Wednesday, March 11, 2015

Problem 1096: Tangent Circles, Common Tangent, Chord, Radius, Center

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

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1096: Tangent Circles, Common Tangent, Chord, Radius, Center.

Posted by Antonio Gutierrez at 5:40 PM
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3 comments:

  1. Peter TranMarch 11, 2015 at 8:21 PM

    http://s9.postimg.org/h8iwun773/Pro_1096.png

    Draw lines per sketch
    We have ∠ ( D₃ O2 O₃)=∠ ( O₃OC)= 30 degrees
    And OO3= R-r
    OO2^2= (R-r)^2-4.r^2= R^2-2.R.r-3.r^2
    OC^2=3/4.OO2^2= ¾(R^2-2.R.r-3.r^2)
    CB^2=R^2-OC^2= ¼(R^2+6.R.r+9.r^2)
    AB^2=4.CB^2=(R+3.r)^2= > AB= R+3.r

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  2. Peter TranMarch 11, 2015 at 10:34 PM

    See revised sketch with more details below.
    http://s22.postimg.org/iruqw8anl/Pro_10961.png

    ReplyDelete
  3. Sumith PeirisNovember 16, 2015 at 4:31 AM

    Let OO2 = h and perpendicular from O to AB be p

    Then < O1O2D1 = 30 and so p = sqrt3/2 h

    Now AB^2 = 4(R^2- p^2)
    and h^2 = (R-r)^2 -4r^2

    So AB^2 = 4R^2 - 3h^2 = 4R^2 - 3(R-r)^2 - 12r^2 = (R+3r)^2

    So AB = R + 3r

    Sumith Peiris
    Moratuwa
    Sri Lanka

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