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

4 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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  3. See diagram here.
    Let O middle of FG and extend EB to intersect circle B in K
    ∆OGH and ∆OEK are SAS similar and DE tangent to B in E => ∠OHG = ∠HKE = ∠OEH
    Hence ∆OGH and ∆HGE are similar => EG = 2x and OE = r = 5x/2 QED

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  4. harvey.littleman@sciences.heptic.frApril 1, 2021 at 6:51 AM

    See the drawing

    In ΔDMF : DM^2=DF^2+FM^2
    (1) DM^2= (r-x/2)^2+x^2
    In ΔAMD : AM^2=AD^2+DM^2 -2xAD (Proposition 13 Euclide Book II)
    r^2= r^2+ DM^2 -2xr
    (2) DM^2=2xr
    (1) and (2) 2xr=(r-x/2)^2+x^2
    2xr=r^2-xr+x^2/4+x^2
    3xr=r^2+5x^2/4
    Dividing by x^2 (x not equal to 0) =>3r/x=5/4+(r/x)^2
    Let y=r/x : y^2-3y+5/4=0
    Delta = 4 and y=5/2 or y=1/2
    Therefore x=2r/5 or x=2r

    Since the only definition of M and H are
    M belongs to the square and circle A,
    and H belongs to the square and Circle B
    Therefore, there are 2 different solutions:
    x=2r/5 (blue square)
    or x=2r (green square)


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