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Clifford's Lemma: What is it?
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Clifford's Lemma
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Four green circles may or may not intersect pairwise. Assume they do and assume they intersect in a sequence, viz., in some ordering, the first meets the second, the second meets the third, the third meets the fourth, and the fourth meets the first. On such an occasion we have four pairs of points of intersection: z1 and w1, z2 and w2, z3 and w3, and z4 and w4, where some pairs may coalesce into a single point in case the corresponding circles touch rather than (more conventionally) intersect.
The applet purports to demonstrate the following fact: if points z1, z2, z3, and z4 are concyclic, then so are the points w1, w2, w3, and w4. The applet shows one of the circles in red, the other in blue. Of course there may be several pairs of red/blue circles.
Now, for the proof. I shall be using complex numbers.
For three distinct complex numbers a, b, c the ratio
From here, four complex numbers (a.k.a. plane points) a, b, c, d are concyclic, provided the quotient of two ratios
| [(a - c)/(b - c)] / [(a - d)/(b - d)] |
is purely real.
This is because two angles in a circle that subtend the same chord are either equal or sum up to 180o, and conversely, two triangles with the common base share a circumcircle, provided their apex angles are either equal or sum up to 180o.Note that the points z1, w1, z4, and w4 lie on the first circle. The points z1, w1, z2, and w2 lie on the second; z2, w2, z3, and w3 lie on the third, while z3, w3, z4, and w4 lie on the fourth circle. We thus have four real numbers
| (1) | [(z1 - z2)/(w2 - z2)] / [(z1 - w1)/(w2 - w1)] |
| (2) | [(z2 - z3)/(w3 - z3)] / [(z2 - w2)/(w3 - w2)] |
| (3) | [(z3 - z4)/(w4 - z4)] / [(z3 - w3)/(w4 - w3)] |
| (4) | [(z4 - z1)/(w1 - z1)] / [(z4 - w4)/(w1 - w4)] |
Taking the product of (1)-(4) we obtain another real number:
| (5) | {[(z1 - z2)/(z3 - z2)] / [(z1 - z4)/(z3 - z4)]} · {[(w1 - w2)/(w3 - w2)] / [(w1 - w4)/(w3 - w4)]} |
from which it follows that the z and w quartets may be only concyclic simultaneously.
It was observed by Gerald Brown that in a case where the four circles are concurrent and pairwise orthogonal there is a particularly simple solution. More accurately, assume the circles are
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(x - a)2 + y2 = a2, (x - b)2 + y2 = b2, x2 + (y - c)2 = c2, x2 + (y - d)2 = d2, |
with a ≠ b and c ≠ d. Any inversion with the center at the origin maps the first two circles into straight lines perpendicular to the x-axis and the other two circles into straight lines perpendicular to the y-axis. The four intersection points form a rectangle, obviously a cyclic figure. Thus the points of intersection (other than the origin) of the circles are also concyclic.
References
- Liang-shin Hahn, Complex Numbers & Geometry, MAA, 1994
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