The applet is supposed to illustrate the following problem:
For a plane triangle A1A2A3, call two circles within the triangle companion incircles if they are the incircles of two triangles formed by dividing A1A2A3 into two smaller triangles by passing a line through one of the vertices and some point on the opposite side. Show that any chain of circles C1, C2, ... such that Cn, Cn+1 are companion incircles for every n consists of at most six distinct circles.
The problem has been posed by Noam D. Elkies in the American Mathematical Monthly (1987, 877). A solution by Jiro Fukuta has been published in 1990 (v. 97, issue 6, pp. 530-531)
Let r be the radius of the incircle of Δ A1A2A3, s be its semiperimeter, and rn be the radius of the nth circle. We may chose indices so that C1, C2 are companion incircles generated by a line through A3, with C1 in the triangle containing A3A1 and C2 in the triangle containing A3A2. Then, successively, the circle Cn is the incircle of two triangles containing the vertex An mod 3. Thus we define An by An = Am , whenever n = m (mod 3). Let an be the angle at An, let an be the length of the side opposite An, and let hn be the altitude from An. We have
an = am, an = am, and hn = hm.
The area of Δ A1A2A3 can be given by rs or anhn/2.
In Theorem 2 of [1], Demir shows that the successive rn's satisfy
1/rn + 1/rn+1 - r/(rnrn+1) = 2/hn+2.
Multiplying by r and subtracting it from 1, we obtain
(r/rn - 1)(r/rn+1 - 1) = 1 - 2r/hn+2.
Multiplying the corresponding equations for n and n+2 and dividing by the equation for n+1 yields
(r/rn - 1)(r/rn+3 - 1) = (1 - 2r/hn+1)(1 - 2r/hn+2)/(1 - 2r/hn).
Since rs = anhn/2, we have 1 - 2r/hn = (s - an)/s, so this equation becomes
where the last equality is well known to students of elementary trigonometry.
To prove the theorem, replace n by n+3 in (1). Since an+3 = an, we have
which suffices.
