Six Circles Theorem (Elkies)
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A Mathematical Droodle
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Copyright © 1996-2018 Alexander BogomolnyThe applet is supposed to illustrate the following problem:
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 A_{1}A_{2}A_{3}, s be its semiperimeter, and r_{n} be the radius of the n^{th} circle. We may chose indices so that C_{1}, C_{2} are companion incircles generated by a line through A_{3}, with C_{1} in the triangle containing A_{3}A_{1} and C_{2} in the triangle containing A_{3}A_{2}. Then, successively, the circle C_{n} is the incircle of two triangles containing the vertex
The area of Δ A_{1}A_{2}A_{3} can be given by rs or a_{n}h_{n}/2.
In Theorem 2 of [1], Demir shows that the successive r_{n}'s satisfy
Multiplying by r and subtracting it from 1, we obtain
Multiplying the corresponding equations for n and n+2 and dividing by the equation for n+1 yields
Since rs = a_{n}h_{n}/2, we have
(1) | (r/r_{n} - 1)(r/r_{n+3} - 1) = (s - a_{n+1})(s - a_{n+2})/[s(s - a_{n})] = tan^{2}a_{n}/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
Because r_{n+3} < r, we may cancel the common factor and obtain
which suffices.
References
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Copyright © 1996-2018 Alexander Bogomolny