Problem:
Point B is on AC with AB=9 and BC=21. Point D is not on AC so that AD=CD, and AD and BD are integers. Let s be the sum of all possible perimeters of △ACD.
Find s.
Solution:
Let AD=CD=a, let BD=b, and let E be the projection of D on AC. It follows that a2−152=DE2=b2−62, or a2−b2=225−36=189. Then (a+b,a−b)=(189,1),(63,3),(27,7), or (21,9), from which (a,b)=(95,94), (33,30),(17,10), or (15,6). The last pair is rejected since b must be greater than 6 . Because each possible triangle has a perimeter of 2a+30, it follows that s=190+66+34+3⋅30=380.
OR
Let (ak,bk) be the possible values for a and b, and let n be the number of possible perimeters of △ACD. Then s=∑k=1n(30+2ak)=30n+∑k=1n[(ak+bk)+(ak−bk)]. But (ak+bk)(ak−bk)=ak2−bk2=189=33⋅7 which has 4 pairs of factors. Thus n=4. Therefore the sum of the perimeters of the triangles is 30⋅4 more than the sum of the divisors of 189 , that is, 120+(1+3+32+33)(1+7)=440. However, this includes the case where D=E, the projection of D on AC, so s=440−60=380.
The problems on this page are the property of the MAA's American Mathematics Competitions