Difference between revisions of "2016 USAJMO Problems/Problem 5"
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\cos^2B & \sin^2B & 0 \\ | \cos^2B & \sin^2B & 0 \\ | ||
\cos^2C & 0 & \sin^2C \\ | \cos^2C & 0 & \sin^2C \\ | ||
| − | \end{vmatrix}=0.</cmath> Expanding, we must prove <cmath>\sin(2A)\sin^2B\sin^2C=\cos^2C\sin^2B\sin(2C)+\sin^2C\cos^2B\sin(2B)</cmath> <cmath>\frac{\sin(2A)}{2}=\sin^2B(1-\sin^2C)\sin(2C)+\sin^2C(1-\sin^2B)\sin(2B)</cmath> (to | + | \end{vmatrix}=0.</cmath> Expanding, we must prove <cmath>\sin(2A)\sin^2B\sin^2C=\cos^2C\sin^2B\sin(2C)+\sin^2C\cos^2B\sin(2B)</cmath> <cmath>\frac{\sin(2A)}{2}=\sin^2B(1-\sin^2C)\sin(2C)+\sin^2C(1-\sin^2B)\sin(2B)</cmath> |
| + | <cmath>\begin{align*} | ||
| + | \frac{\sin(2A)+\sin(2B)+\sin(2C)}{2}&=\sin^2B\sin(2C)+\sin^2C\sin(2B)\\ | ||
| + | &=2\sin B\sin C(\sin B\cos C+\cos B\sin C) \\ | ||
| + | &=2\sin B\sin C\sin A.\end{align*}</cmath> | ||
| + | |||
| + | Let <math>x=e^{iA}, y=e^{iB}, z=e^{iC},</math> such that <math>xyz=-1.</math> The left side is equal to <cmath>\frac{x^2+y^2+z^2-\frac{1}{x^2}-\frac{1}{y^2}-\frac{1}{z^2}}{4i}.</cmath> The right side is equal to | ||
| + | <cmath>\begin{align*} | ||
| + | 2\cdot \frac{x-\frac{1}{x}}{2i}\cdot \frac{y-\frac{1}{y}}{2i}\cdot \frac{z-\frac{1}{z}}{2i}&=\frac{xyz-\frac{1}{xyz}-\frac{xy}{z}-\frac{yz}{x}-\frac{xz}{y}+\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}}{-4i}\\ | ||
| + | &=\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-x^2-y^2-z^2}{-4i},\end{align*}</cmath> which is equivalent to the left hand side. Therefore, the determinant is <math>0,</math> and <math>O,P,Q</math> are collinear. <math>\blacksquare</math> | ||
Revision as of 15:09, 21 April 2016
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. 
