Wednesday, March 25, 2015

Problem 1103: Right Triangle, Incircle, Inscribed Circle, Radius, Geometric Mean, Sangaku, Japanese, Metric Relations

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Online Math: Geometry Problem 1103: Right Triangle, Incircle, Inscribed Circle, Radius, Geometric Mean, Sangaku, Japanese, Metric Relations.

Posted by Antonio Gutierrez at 2:23 PM
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3 comments:

  1. Peter TranMarch 25, 2015 at 6:54 PM

    http://s25.postimg.org/815s5zg3j/pro_1103.png
    Draw lines per sketch
    We have ∠ (OED1)= ∠ (OEO2)= ∠ (OFO1)= ∠ (ODO2)= 45 degrees
    And ∠ (O1FD)= ∠ (FDO2)= ∠ (O1EO2)= 90 degrees
    Calculate O1O2^2= O1E^2+O2E^2= 2. r1^2+ 2.r2^2
    O1O2^2= FD^2+(DO2-FO1)^2
    Replace FD^2= 2.r^2 , DO2= r2.sqrt(2) , FO1= r1.sqrt(2) in above expression
    We get r^2=2.r1.r2

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  2. PravinMarch 27, 2015 at 10:42 AM

    A/2 + C/2 = Pi/4,
    1 + tan C/2 = 1 + tan ( Pi/4 - A/2) = 1 + [(1 - tan A/2) /(1 + tan A/2)] = 2/(1 + tan A/2)
    Let L, M be the projections of O1, O2 on AC resp..
    C, O2, O are collinear and r/r2 = EC/ MC = 1 + EM/MC = 1 + O2M/MC = 1 + tan C/2 = 2/(1 + tan A/2)
    A, O1, O are collinear and r/r1 = AE/AL = 1 + LE/AL = 1 + r1/AL = (1 + tan A/2)
    Hence (r/r2)(r/r1) = 2, r^2 = 2.r1.r2

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  3. Sumith PeirisNovember 23, 2015 at 1:00 AM

    FODB is a square of side r so easily

    r = (c+a-b)/2.......(1)
    From similar Tr. s (r-r1)/r = r1/(c-r)......(2)

    Eliminating r we have

    r1 = (c+a-b)/(c+b-a)/4c = {c^2 - (a-b)^2}/4c

    Simplifying using c^ 2 = b^ 2 - a^2 we have

    r1 = a(b-a)/2c and similarly
    r2 = c(b-c)/2a

    So 2r1r2 = (b-a) (b-c)/2.....(3)

    Now from (1),
    r ^2 = (c+a-b)^2 / 4 which simplifies to (b-a)(b-c) /2 which from (3) is therefore 2r1r2

    Sumith Peiris
    Moratuwa
    Sri Lanka

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