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 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.