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By William J. M'Clelland

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A treatise on the geometry of the circle and some extensions to conic sections by the method of reciprocation, with numerous examples.

Leopold is overjoyed to submit this vintage publication as a part of our huge vintage Library assortment. a few of the books in our assortment were out of print for many years, and for that reason haven't been obtainable to most people. the purpose of our publishing software is to facilitate quick entry to this enormous reservoir of literature, and our view is this is an important literary paintings, which merits to be introduced again into print after many many years.

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Extra resources for A treatise on the geometry of the circle and some extensions to conic sections by the method of reciprocation, with numerous examples.

Sample text

45 AB are given lines. Hence any vertex (C) BO, CA, and found by describing a segment of a circle upon PQ as centre and containing an angle equal to C, and with is OC is as radius describing circle. the required position of C. Again in the triangle BOG 00 a maximum, and is is Where when these circles meet EG is a maximum therefore a diameter of the Then OPC is a right angle. Hence the triangle of given species escribed to a given one is that whose sides are perpendicular to OP, OQ, OR. circle OPQO.

Therefore if the three perpendi- l is a maximum for the point of intersection Theorem. If a right line be divided into any two and b the sum of their squares is a minimum a parts when the line is bisected. For (Euc. II. 9, 10) 15. minimum when a b is minimum, because that is when a b. a+b COR. The sum of the squares of the segments of a line is a minimum when the segments are equal. Hence a2 +6 2 16. is constant is ; If a right Problem. -+... is are minimum p y known quantities. Let the segment a be divided into a equal parts part = is therefore a/a M= a2 \a/ a 2 any when ; each and the sum of squares of the parts Similarly if the segment b be divided into /3 equal parts 2 the sum of squares of the subdivisions = 6 //3 and so on.

Point 0, the triangles species, and BOG, CO A, and AOB are given in are therefore completely determined, since For other solutions see " Mathematics from the Educational Times" Vol. , p. 29, by D. Biddle and Rev. T. C. Simmons. * MAXIMUM AND MINIMUM. ^ 45 AB are given lines. Hence any vertex (C) BO, CA, and found by describing a segment of a circle upon PQ as centre and containing an angle equal to C, and with is OC is as radius describing circle. the required position of C. Again in the triangle BOG 00 a maximum, and is is Where when these circles meet EG is a maximum therefore a diameter of the Then OPC is a right angle.

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