Monday, May 19, 2008

Geometry Problem 86



See complete Problem 86
Intouch and Extouch Triangles, Area. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

6 comments:

  1. AF=AE=p-a=BM=CH
    BF=BD=p-b=AM=CG
    CD=CE=p-c=BG=AH
    [DEF]=[ABC]-0.5(p-a)²sinA-0.5(p-b)²sinB
    -0.5(p-c)²sinC
    [MGH]=[ABC]-0.5(p-b)(p-c)sinA-0.5(p-a)(p-b)sinC
    -0.5(p-a)(p-c)sinB
    sinA=a/(2R);sinB=b/(2R);sinC=c/(2R)
    with a "little" algebra
    S_i=S_e

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  2. In the solution of pr. 86 posted by Anonymous, could anyone explain me why is it true that
    AF = BM = CH = p - a ?
    Thanks for the help.
    Nilton Lapa

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    Replies
    1. To Nilton,
      Problem 86, See

      Semiperimeter s, Incircle and Excircles
      " or

      Semiperimeter Index
      " or

      use "two tangent segments to a circle from an external point are congruent."

      Delete
  3. To Antonio: thanks for the hints. My doubt is solved now.

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  4. of the above solution, why it can be said that the triangle DEF = triangle MGH??

    ReplyDelete
  5. In the solution of pr. 86 posted by Anonymous, could anyone please explain me how to do this ''little'' algebra?

    ReplyDelete