Difference between revisions of "1987 AIME Problems/Problem 9"
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[[Triangle]] <math>ABC</math> has [[right angle]] at <math>B</math>, and contains a [[point]] <math>P</math> for which <math>PA = 10</math>, <math>PB = 6</math>, and <math>\angle APB = \angle BPC = \angle CPA</math>. Find <math>PC</math>. | [[Triangle]] <math>ABC</math> has [[right angle]] at <math>B</math>, and contains a [[point]] <math>P</math> for which <math>PA = 10</math>, <math>PB = 6</math>, and <math>\angle APB = \angle BPC = \angle CPA</math>. Find <math>PC</math>. | ||
− | + | <asy> | |
unitsize(0.2 cm); | unitsize(0.2 cm); | ||
Line 17: | Line 17: | ||
draw(C--P); | draw(C--P); | ||
− | label(" | + | label("$A$", A, NW); |
− | label(" | + | label("$B$", B, SW); |
− | label(" | + | label("$C$", C, SE); |
− | label(" | + | label("$P$", P, NE); |
− | + | </asy> | |
== Solution == | == Solution == |
Revision as of 17:44, 5 February 2024
Contents
[hide]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.