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. 1883 Excerpt: ... (By Professor Hudson, M.A.)--Prove (1) that, if the curve r = a + b cos fl have a point of inflexion, a must lie between b and lb; and (2) trace the change in form of the curve, as a changes from 0 to 2A. 7197. (By the Editor)--With two given sides, construct a triangle such that the square of the sine of half the angle contained by the sides may be equal to the squares of the sines of the other two semi-angles. II. Solution by M. Collins, B.A.; J. O'regan; and others. The triangle is of the form in part (3) of Quest. 6287. Thus, if e be the third or required side, Q the centre, r the radius of the inscribed circle, and O the centre of the circumscribed circle, we have r (a + b + c) = ah sin C, which differentiated gives (since r is a maximum, and therefore dr = 0) rdc = ab cos C. dG; also c2 = a2 + As--lab cos C; whence, differentiating, c. d = ab sin C. dC; and from these equations we obtain r = ccot C = 2BD. cot BOD-2OD; hence DE = DE', therefore E'F = 2OG, that is, r'= radius of circle escribed to side c = 2R = diameter of circumscribed circle. abe, '. .._-_.-_-, _., 4A2 Now E = therefore abe(s-c) = 2a2=2a(---) (s-a) (3-e), and as e=2s--a--b, we get 2s3--2 (a + b) s2 + -A(- + b) = 0, an equation which determines -, and therefore the construction of the triangle. It is easy to see that the line DQ, which joins the middle D of the base to the centre Q of the inscribed circle, would pass through the point of intersectien of the three altitudes of the required Aabc when its inscribed circle (or its radius QE) is a maximum. Solution by the Rev. J. L. Kitchin, M.A.; D. Edwardes; and others. We have 6878. (By T. C. Simmons, M.A.)--If A, B are two points in the plane of a given circle; give a geometrical construction for f...