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)
Let 
such that 
The left side is equal to ![\[\frac{x^2+y^2+z^2-\frac{1}{x^2}-\frac{1}{y^2}-\frac{1}{z^2}}{4i}.\]](http://latex.artofproblemsolving.com/0/d/d/0dd8f654162cb9241bd30e5e9aa299525d6d18d0.png)
The right side is equal to
which is equivalent to the left hand side. Therefore, the determinant is 
and 
are collinear. 
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
See also
| 2016 USAJMO (Problems • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAJMO Problems and Solutions | ||
