Perimeter of a Right Triangle

Date: 08/14/2001 at 21:47:06
From: Hanul Oh
Subject: High School mathematics contest

What is the perimeter of a right triangle with hypotenuse 65 that can 
be circumscribed about a circle with radius 12?

Thank you.

Date: 08/15/2001 at 12:54:13
From: Doctor Greenie
Subject: Re: High School mathematics contest

Hello, Hanul -

Here is a rough picture of the problem (I have only drawn the radii of 
the circle - I didn't want to try to draw a circle using typed text).

    A
    *
    *   *
    *       *    D
    *           *
    *          *    *
    *         *         *
    *        *              *
  F ********* O                 *
    *       *                       *
    *       *                           *
    *       *                               *
    *********************************************
    C       E                                   B

The right triangle is ABC; the circle has center O and is tangent to 
triangle sides AB, BC, and CA at points D, E, and F, respectively.  
AB has length 65. OD, OE, and OF are radii of the circle and so have 
length 12. Because the angle at C is a right angle, CE and CF also 
have length 12.

We know, with circle O being inscribed in triangle ABC, that AD = AF 
and BD = BE. Let AF = x; then AD = x also. Because we are given that 
AB = 65, BD is (65-x); then BE is (65-x) also.

You can solve the problem by finding the area of triangle ABC by two 
different methods and equating the expressions:

(1) area = (1/2) base * height = (1/2)(AC)(BC) = (1/2)(x+12)(77-x)

(2) area = (radius of inscribed circle) * (semiperimeter of triangle)
         = (12)(1/2)[(65)+(x+12)+(77-x)]

So you have

    (1/2)(x+12)(77-x) = (12)(1/2)[(65)+(x+12)+(77-x)]

Solve this equation for x (it's not as ugly as it looks); then you 
will have all the information you need to find the perimeter of the 
triangle.

Write back if you have further questions on this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/