Problem:
Triangle ABC0​ has a right angle at C0​. Its side lengths are pairwise relatively prime positive integers, and its perimeter is p. Let C1​ be the foot of the altitude to AB, and for n≥2, let Cn​ be the foot of the altitude to Cn−2​B​ in △Cn−2​Cn−1​B. The sum ∑n=1∞​Cn−1​Cn​=6p. Find p.
Solution:
Let a=BC0​,b=AC0​, and c=AB. Then C0​C1​=cab​, and △BC0​C1​ is similar to △BAC0​ with ratio of similarity ABBC0​​=ca​. Furthermore, for n≥2, △BCn−1​Cn​ is similar to △BCn−2​Cn−1​ with the same ratio, so Cn−1​Cn​=cnban​. The sum of all lengths Cn​Cn−1​ is
n=1∑∞​cnban​=1−ca​cab​​=c−aab​
This is 6p, so ab=6(c−a)(a+b+c)=6(c2−a2+bc−ab)=6(b2+bc−ab), from which 6c=7a−6b. Squaring both sides gives 36c2=36(a2+b2)=49a2−84ab+ 36b2, which implies that 13a−84b=0. Because a and b are relatively prime, it follows that a=84 and b=13. Thus c=85, and p=84+13+85=182​.
The problems on this page are the property of the MAA's American Mathematics Competitions