Thursday, June 14, 2012

Problem 764: Triangle, Inradius, Altitudes, Harmonic Mean

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

Click the figure below to see the complete problem 764.

Online Geometry Problem 764: Triangle, Inradius, Altitudes, Harmonic Mean.

4 comments:

  1. Let S be the area of triangle ABC,and a,b,c be the lengths of the sides
    By the properties of incenter
    S = (1/2)(r)(a+b+c)
    Now we consider the area of the triangle and the altitudes
    S = (Ha)(a)/2 = (Hb)(b)/2 = (Hc)(c)/2
    Rearranging, a = 2S/(Ha) , b = 2S/(Hb), c = 2S/(Hc)
    Hence,
    S = (1/2)(r)(2S)[1/(Ha)+1/(Hb)+1/(Hc)]
    1 = r [1/(Ha)+1/(Hb)+1/(Hc)]

    q.e.d.

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  2. http://img850.imageshack.us/img850/6278/problem764x.png

    Denote S(XYZ)= area of triangle XYZ
    Draw additional lines per attached sketch.
    We have ∆BEB’ similar to ∆IFB’
    So IF/BE=r/hB =B’I/B’B= S(IAC)/S(ABC)
    Similarly r/hA=A’I/A’A=S(IBC)/S(ABC) and r/hC=C’I/C’C=S(IAB)/S(ABC)
    But S(ABC)=S(IAC)+S(IAB)+S(IBC)
    So r/hA+r/hB+r/hC= 1 or 1/r=1/hA+1/hB+1/hC

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  3. Let BI extended meet AC at E.
    BI:IE = (a + c): b
    So r/h_b = IE/BE = b/(a + b +c)
    Similarly r/h_c = c/(a + b +c) and r/h_a = a/(a + b +c)
    Hence by adding
    r/h_b + r/h_c + r/h_a = 1,
    1/h_b + 1/h_c + 1/h_a = 1/r

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  4. Result follows easily by noting that double the area of Tr. ABC = r(a+b+c) = a(ha) = b(hb) = c(hc) and eliminating a, b, c

    Sumith Peiris
    Moratuwa
    Sri Lanka

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