2012 USAMO Problems/Problem 5
Problem
Let be a point in the plane of triangle , and a line passing through . Let , , be the points where the reflections of lines , , with respect to intersect lines , , , respectively. Prove that , , are collinear.
Solution
By the sine law on triangle , so
Similarly, Hence,
Since angles and are supplementary or equal, depending on the position of on , Similarly,
By the reflective property, and are supplementary or equal, so Similarly, Therefore, so by Menelaus's theorem, , , and are collinear.
Solution 2, Barycentric (Modified by Evan Chen)
We will perform barycentric coordinates on the triangle , with , , and . Set , , as usual. Since , , are collinear, we will define and .
Claim: Line is the angle bisector of , , and . This is proved by observing that since is the reflection of across , etc.
Thus is the intersection of the isogonal of with respect to with the line ; that is, \[ B' = \left( \frac pk \frac{b^2}{\ell}
: \frac{b^2}{\ell} : \frac{c^2}{q} \right). \]
Analogously, is the intersection of the isogonal of with respect to with the line ; that is, \[ A' = \left( \frac{p}{\ell} \frac{b^2}{k}
: \frac{b^2}{k} : \frac{c^2}{q} \right). \]
The ratio of the first to third coordinate in these two points is both , so it follows , , and are collinear.
~peppapig_
See also
2012 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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