Proposed Problem

Click the figure below to see the complete problem 315.

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Complete Problem 315

Level: High School, SAT Prep, College geometry

## Sunday, July 5, 2009

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

Labels:
circle,
common tangent,
geometric mean,
tangent

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Let point of tangent on circles be C',B',A'.

ReplyDeleteNow 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.

Hi shailesh.......

ReplyDeleteby equating those two equations for C'A' you will get a quadratic equation which has complex roots.....

just verify....

draw BA1perpendicular to a, CB1 perpendicular to b

ReplyDeleteright triangles AA1B and BB1C are similar (angle ABB1=angleBCB1) => a-b/a+b=b-c/b+c =>b'2=ab

Actually you get:

ReplyDelete2b^2 = 2ac

=> b^2 = ac

=> b = sqrt(ac)

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

ReplyDeleteBy 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