Difference between revisions of "1987 AIME Problems/Problem 9"

Revision as of 15:23, 16 June 2008

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

$AB^2 = 36 + 100 + 60 = 196, BC^2 = 36 + x^2 + 6x, CA^2 = 100 + x^2 + 10x$

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

$x^2 + 10x + 100 = x^2 + 6x + 36 + 196$

and

$4x = 132 \Longrightarrow x = 033$

@@ Line 14: / Line 14: @@
 and
-<cmath>4x = 132 \Longrightarrow x = 33</cmath>
+<cmath>4x = 132 \Longrightarrow x = 033</cmath>
 == See also ==
 {{AIME box|year=1987|num-b=8|num-a=10}}
 [[Category:Intermediate Geometry Problems]]

1987 AIME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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Difference between revisions of "1987 AIME Problems/Problem 9"

Revision as of 15:23, 16 June 2008

Problem

Solution

See also