on the circumcircle of 
.
be non-negative real numbers such that 
Prove that
![$ [AMR]=\frac{MR}{MK}\cdot\frac{AM}{AB}\cdot\frac{BK}{BC}[ABC]$](http://latex.artofproblemsolving.com/3/d/4/3d478873eaf22f1bbd7c298919b1d36c75c9c00c.png)
. We will find the 
factor in the expression for ![$ [ALF]$](http://latex.artofproblemsolving.com/e/0/9/e096650c8d8d67f81156af1e2958cb92f34ba3c5.png)
. By AM-GM, 
. We similarly maximize the other factors, and the answer follows.
![\[\frac{[AME]}{[ABC]}=\frac{[AME]}{[AMK]}\times\frac{[AMK]}{[ABK]}\times\frac{[ABK]}{[ABC]}=\frac{ME}{MK}\times\frac{AM}{AB}\times\frac{BK}{BC}.\]](http://latex.artofproblemsolving.com/c/8/b/c8b28ed74502526868a67b3025a6633d75400a27.png)
Hence, by AM-GM we have ![\[\sqrt[3]{\frac{[AME]}{[ABC]}}\le\frac{ME/MK+AM/AB+BK/BC}{3}.\]](http://latex.artofproblemsolving.com/a/b/0/ab09503bbd12fb2c9f008454f0fbbc6563bbfdee.png)
![\[\frac{[CKE]}{[ABC]}=\frac{KE}{MK}\times\frac{KC}{BC}\times\frac{MB}{AB}\]](http://latex.artofproblemsolving.com/0/e/2/0e2db2b84d18623536d3f96184c1bed6fa1fc999.png)
and by AM-GM ![\[\sqrt[3]{\frac{[CKE]}{[ABC]}}\le\frac{KE/MK+MB/AB+KC/BC}{3}.\]](http://latex.artofproblemsolving.com/2/9/d/29dd316a43ad821e625f5327b33370d1013c2e44.png)
![\[\sqrt[3]{[AME]}+\sqrt[3]{[CKE]}\le\sqrt[3]{ABC}.\]](http://latex.artofproblemsolving.com/4/7/5/4758e9db9a77c581223c178dbc6595a6e92233b2.png)
We repeat the same procedure for the other pairs. If we denote the areas of the six smaller triangles as 
we have ![$\sum \sqrt[3]{A_i}\le 3\sqrt[3]{[ABC]}.$](http://latex.artofproblemsolving.com/3/9/5/395c775da1f7b3ef367fdab422071664bd953f66.png)
![\[\sum\sqrt[3]{A_i}\ge 6\left(\prod\sqrt[3]{A_i}\right)^{\frac{1}{6}}.\]](http://latex.artofproblemsolving.com/8/a/6/8a69a07bda94464c39367f6407dcdaca555ce401.png)
![$\sqrt[3]{[ABC]}\le 2\left(\prod\sqrt[3]{A_i}\right)^{1/6}.$](http://latex.artofproblemsolving.com/6/c/d/6cdf8ecf66b1bb7cfc4bc39bc282ff12af0acf91.png)
Cubing gives ![\[\left(\prod A_i\right)^{\frac{1}{6}}\le\frac{1}{8}[ABC].\]](http://latex.artofproblemsolving.com/0/a/3/0a35f77f6f068a1f2ac8a96880bcecdc5d16d153.png)
We are done.
choose any points 
and 
If 
and 
denote the areas of the triangles 
and 
respectively, show that![\[ E \geq 8 \cdot \sqrt [6]{E_1 E_2 E_3 E_4 E_5 E_6}.
\]](http://latex.artofproblemsolving.com/e/c/0/ec080b923fa671003c18e623350f6bfdb70614c7.png)
