First assume that the six points are such that the chords AD, BE, CF concur in point P. Observe the presence of three pairs of similar triangles. (In the diagram one pair is highlighted and pairs of equal angles are shown. One pair is of central angles; two pairs are inscribed subtended by the same arc.)
Multiplying these together gives us the desired result.
AB · CX · EF = BC · XE · FA.
Lemma 2
Let two externally tangent circles, (P) and (Q), be internally tangent to circle (C) at points A and B respectively. If the radii of circles (C), (P), and (Q) are R, p, and q, respectively, then
AB² = 4R² f(p)·f(q),
where f(x) = x / (R - x).
Proof
Triangles APM and ACD are isosceles with a common base angle at A. It follows that their other base angles (at M and D) are also equal. Therefore, CD||MP. Similarly, CE||MQ. But PMQ is a straight line and then so is DCE.
Observe also that, as inscribed angles subtended by the same arc,
∠EBA = ∠EDA and
∠BAD = ∠BED.
It follows there are three pairs of similar triangles: APM/ACD, BQM/BCE, and ABM/EDM which gives the following proportions:
AB/DE = AM/EM = BM/DM.
But DE = 2R. Thus we have
| AB/2R · AB/2R | = AM/EM · BM/DM |
| | = AM/EM · BM/DM |
| | = AM/DM · BM/EM |
| | = AP/CP · BQ/CQ |
| | = p(R - p) · q(R - q). |
Proof of the Seven Circles Theorem
Assume the points of tangency of circle C with circles Sk are Ak, and that rk denotes the radius of Sk, k = 1, 2, ..., 6. By Lemma 2,
(AkAk+1)² = 4R² f(rk)f(rk+1),
k = 1, 2, ..., 6, and r7 = r0. It follows that
(A1A2·A3A4·A5A0)² = 64R6 f(r0)f(r1)f(r2)f(r3)f(r4)f(r5) = (A0A1·A2A3·A4A5)²
which of course implies
A1A2·A3A4·A5A0 = A0A1·A2A3·A4A5
and by Lemma1 the three chords A0A3, A1A4, A2A5 are concurrent.
The applet shows that the concurrency takes place in other configurations as well. For example, circles Sk may all be tangent to circle C externally. The above proof goes through in this case if the function f(x) is redefined: f(x) = x/(R + x).
In a special case where circles S1, S3, S5 expand into straight lines, the configuration becomes that of the incircle of a triangle with three additional circles inscribed into curvilinear triangles cut off by the incircle. If the radii of the latter shrinks to 0, the remaining points of tangency are none other than the intouch points and the point of concurrency is the point of Gergonne.
As can be grasped from the diagrams below, the theorem holds in a more general setting where the points of tangency Ak do not form a convex hexagon.
References
- H. S. M. Coxeter, Problem 527, The Mathematics Student 28 (1980) 3.
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Copyright © 1996-2012 Alexander Bogomolny