2015 USAJMO Problems/Problem 5
Contents
Problem
Let be a cyclic quadrilateral. Prove that there exists a point on segment such that and if and only if there exists a point on segment such that and .
Solution 1
Note that lines are isogonal in , so an inversion centered at with power composed with a reflection about the angle bisector of swaps the pairs and . Thus, so that is a harmonic quadrilateral. By symmetry, if exists, then . We have shown the two conditions are equivalent, whence both directions follow
Solution 2
All angles are directed. Note that lines are isogonal in and are isogonal in . From the law of sines it follows that
Therefore, the ratio equals
Now let be a point of such that . We apply the above identities for to get that . So , the converse follows since all our steps are reversible.
Beware that directed angles, or angles , are not standard olympiad material. If you use them, make sure to define and indicate what they mean. [Mentioned in Evan Chen's Geometry Olympiad book.]
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See Also
2015 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |