If A is obtuse, AO is arithmetically - 2R cos A, being on the side of A remote from BC. OL = BL tan LBO = AB cos B cot C = 2R cos B cos C. If B or C is obtuse, OL is arithmetically - 2R cos B cos C. COR. The distances of O from the sides are inversely as those of S. 213. The distances of the centre of gravity from the sides and vertices of the triangle. See fig., Art. 192. Since GD = AD, :. the distance of G from BC S =AL=3b sin C = . -R sin B sin C......(1). α Also, producing AD to A', so that A'DAD, ABA'C is a parallelogram, and the angle ABA' is the supplement of A; .. 4 AD2 = A A'2 = b2 + c2 + 2bc cos 4............(2), = 2b2+2c2 - a2. .. GA=AD = √(2b2 + 2c2 - a3) ..(3). ...(4). COR. The distances of G from the sides are inversely as the 214. Expressions for the radius (p) of the Cosine-Circle; for the distances of its centre from the sides, and for its intercepts on the sides. See fig., Art. 193. Draw KX, KY, KZ perpendiculars on the sides. Then ·· the antiparallels through K, which are diameters of the cosine-circle, cut BC at an inclination A, .. the intercepts on the sides are, respectively, 2p cos A, 2p cos B, 2p cos C, and are .. proportional to the cosines of the angles. [Hence the name of the circle.] And the perpendiculars on the sides are, respectively, p sin A, p sin B, p sin C, .. KX: KY: KZ = sin A: sin B: sin C If now x, y, z be the perpendiculars on the sides from any point within the triangle, it follows, as in Art. 208, that xa+yb + zc= 2. (area of triangle) = 28. This gives the distances of K from the sides. .(2). Let CO, 40, BQ subtend equal angles w at A, B, C respectively. Then clearly the angles ANB, BOC, COA must be the supplements of A, B, C. If again BO', C'N', AN' subtend equal angles w' at A, B, C respectively, the same equation (1) will give w'. If NX, NY, NZ be perpendiculars on the sides, since circles go round AYZO, BZXQ, CXYN, it follows that XYZ is similar to CAB, and has 2 and w for Brocard point and Brocard angle respectively. And since NX: NC = sin w, .. sin w is the ratio of similitude of XYZ to CAB. Now b sin2 w NX= NC sin w = c sin2 w If 'X' be perpendicular to BC, 'X' = .. QX. Q'X' =QY. Q'Y' =QZ. Q'Z' = 4R2 sin1 w. .. ('N)2= (radius of circle XX' YY' ZZ')2 - 4R2 sin1 w = R2 sin2 w (1 − 4 sin2 w) = R2 sin2 w (cos2 W 3 sin2 w). .. 'N=R sin 2w (1-3 tan2 w)(3). 216. Since B and C are inclined to BC at equal angles w, therefore if their point of intersection D' is joined to D (the middle point of BC), we have and Comparing this with Art. 215 (1), and Art. 214 (5), we have KX = DD', KY = EE', KZ = FF", COR. (Lemoine-Radius)=(circumradius)+(cosine-radius)". .. Lemoine-Radius = R sec w. 2 .. the sides of A,B,C, and of A,B,C, (where the Lemoine circle cuts ABC) are to those of ABC in the ratio 1 sec w to 1. 217. The distance between the Circumcentre and Incentre. By Prop. IX. in the last chapter, the rectangle of the segments of any chord of the circumcircle drawn through the incentre 2Rr. Draw through I the diameter DSIE of the circumcircle. Then Similarly DI. IE + SI2 = DS2, SI2 = DS2 - DI.IE = R2 – 2Rr. SIR2+2Rr1. 218. The distance between the Circumcentre and Orthocentre. SO2 = SA? + 04 −2SA. 04 cos SAO, = R2 + 4R2 cos2 A-4R2 cos A cos (C – B) = = R2 - 4R2 cos A {cos (C + B) + cos (C – B)} = COR. Since the Nine-Points Centre T bisects SO, .. ST2 = r2 (1 − 8 cos A cos B cos C), where 7 = nine-points radius 219. The distance between the nine-points centre and incentre. Drop perpendiculars from T on BC, and from I on this perpendicular. Let AI cut the circumcircle in U. Then, by Prop. IX. of last Chapter, UI. IA = 2Rr = 4Tr; and by Prop. XI, Cor. II, ▲ DPL=2. ¿UIX = 20 say. .. T'I (T cos 20 − r)2 + (τ sin 20 – DX)2 = 72 + r2 − 2tr cos 20 – DX (27 sin 20 – DX) |