Sunday, July 5, 2009

Problem 315: Three tangent circles, Common external tangent line, Geometric Mean

Proposed Problem
Click the figure below to see the complete problem 315.

 Problem 315: Three tangent circles, Common external tangent line, Geometric Mean.
See also:
Complete Problem 315
Level: High School, SAT Prep, College geometry

5 comments:

  1. Let point of tangent on circles be C',B',A'.
    Now it can be proved that
    C'B' = 2*sqrt(cb).
    B'A' = 2*sqrt(ab)
    C'A' = sqrt((c+2*b+a)^2-(a-c)^2)
    C'A' = C'B' + B'A'
    solving above will give you b^2=a*c.

    ReplyDelete
  2. Hi shailesh.......
    by equating those two equations for C'A' you will get a quadratic equation which has complex roots.....
    just verify....

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  3. draw BA1perpendicular to a, CB1 perpendicular to b
    right triangles AA1B and BB1C are similar (angle ABB1=angleBCB1) => a-b/a+b=b-c/b+c =>b'2=ab

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  4. Actually you get:
    2b^2 = 2ac
    => b^2 = ac
    => b = sqrt(ac)

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  5. Let the point M be the point of tangency with the circle B. Connect M with E and F.

    By problem 278 we can find that :

    EM^2= 4b^2a/(a+b)

    and

    MF^2 = 4b^2c/(b+c)

    Because EMF is a right triangle by thales we can apply pythagoras :

    MF^2+EM^2=EF^2
    4b^2c/(b+c)+4b^2a/(a+b)=4b^2
    c/(b+c)+a/(a+b)=1
    c(a+b)+a(b+c)=(b+c)(b+a)
    ac+bc+ab+ac=b^2+bc+ab+ac

    ac=b^2

    ReplyDelete