Monday, February 9, 2015

Geometry Problem 1081: Equilateral Triangle, Inscribed Circle, Inradius, Tangent Circles, Radius, Tangent Line, Sangaku Japanese Problem

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 1081: Equilateral Triangle, Inscribed Circle, Inradius, Tangent Circles, Radius, Tangent Line, Sangaku Japanese Problem.

Posted by Antonio Gutierrez at 4:12 PM
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3 comments:

  1. Peter TranFebruary 9, 2015 at 8:14 PM

    http://s14.postimg.org/7bjgzefe9/pro_1081.png
    Draw lines and points per attached sketch
    We have OB= 2.r , OE= r, DB= r
    Triangle ODF is equilateral
    Along OD we have OD= OB-DB => a+ b= 2r-2a => 3a+b= 2r …. (1)
    Along OE we have OF=OE-FE => a+b=r-b => a+2b=r.. ( 2)
    Along OF we have OF=2b+4c=r ……(3)
    From (1) and ( 2) we have a= 3/5 r and b= r/5
    Replace value of b in (3) we get c= r/10

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  2. Jacob HA (EMK2000)February 9, 2015 at 8:38 PM

    Consider the altitude of the triangle, we have
    3a + 4b + 4c = 3r

    Consider thr radius of the large circle C1,
    2a + b = 5b + 4c (=R)
    a − 2b − 2c = 0

    Consider the radius of the in-circle,
    3b + 4c = r

    Summarizing, we have
    3a + 4b + 4c = 3r
    a − 2b − 2c = 0
    3b + 4c = r

    Hence,
    a=3r/5, b=r/5, r/10.

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  3. PravinFebruary 9, 2015 at 9:02 PM

    Easy to observe that
    r = 3b + 4c,
    b + 2a = R = r + 2b and thus r = 2a - b,
    2r = b + 3a.
    Add the last two equations:to get 3r = 5a or a = 3r/5
    Next b = 2a - r = 6r/5 - r = r/5,
    c = r/4 - 3b/4 = r/4 - 3r/20 = 2r/20 = r/10

    ReplyDelete

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