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 16:09, 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 prove that the points
and
are collinear.
Solution
We will use barycentric coordinates with respect to The given condition is equivalent to
Note that
Therefore, we must show that
Expanding, we must prove
Let such that
The left side is equal to
The right side is equal to
which is equivalent to the left hand side. Therefore, the determinant is
and
are collinear.