See complete Problem 143

Four Triangles, Incircle, Tangent and Parallel to Side, Circumradii. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Friday, July 25, 2008

### Elearn Geometry Problem 143

Labels:
circle,
circumradius,
incircle,
parallel,
similarity,
tangent,
triangle

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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.