Sunday, October 7, 2012

Problem 812: Four Tangent Circles, Common Tangent Line, Diameter, Area, Collinear Centers

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 812.

Online Geometry Problem 812: Four Tangent Circles, Common Tangent Line, Diameter, Area, Collinear Centers.



1 comment:

  1. Let AB = 2·r1, BC = 2·r2, CD = 2·r3.

    ***
    First of all,

    a^2 = (r1+r2)^2 − (r1−r2)^2 = 4·r1·r2
    b^2 = (r2+r3)^2 − (r2−r3)^2 = 4·r2·r3

    ab = 4·r2·√(r1·r3) ...(1)

    ***
    On the other hand,

    (a+b)^2 = (r1+2·r2+r3)^2 − (r1−r3)^2
    = 4·r2^2 + 4·r1·r2 + 4·r2·r3 + 4·r1·r3
    = a^2 + b^2 + 4(r2^2 + r1·r3)

    ab = 2(r2^2 + r1·r3) ...(2)

    ***
    Comparing (1) & (2),

    r2^2 + r1·r3 = 2·r2·√(r1·r3)
    [r2 − √(r1·r3)]^2 = 0
    r2 = √(r1·r3)
    r2^2 = r1·r3 ...(3)

    ***
    By (2) & (3), we have

    ab = 4·r1·r3

    ***
    Hence,

    2·S + π(r1^2 + r2^2 + r3^2) = π(r1+r2+r3)^2

    S = π(r1·r2 + r2·r3 + r1·r3)
    = π/4 (a^2 + b^2 + ab)

    ReplyDelete