Longueyhigginsintell2.Pdf
. .IB A ,1, " , , ,, " , , , , ," , ,, , , ,, , " , , , , , , , , , L, , " , , , , , , , , , , " , , , ," , , , , , ,, F A," , , , , , , , , , I , , ,, B D A' C *GB ' , , *' 'C E, and F denote the feet of the altitudes AH, BH, and CH; ",', GA,' see Figure 6. Then, HDCE, for example,is a cyclic quadri- I " " ',' , lateral;hence, LBHD = C. Therefore, " , I "':', " HD = BD cot C = ABcosBcot C. " , ',,~ " BecauseAB = 2Rsin C, this givesus .. IA H = 2R(cos B cas C, cosC cosA, cosA cosB). I' The coordinates of the point L, which is the reflection of i erties of two pencils of rays having the same cross-ratio, H in 0 (i.e., L = 2 X 0 - H), can therefore be written [ b~t with one ray in common. The proof is too long to be L = 2R(a - {3'Y,(3 - 'Ya,'Y - a{3), (2) given here. The fourth and shortest proof goes as follows. The trilinear coordinates of a point P in the plane of a where a, {3, and 'Y stand for cos A, cos B, and cos C, re- triangle ABC may be defmed [4, 10] as the lengths (x, y, z) spectively (see also [4] and [5]). I of the perpendicularsfrom P to the three sidesBC, CA, and What are the coordinatesof the Gergonnepoint G?Now, I AB. By convention, x, y, and z are positive when Plies in- from Figure 3, we see that XC = r cot(iC), and therefore side the triangle. for X, we have Thus, the incenter I, being equidistant from the three y = XC sin C = 2r COS2(lC). sides, has coordinates 2 Similarly, I = r(1, 1, 1), (1) z = XB sin B = 2r cos2(lB).
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