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−2B in △Cn−2Cn−1B. The sum ∑n=1∞Cn−1Cn=6p. Find p.
Solution:
Let a=BC0,b=AC0, and c=AB. Then C0C1=cab, and △BC0C1 is similar to △BAC0 with ratio of similarity ABBC0=ca. Furthermore, for n≥2, △BCn−1Cn is similar to △BCn−2Cn−1 with the same ratio, so Cn−1Cn=cnban. The sum of all lengths CnCn−1 is
n=1∑∞cnban=1−cacab=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