Sum of Radii (Posted on 2008-03-12)

     SX = |OOX|  if the line segment OOX intersects
                 the interior of triangle ABC,

        = -|OOX| otherwise.

	Submitted by Bractals
Rating: 3.6667 (3 votes)
Solution:	(Hide)
The proof is a little different for acute, obtuse, and right triangles. I will do it for acute and obtuse triangles and leave right triangles for the reader. To make the proofs easier, add line segments OA, OB, OC, OAOB, OBOC, and OCOA. Let a, b, and c be the sides of triangle ABC opposite vertices A, B, and C respectively. Let s = (a + b + c)/2 be the semiperimeter. Acute Triangle: The area of triangle ABC is rs = a/2 |OOA| + b/2 |OOB| + c/2 |OOC| (1) Applying Ptolemy's theorem to cyclic quadrilateral AOCOOB gives |OBOC||OA| = |AOC||OOB| + |AOB||OOC| or a/2 R = c/2 |OOB| + b/2 |OOC| (2) Similarly for cyclic quadrilaterals BOAOOC and COBOOA b/2 R = a/2 |OOC| + c/2 |OOA| (3) c/2 R = b/2 |OOA| + a/2 |OOB| (4) Adding equations (1)-(4) and dividing by s gives r + R = |OOA| + |OOB| + |OOC| (5) Obtuse Triangle: WOLOG let angle C be obtuse. The area of triangle ABC is rs = a/2 |OOA| + b/2 |OOB| - c/2 |OOC| (6) Applying Ptolemy's theorem to cyclic quadrilateral AOOCOB gives |AOC||OOB| = |OBOC||OA| + |AOB||OOC| or a/2 R = c/2 |OOB| - b/2 |OOC| (7) Applying Ptolemy's theorem to cyclic quadrilateral BOAOCO gives |BOC||OOA| = |OCOA||OB| + |BOA||OOC| or b/2 R = c/2 |OOA| - a/2 |OOC| (8) Applying Ptolemy's theorem to cyclic quadrilateral COBOOA gives |OAOB||OC| = |COB||OOA| + |COA||OOB| or c/2 R = b/2 |OOA| + a/2 |OOB| (9) Adding equations (6)-(9) and dividing by s gives r + R = |OOA| + |OOB| - |OOC| (10) Combining equations (5) and (10) gives r + R = SA + SB + SC

	Subject	Author	Date
	Triangle trigonometry	Chesca Ciprian	2008-03-13 16:11:17