1987 AIME Problems/Problem 9

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Problem

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

AIME 1987 Problem 9.png

Solution

Let $PC = x$.

Since the three angles $\angle APB$, $\angle BPC$ and $\angle CPA$ are equal, 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

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

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

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

and

$4x = 132$ so $x = 033$.

See also

1987 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AIME Problems and Solutions