1996 USAMO Problems/Problem 5
Problem
Let be a triangle, and an interior point such that , , and . Prove that the triangle is isosceles.
Solution
Solution 1
Clearly, and . Now by the Law of Sines on triangles and , we have and Combining these equations gives us Without loss of generality, let and . Then by the Law of Cosines, we have
Thus, , our desired conclusion.
Solution 2
By the law of sines, and , so .
Let . Then, . By the law of sines, .
Combining, we have . From here, we can use the given trigonometric identities at each step:
The only acute angle satisfying this equality is . Therefore, and . Thus, is isosceles.
Solution 3
If then by Angle Sum in a Triangle we have . By Trig Ceva we have Because is monotonic increasing over , there is only one solution to the equation. We claim it is , which will make isosceles with .
Notice that as desired.
See Also
1996 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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