2016 USAJMO Problems/Problem 5
Problem
Let 
be an acute triangle, with 
as its circumcenter. Point 
is the foot of the perpendicular from 
to line 
, and points 
and 
are the feet of the perpendiculars from to the lines 
and 
, respectively.
Given that ![\[AH^2=2\cdot AO^2,\]](http://latex.artofproblemsolving.com/e/d/1/ed135ba8f97efbea1d411b874540ea32498b9919.png)
prove that the points 
and are collinear.
Solution
We will use barycentric coordinates with respect to 
The given condition is equivalent to 
Note that ![\[O=(\sin(2A):\sin(2B):\sin(2C)), P=(\cos^2B,\sin^2B,0), Q=(\cos^2C,0,\sin^2C).\]](http://latex.artofproblemsolving.com/4/1/4/41485f502669b8baf19b38236435b7bbf69d940c.png)
Therefore, we must show that ![\[\begin{vmatrix} \sin(2A) & \sin(2B) & \sin(2C) \\ \cos^2B & \sin^2B & 0 \\ \cos^2C & 0 & \sin^2C \\ \end{vmatrix}=0.\]](http://latex.artofproblemsolving.com/6/2/f/62fab89cbbc97e4fb78e08613d17dade9ab89ac7.png)
Expanding, we must prove ![\[\sin(2A)\sin^2B\sin^2C=\cos^2C\sin^2B\sin(2C)+\sin^2C\cos^2B\sin(2B)\]](http://latex.artofproblemsolving.com/1/c/6/1c69467b07ddf1d776b4afb234bdc30f40189bc9.png)
![\[\frac{\sin(2A)}{2}=\sin^2B(1-\sin^2C)\sin(2C)+\sin^2C(1-\sin^2B)\sin(2B)\]](http://latex.artofproblemsolving.com/5/c/a/5cae51d657fef6da0c0ccd03951fb9b174138479.png)
(to be finished)
