Difference between revisions of "2016 USAJMO Problems/Problem 5"
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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}\ | 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> | &=\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> | ||
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+ | {{MAA Notice}} | ||
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+ | ==See also== | ||
+ | {{USAJMO newbox|year=2016|num-b=4|num-a=6}} |
Revision as of 15:32, 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.
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 |