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1987 AIME Problems/Problem 9

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Problem

Triangle $ABC$ has right angle at $B$ , and contains a point $P$ for which $PA = 10$ , $PB = 6$ , and $\angle APB = \angle BPC = \angle CPA$ . Find $PC$ .

Solution

Let $PC = x$ . Since $\angle APB = \angle BPC = \angle CPA$ , each of them is equal to $120^\circ$ . By the Law of Cosines applied to triangles $\triangle APB$ , $\triangle BPC$ and $\triangle CPA$ at their respective angles $P$ , remembering that $\cos 120^\circ = -\frac12$ , we have

Then by the Pythagorean Theorem, $AB^2 + BC^2 = CA^2$ , so

and

$4x = 132 \Longrightarrow x = 033$

See also

1987 AIME (Problems • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1987_AIME_Problems/Problem_9"

Category: Intermediate Geometry Problems

Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.

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