Problem 129 An introduction to modern geometry Shively Levi

Problem 12.9 - An introduction to modern geometry, Shively, Levi S

Let C be any point on the line segment AB between A and B, and let semicircles be drawn on the same side of AB, with AB, AC and CB as diameters. Then the figure whose boundary consists of these semicircles is called arbelos or a shoemaker’s knife. Let the radii of the semicircles on AB, AC, CB be r, r₁, r₂, respectively. Draw the perpendicular to AB at C, intersecting the largest semicircle at D (with the diameter AB). Also let E and F be the points of contact of the direct common tangent to the two semicircles on AC and CB, respectively. The following properties are easily verified:

a) The perimeter of the arbelos is equal to the circumference of the circle on AB as diameter.

b) The area of the arbelos is equal to the area of the circle on CD as diameter.

c) The segment EF of the common tangent is equal to CD, and these two lines bisect each other at H. Thus, ECFD is a rectangle, and the center of its circumscribing circle is H.

d) The points A, E, D are collinear, as are also B, F, D. We now prove:

e) If a circle be drawn tangent to CD and to the semicircles on AB and AC as diameters, and another be drawn tangent to CD and to the semicircles on AB and CB as diameters, these two circles are equal.

Let the first of these circles touch the line CD at P, the semicircle on AC at Q, and the semicircle on AB at R. Let the first of these circles touch the line CD at P, the semicircle on AC at Q, and the semicircle on AB at R; and let SP be the diameter through P. Then it is easy to show that the triads A, Q, P; A, S, R; B, P, R; and C, Q, S are collinear, and that, if T be the point of intersection of AS and CD, the lines BT and CS are parallel. From this it follows that: SP/AC = ST/AT = CB/ AB, and SP = r₁r2/r.

The symmetry shows the diameter of the other inscribed circle to have the same length.

Demonstrate

1. The lines RB and AB are antiparallel with respect to CD and AR, and the points A, R, P, C are concyclic.

2. The point B has equal powers with respect to the circles on SP and AC as diameters, from which we conclude that the common tangent to these circles at Q passes through B.

Solution

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