Difference between revisions of "1987 AIME Problems/Problem 9"
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<cmath>4x = 132 \Longrightarrow x = \boxed{033}.</cmath> | <cmath>4x = 132 \Longrightarrow x = \boxed{033}.</cmath> | ||
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This is the Fermat point of the triangle. | This is the Fermat point of the triangle. | ||
Latest revision as of 00:57, 13 May 2020
Contents
Problem
Triangle has right angle at , and contains a point for which , , and . Find .
Solution
Let . Since , each of them is equal to . By the Law of Cosines applied to triangles , and at their respective angles , remembering that , we have
Then by the Pythagorean Theorem, , so
and
Note
This is the Fermat point of the triangle.
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.