Problem:
Triangle ABC has positive integer side lengths with AB=AC. Let I be the intersection of the bisectors of ∠B and ∠C. Suppose BI=8. Find the smallest possible perimeter of △ABC.
Solution:
Let M denote the midpoint of side BC. Note that A,I, and M are collinear. Set a=AB, and b=BM. Note that a>BI=8. Then
cos(∠ABM)=ABBM=ab and cos(∠IBM)=BIBM=8b.
The double-angle formula for cosine yields
ab=2⋅82b2−1 or ab2−32b−32a=0
Solving for a yields a=b2−3232b. Because BC=2b must be an integer, let c=2b so that a=c2−12864c. This shows that c2>128, so c>11, and c=2BM< 2BI=16. Testing c=12,13,14, and 15, it follows that a is an integer only when c=12 and a=48, and the perimeter of △ABC is 48+48+12=108.
The problems on this page are the property of the MAA's American Mathematics Competitions