Friday, August 7, 2009

Problem 336. Two equal circles, a Common Tangent and a Square

Proposed Problem
Click the figure below to see the complete problem 336 about Two equal circles, a Common Tangent and a Square.

Problem 336. Two equal circles, a Common Tangent and a Square.
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Complete Problem 336
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 10:38 AM
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2 comments:

  1. Extend FM to meet AC in N. Also extend DA & EB to meet the two circles in D' & E' respectively.
    Now, AN = r - x/2 & NM = r - x. From triangle MAN, (r-x/2)^2+(r-x)^2=r^2 from which we get x=2r/5 or x=2r. The former is the side of square FGHM while the latter is the side of DEE'D'
    vihaan: vihaanup@gmail.com

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  2. Extend MH so that it meets the radii AD and BE at points P and Q
    Let PM = HQ = y

    Then we have two equations:

    (r-x)^2 + y^2 = r^2 ..... (1)

    and

    2y + x = 2r ..........(2)

    Eliminate y between the two equations to get a quadratic which resolves to
    x = (2/5)*r or x = 2r.
    x = 2r is absurd.
    So x = (2/5)*r is the answer
    which will

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