Problem:
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 8:7. Find the minimum possible value of their common perimeter.
Solution:
Let the length of the base of one of the triangles be 8a and let the length of the base of the other triangle be 7a, for some positive integer a. Because these two triangles have the same area, the lengths of the corresponding altitudes must be 7h and 8h. Because the perimeters of the triangles are equal, it follows that
8a+216a2+49h2=7a+2449a2+64h2, or a+216a2+49h2=2449a2+64h2
Squaring both sides of the last equation and simplifying gives
a2+64a2+196h2+4a16a2+49h2=49a2+256h2, or a16a2+49h2=15h2−4a2
Squaring both sides of this equation and simplifying yields
a2(16a2+49h2)=225h4−120a2h2+16a4, or 225h2=169a2.
Thus h=1513a, and the common perimeter is
8a+216a2+49h2=8a+15218a
Because the triangles are integer-sided, the equal sides must also be integers.
For the triangle with base 8a, the equal side length is
16a2+49h2=16a2+22549⋅169a2=15109a,
which is an integer only if 15∣a.
For the triangle with base 7a, the equal side length is