The Mathematician Volume 2 (Paperback)

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1856 edition. Excerpt: ...(B). If P P, be two points in an ellipse whose centre is O, and PiQ, P2Q2 perpendiculars to the major axis meeting the circle described on this axis in Q, and Q, respectively, the ratios of the triangle PiOP, to the triangle QiOQj) of the sector PiOP2 to the sector QiOQ2, and of the segment upon Pi Pj to the segment upon QiQj, are each equal to the ratio of the minor axis to the major. Since sector QiOQj = a'(0---0i), we have (15) sector PPj = 106(0, -00 (17) If 0 = 0 and 0j = 2tt, the sector becomes the whole ellipse, and we have the usual formula (irab) for its area. From (13,17), segment upon P = ab (0, --0i)--sin(0j--0 )) (18) Let Q Q2..., Q be the angles of a regular polygon of re sides, inscribed in the circle, and P P2... Pn the points in which the perpendiculars on the major axis from Q Q, ....Qa intersect the ellipse. Then will 2r 0m+i--0m =--, and hence (17) sector P, OP, = sector P2OP3 =.. = sector P _iOP-=sector P OP, =--re....(19) From which we see that (C). Whenever n is such that a regular polygon of re sides can be described geometrically, an ellipse can be divided geometrically into re equal sectors, in an indefinite number of ways. The following construction gives one of these ways: (D). Upon AB, the major axis of the ellipse, describe a circle, in which inscribe a regular polygon Q( Q2..l.Qnof the proposed number (re) of sides; from Q Qj....Qn demit perpendiculars upon AB meeting the ellipse in P P2.... P, and join P P2...P and the centre O; the ellipse is divided into re equal sectors PPj, P2OPs....P _iOPn, P OP, . 2w Since QfiB or 0, may obviously have any value between 0 and--, the number of ways in which this division can be...