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The applet demonstrates a curious property of a geometric wonder, arbelos - the Shoemaker's knife.

Discussion

Arbelos is a geometric shape bounded by three semicircles. In the applet these are AC, CB and AB. Assume M is the midpoint of the semicircle on the diameter AC, N that of the semicircle CB. A circle is inscribed in the arbelos tangent to the semicircles in points E, F, and D. Then the points A, C, F, D are concyclic and lie on a circle centered at M. Similarly, points B, C, E, D are concyclic with center at N.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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The statement is due to L. Bankoff, the proof below is a variant of a proof from [Woo].

### Proof

Perform an inversion with center A and power AC·AB. This will invert B into C and vice versa. Hence the circle on the diameter CB is stationary. The big semicircle inverts into the line perpendicular to AB at B, and the left semicircle into a line perpendicular to AB at C. The incircle of the arbelos that is tangent to all three semicircles inverts into the circle tangent to the two vertical lines and the right semicircles. The image of F is N. The images D' and E' of the points D and E lie on the vertical lines and, therefore form a diameter D'E' of the inversive image of the incircle. By Archimedes' Proposition 1, points B, N, and D' are collinear. The image of the lines through these three points is a circle through C, F and D. Since this circle is the image of a straight line it passes through A.

It remains to be shown that the center of that circle lies in M. However, since inversion preserves angles and the angle NBA is 45°, the circle forms a 45° angle with AB as well. Thus the claim follows.

The proof of collinearity of points B, C, E and D is similar and uses the inversion with center at B.

### Remark

The statement, as noted, by P. Y. Woo, yields a simple straightedge and compass construction of the incircle of an arbelos: find points M and N and draw circles with radii MC and NC to obtain points E, F, D. Woo also showed that the same diagram yields other constructions of the incircle.

It also clear from the diagram (and, even more so, from the construction) that the points C, N and E' are collinear. The line they lie on is the inversive image of a circle through the center of inversion A. Thus the points A, B, F, E are concyclic. The center of their circle is the point L1 symmetric to L in AB.

### References

1. L. Bankoff, The Marvelous Arbelos, in The Lighter Side of Mathematics, R.K.Guy and R.E.Woodrow, eds, MAA, 1994
2. P. Y. Woo, Simple Constructions of the Incircle of an Arbelos, Forum Geometricorum, V. 1 (2001), 133-136