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
8​a+216a2+49h2​=7a+2449​a2+64h2​, or a+216a2+49h2​=2449​a2+64h2​​
Squaring both sides of the last equation and simplifying gives
a2+64a2+196h2+4​a16a2+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 perimeter p is an increasing function of a, it must attain its minimum for the smallest acceptable value of a. The triangle are integersided, and therefore the value of p must also be an integer. Because 218 and 15 have no common factors, the smallest value of a for which p is an integer is 15. Thus the value requested is