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 $PC$.

AIME 1987 Problem 9.png

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

\[4x = 132 \Longrightarrow x = \boxed{033}.\]

Note

This is the Fermat point of the triangle.

See also

1987 AIME (ProblemsAnswer KeyResources)
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

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