Difference between revisions of "2012 USAMO Problems/Problem 5"
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<cmath>\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} = 1,</cmath> | <cmath>\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} = 1,</cmath> | ||
so by [[Menelaus'_Theorem|Menelaus's theorem]], <math>A'</math>, <math>B'</math>, and <math>C'</math> are collinear. | so by [[Menelaus'_Theorem|Menelaus's theorem]], <math>A'</math>, <math>B'</math>, and <math>C'</math> are collinear. | ||
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==See also== | ==See also== | ||
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{{USAMO newbox|year=2012|num-b=4|num-a=6}} | {{USAMO newbox|year=2012|num-b=4|num-a=6}} | ||
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 13:01, 18 June 2020
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.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.