## 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.

Complete Problem 315
Level: High School, SAT Prep, College geometry

1. 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.

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

3. 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

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

5. 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