2010 AIME II Problems/Problem 12

Problem 12

Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is . Find the minimum possible value of their common perimeter.

Solution

Let the first triangle has side lengths , , ,

and the second triangle has side lengths , , ,

where .

Equal perimeter:

Equal Area:

$\begin{array}{ccc} 7c(\sqrt{a^2-(7c)^2})&=&8c(\sqrt{b^2-(8c)^2})&{}\\ 7(\sqrt{(a+7c)(a-7c)})&=&8(\sqrt{b+8c)(b-8c)})&{}\\ 7(\sqrt{(a-7c)})&=&8(\sqrt(b-8c)})&\text{---(Note that} a+7c=b+8c)\\ 49a-343c&=&64b-512c&{}\\ 49a+169c&=&64b&{}\\ 49a+169(a-b)&=&64b&\text{---(Note that} c=a-b)\\ 218a&=&233b&{}\\ \end{array}$ (Error compiling LaTeX. Unknown error_msg)

Since and are integer, the minimum occurs when , , and

Perimeter