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

Posted by Antonio Gutierrez at 3:48 PM
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6 comments:

  1. shaileshSeptember 4, 2009 at 4:29 AM

    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. AnonymousSeptember 24, 2009 at 12:18 AM

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

    ReplyDelete
  3. AnonymousNovember 20, 2009 at 7:05 AM

    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

    ReplyDelete
  4. AnonymousDecember 30, 2010 at 10:43 AM

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

    ReplyDelete
  5. ArifFebruary 9, 2018 at 8:14 AM

    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
  6. MarcoApril 9, 2023 at 1:55 AM

    Let the other tangent point from A, B & C are X, Y & Z respectively in which XY=x & YZ=y
    By intercpt theorem, (b+c)/(a+b)=y/x---------(1)

    Then by using area,
    [ACZX]=[ABYX]+[BCZY]
    (c+a)*(x+y)/2=(b+c)*y/2+(a+b)*x/2
    cx+cy+ax+ay=by+cy+ax+bx
    cx+ay=by+bx
    y/x=(b-c)/(a-b)---------(2)
    By equating (1) & (2)
    (b-c)/(a-b)=(b+c)/(a+b)
    ab+b^2-ac-cb=ab+ac-b^2-bc
    2b^2=2ac
    b=sqrt(ac)

    ReplyDelete

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