Geometry Problem

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

Click the figure below to see the complete problem 812.

## Sunday, October 7, 2012

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

Labels:
area,
circle,
common tangent,
diameter

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Let AB = 2·r1, BC = 2·r2, CD = 2·r3.

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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)

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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)

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Comparing (1) & (2),

r2^2 + r1·r3 = 2·r2·√(r1·r3)

[r2 − √(r1·r3)]^2 = 0

r2 = √(r1·r3)

r2^2 = r1·r3 ...(3)

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By (2) & (3), we have

ab = 4·r1·r3

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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)

How did you simplify S = π(r1·r2 + r2·r3 + r1·r3) to equal π/4 (a^2 + b^2 + ab)?

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