Monday, May 19, 2008

Elearn Geometry Problem 82



See complete Problem 82
Area of the Contact Triangle, Inradius, Circumradius. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

3 comments:

  1. S1 = Tr.IDE + Tr.IEF + Tr.IFD
    = [r^2*S/bc + r^2*S/ca + r^2*S/ab] by previous problem.
    Thus, S1= S*r^2[1/bc + 1/ca + 1/ab]
    and S = abc/4R where a,b & c have their usual meaning.
    Hence. S1 = (abcr^2/4R)[1/bc + 1/ca + 1/ab]
    = (r^2/4R)(a+b+c)
    = (r/2R)[r*(a+b+c)/2]
    We know that S = Inradius*Semisum of sides
    = [r*(a+b+c)/2]
    Hence, S1 = (r/2R)* S or S1/S = r/2R
    Ajit: ajitathle@gmail.com

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  2. Alternative solution to problem 82.
    It was proved in problem 081 that S(DIE) = (a.r*2)/(4R). Similarly, S(DIF) = (b.r*2)/(4R) and S(EIF) = (c.r*2)/(4R). The area S1 of tr. DEF is the sum of those three areas, so S1 = (a+b+c).r*2/(4R).
    We also know that the area of tr. ABC is given by S = (a+b+c).r/2. Hence
    S1/S = ((a+b+c).r*2).2)/(4R.(a+b+c).r) = r/2R

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  3. Reference my proof of Problem 81,

    S(DEF) = ar^2/4R + br^2/4R + cr^2/4R

    Hence S1 = (r/2R)(ra/2 + rb/2 + rc/2) = (r/2R).S

    Sumith Peiris
    Moratuwa
    Sri Lanka

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