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
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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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