See complete Problem 143
Four Triangles, Incircle, Tangent and Parallel to Side, Circumradii. Level: High School, SAT Prep, College geometry
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Friday, July 25, 2008
Elearn Geometry Problem 143
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Since the tangents DE.HM & FG are parallel to corresponding sides of Tr.ABC, we've MH = EF, MD = GF & DE = HG. and Tr.BDE similar to Tr.ABC; hence R2/R = BE/BC. Likewise R1/R = MH/BC & R3/R = CF/BC.
ReplyDeleteSo R1/R + R2/R + R3/R = BE/BC + MH/BC + FC/BC
(R1 + R2 + R3)/R = BE/BC + EF/BC + FC/BC
= (BE + EF + FC)/BC
= BC/BC = 1
or R = R1 + R2 + R3
Ajit: ajitathle@gmail.com
How do you prove that "MH = EF, MD = GF & DE = HG." see line 1
DeleteWe use S = 2 R^2 sin A sin B sin C
ReplyDelete& the result of Problem 142:
√S=√S1 + √S2 + √S3
Each of the triangles AMH, BED, CGF is equiangular to triangle ABC
Si = 2 Ri^2 sin A sin B sin C (i=1,2,3)
Denote 2 sin A sin B sin C = k
R1 + R2 + R3 = √(S1/k) + √(S2/k) + √(S3/k)
= (1/√k)(√S1 + √S2 + √S3) = (1/√k)√S
= √(S/k) = R
Solution to problem 143.
ReplyDeleteLet be P, P1, P2, P3 the perimeters of triangles ABC, AMH, DBE and GFC, respectively. As proved in problem 141, P = P1 + P2 + P3.
Triangles AMH and ABC are similar, with ratio (P1)/P. Triangles DBE and ABC are similar, with ratio (P2)/P. Triangles GFC and ABC are similar, with ratio (P3)/P.
So (R1)/R + (R2)/R + (R3)/R = (P1)/P + (P2)/P + (P3)/P = (P1 + P2 + P3)/P = P/P = 1.
Thus R1 + R2 + R3 = R.