Difference between revisions of "2012 USAJMO Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | By the [[Law_of_Sines| | + | By the [[Law_of_Sines|law of sines]] on triangle <math>AB'P</math>, |
<cmath>\frac{AB'}{\sin \angle APB'} = \frac{AP}{\sin \angle AB'P},</cmath> | <cmath>\frac{AB'}{\sin \angle APB'} = \frac{AP}{\sin \angle AB'P},</cmath> | ||
so | so |
Revision as of 17:55, 23 February 2019
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 law of sines 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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
2012 USAJMO (Problems • Resources) | ||
Preceded by Problem 5 |
Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.