Difference between revisions of "1974 USAMO Problems/Problem 5"
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 18:57, 3 July 2013
Problem
Consider the two triangles and shown in Figure 1. In , . Prove that .
Solutions
Solution 1
We rotate figure by a clockwise angle of about to obtain figure :
Evidently, is an equilateral triangle, so triangles and are congruent. Also, triangles and are congruent, since they are images of each other under rotations. Then Then by symmetry,
But is composed of three smaller triangles. The one with sides has area . Therefore, the area of is Also, by the Law of Cosines on that small triangle of , , so by symmetry, Therefore But the area of triangle is . It follows that , as desired.
Solution 2
Rotate degrees clockwise about to get . Observe that is equilateral, which means . Also, are collinear because and . The resulting has side lengths and the angle opposite side has magnitude . If we perform the rotation about points and , we get two triangles. One has side lengths and the angle opposite side has magnitude , and the other has side lengths and the angle opposite side has magnitude . These three triangles fit together because . The result is an equilateral triangle of side length .
Solution 3
As in the first solution, we rotate and establish that .
Let and be points on and , respectively, such that lies on and . We note that . The rotation then takes to , so . It follows that , , .
Since and , is cyclic. By Ptolemy's theorem, Finally, , as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1974 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
- <url>viewtopic.php?t=33494 Discussion on AoPS/MathLinks</url>
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