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Six Circles Theorem (Elkies): What is this about?
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| Buy this applet What if applet does not run? |
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 thew incircle of 
A1A2A3, s be its semiperimeter, and rn be the radius of the n
| an = am, an = am, and hn = hm. |
The area of 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) | (r/rn - 1)(r/rn+3 - 1) = (s - an+1)(s - an+2)/[s(s - an)] = tan2an/2, |
where the last equality is well known to students of elementary trigonometry.
To prove the theorem, replace n by n+3 in (1). Since
| (r/rn - 1)(r/rn+3 - 1) = (r/rn+3 - 1)(r/rn+6 - 1). |
Because rn+3 < r, we may cancel the common factor and obtain
| rn = rn+6, |
which suffices.
References
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