1999 AIME Problems/Problem 12

Revision as of 17:22, 14 October 2007 by Azjps (talk | contribs) (solution)

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 $2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}$.

See also

1999 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AIME Problems and Solutions