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set of equal paper rectangles and can be generalized so as to insert n means between 2 given segments.11 4.4 The Cissoid of Diodes Diodes (ca. 180 B.C.) invented the dssoid in order to solve the duplication problem. A general cissoid may be defined as follows: Let C\ and C2 be 2 given curves, and let O be a fixed point. Let P\ and P2 be the intersections of a variable line through O with the given curves. The locus of P on this line such that OP = OP2 - OP\ = P\P2 is called the cissoid of C\ and C2 for the pole O. If Ci is a circle, C2 a tangent to Ci at point A, and O is the point on Ci diametrically opposite A, then the cissoid of Ci and C2 for the pole O is the dssoid of Diodes. (a) Taking O as origin and OA as the positive x-axis, show that the Cartesian equation of the cissoid of Diodes is y2 = x3l(2a - x), where a is the radius of Ci. Show that the corresponding polar equation is r = 2a sin 0 tan 0. (b) On the positive y-axis, lay off OD = n(OA). Draw DA to cut the cissoid in P. Let OP cut line C2 in Q. Show that (AQ)3 = n(OA)3. When n = 2, we have a solution of the duplication problem. (c) Newton has shown how the cissoid of Diodes may be generated by a carpenter’s square. Let the outside edge of the square be ACB, AC being the shorter arm. Draw a line MN and mark a point R at distance AC from MN. Move the square so that A always lies on MN and BC always passes through R. Show that the midpoint P of AC describes a cissoid of Diodes. (d) What is the cissoid of 2 concentric circles with respect to their common center? Of a pair of parallel lines with respect to any point not on either line? (e) If Ci and C2 intersect in P, show that OP is a tangent at O to the cissoid of Ci and C2 for the pole O. 11 For a more recent mechanical approach, see George E. Martin, “Duplicating...