Summer Problem Solving Marathon Solution #24

The first perpendicular is drawn from point C to point D. The second is drawn from point D to point E. Let points F, G, H, and so on be the ending points of the further perpendiculars.

CDE forms a right triangle, as does DEF. Furthermore, CD is parallel to EF (because both meet OB at a right angle). Thus the triangles CDE and DEF are similar. In fact, we form an infinite sequence of similar triangles -- CDE, DEF, EFG, FGH, and so on.

CD is the hypotenuse of CDE, and DE is the hypotenuse of DEF. CD is of length a, and DE is of length b. So the ratio of similarity between DEF and CDE is b/a. Side EF of DEF corresponds to side DE of CDE. So EF is of length b(b/a) = b²/a.

EF is the hypotenuse of EFG, and DE is the hypotenuse of DEF. So the ration of similarity between EFG and DEF is (b²/a)/b = b/a. In fact, we can easily see that the ratio of similarity between consecutive similar triangles in our sequence is always b/a. So side FG is EF(b/a) = (b²/a)(b/a) = b³/a².

Generalizing, we get a sequence of lengths:

a, b, b²/a, b³/a², b⁴/a³, ...

This is an infinite sequence with first term a and ratio b/a. Its sum is then:

a/(1 - b/a) = a((a-b)/a) = a²/(a-b)