Friday, July 25, 2008

Elearn Geometry Problem 143



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.

3 comments:

  1. 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.
    So 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

    ReplyDelete
  2. We use S = 2 R^2 sin A sin B sin C
    & 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

    ReplyDelete
  3. Solution to problem 143.
    Let 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.

    ReplyDelete