Another Isosceles Triangle

Date: 10/27/1999 at 02:49:07
From: Imaad Dalal
Subject: Geometry

A triangle has sides of length 29, 29, and 40 cm. Find another 
isosceles triangle with the same perimeter and area that also has 
sides of integral length.

I tried creating a pair of simultaneous equations but that did not 
work. I then tried to work with the formula for area of the triangle, 
for instance, double the height and halve the base so that the area 
remains the same. That changes the perimeter. Help!

Date: 10/27/1999 at 05:19:55
From: Doctor Pete
Subject: Re: Geometry

Hi,

This is a fairly tricky problem. Given an isosceles triangle with 
sides a and base 2b, the perimeter is

     P = 2a + 2b

and the area is

     A = b Sqrt[a^2 - b^2]

(you can check these). Then for a = 29, b = 20, corresponding to the 
given triangle, we have the system of equations

     a + b = 49,
     b Sqrt[a^2 - b^2] = 420.

(Remember that b is half the base.) Substitution gives

       b Sqrt[(49-b)^2 - b^2]

     = b Sqrt[49(49-2b)]

     = 7b Sqrt[49-2b] = 420.

Dividing by 7 and squaring both sides gives

     b^2 (49 - 2b) = 60^2
or
     2b^3 - 49b^2 + 3600 = 0.

Now here comes the tricky part: we already know that b = 20 is a 
solution to this equation, so this cubic polynomial has a factor of 
(b-20). Dividing this factor out gives a quadratic,

     2b^2 - 9b - 180 = 0.

From here, it is straightforward to factor it completely: We have

     (2b + 15)(b - 12) = 0

from which it follows that b = 12 is another solution. The resulting 
isosceles triangle has sides a = 49 - 12 = 37, 2b = 24.

Finally, some questions for you: Are there non-isosceles triangles 
that also satisfy the conditions of the problem? How would you go 
about finding such a triangle or triangles? Also, does the value 
b = -15/2 have any geometric significance? And, can you find other 
pairs of integers for which there is only one other isosceles triangle 
with integer sides and the same area and perimeter?

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