1999 AIME Problems/Problem 12

Problem

The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P_{},$ and its radius is 21. Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle.

Solution

File:1999 AIME-12.png

Let $Q$ be the tangency point on $\overline{AC}$ , and $R$ on $\overline{BC}$ . By the Two Tangent Theorem, $AP = AQ = 23$ , $BP = BR = 27$ , and $CQ = CR = x$ . Using $rs = A$ , where $s = \frac{27 \cdot 2 + 23 \cdot 2 + x \cdot 2}{2} = 50 + x$ , we get $(21)(50 + x) = A$ . By Heron's Formula, $A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(50+x)(x)(23)(27)}$ . Equating and squaring both sides,

$$ (Error compiling LaTeX. Unknown error_msg)\begin{eqnarray*} [21(50+x)]^2 &=& (50+x)(x)(621)\\ 441(50+x) &=& 621x\\ 180x = 441 \cdot 50 \Longrightarrow x $=$ \frac{245}{2} $$ (Error compiling LaTeX. Unknown error_msg)

We want the perimeter, which is $2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}$ .

1999 AIME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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1999 AIME Problems/Problem 12

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