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1999 AIME Problems/Problem 12

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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.

Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See also

Solution

pathpen = black + linewidth(0.65); pointpen = black;pair A=(0,0),B=(50,0),C=IP(circle(A,23+245/2),circle(B,27+245/2)), I=ince...

Solution 1

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,

$\begin{eqnarray*} [21(50+x)]^2 &=& (50+x)(x)(621)\\441(50+x) &=& 621x\\180x = 441 \cdot 50 &\Longrightarr...$

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

Solution 2

Let the incenter be denoted $I$ . It is commonly known that the incenter is the intersection of the angle bisectors of a triangle. So let $\angle ABI = \angle CBI = \alpha, \angle BAI = \angle CAI = \beta,$ and $\angle BCI = \angle ACI = \gamma.$

We have that $\begin{eqnarray*} \tan \alpha & = & \frac {21}{27} \\\tan \beta & = & \frac {21}{23} \\\tan \gamma & = &a...$ So naturally we look at $\tan \gamma.$ But since $\gamma = \frac \pi2 - (\beta + \alpha)$ we have $\begin{eqnarray*} \tan \gamma & = & \tan\left(\frac \pi2 - (\beta + \alpha)\right) \\& = & \frac 1{\tan(\alph...$ Doing the algebra, we get $x = \frac {245}2.$

The perimeter is therefore $2\cdot\frac {245}2 + 2\cdot 23 + 2\cdot 27 = \boxed{345}.$

See also

1999 AIME (Problems • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1999_AIME_Problems/Problem_12"

Category: Intermediate Geometry Problems

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

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