Thursday, June 14, 2012

Problem 765: Triangle, Exradii, Inradius, Three Exradius, Harmonic Mean

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 765.

Online Geometry Problem 765: Triangle, Inradius, Exradii, Three Exradius, Harmonic Mean.

3 comments:

  1. http://img43.imageshack.us/img43/9397/problem765.png

    Denote S(XYZ)= area of triangle XYZ
    Draw additional lines per attached sketch.
    We have ∆AA’F similar to ∆AIE
    So IE/A’F=r/rA =AI/AA’= S(IC’B’)/S(A’B’C’)
    Similarly r/rB=S(IA’C’)/S(A’B’C’) and r/rC=S(IA’B’)/S(A’B’C’)
    But S(A’B’C’)=S(IA’C’)+S(IA’B’)+S(IC’B’)
    So r/rA+r/rB+r/rC= 1 or 1/r=1/rA+1/rB+1/rC

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  2. Let S be the area of the triangle and X = (1/2)(a+b+c), where a,b,c are the side lengths of the triangle.
    By the properties of the exradii, S=(Ra)(X-a) ; S=(Rb)(X-b) ; S=(Rc)(X-c).
    Therefore,
    S/(Ra)=(X-a)
    S/(Rb)=(X-b)
    S/(Rc)=(X-c)
    By summing up, S[(1/Ra)+(1/Rb)+(1/Rc)]=3X-(a+b+c) = X.
    By the properties of the inradii, S=RX => X =S/R
    So,
    S/R = S[(1/Ra)+(1/Rb)+(1/Rc)]
    1/R = (1/Ra)+(1/Rb)+(1/Rc)

    q.e.d.

    Now,

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  3. Let BC touch incircle at X and excircle opp to B at Y
    Triangles BIX and BI₂Y are similar
    r/r₂ = AX/AY = (s - b)/s
    Similarly r/r₃ = (s - c)/s and r/r₁= (s - a)/s
    r/r₁ + r/r₂ + r/r₃
    =(s - a)/s + (s - b)/s + (s - c)/s
    = [3s - (a + b + c)] = (3s - 2s) /s = 1
    Hence 1/r₁ + 1/r₂ + 1/r₃ = 1/r

    ReplyDelete