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 Three Circles (Posted on 2004-07-28)
Three circles of radius 6, 7, and 8 are externally tangent to each other. There exists a smaller circle tangent to all three (in the space created between the three original circles).

What is the radius of this smallest circle?

 No Solution Yet Submitted by ThoughtProvoker Rating: 3.0000 (3 votes)

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 simple yet elegant Comment 7 of 7 |

Make a triangle with the centres of each of the 3 large circles. This triangle has sides of dimension 13, 14 & 15 units.

The small centre circle has radius R.

Divide the large triangle into 3 smaller triangles, with the common corner being the centre of the small circle. This makes the sides of the smaller triangles:

T1 : 13, 7+R, 6+R

T2 : 14, 8+R, 6+R

T3 : 15, 8+R, 7+R

Using area = sqrt[s(s-a)(s-b)(s-c)],

where a,b,c are sides of triangle and s=(a+b+c)/2

Area large triangle = sqrt(21.6.7.8) = 84

Area T1 = sqrt[(13+R).R.6.7] = sqrt(42R^2 + 546R)

=> Area T1 ~ sqrt(546R) for small R

&   Area T2 = sqrt[(14+R).R.6.8] ~ sqrt(672R)

&   Area T3 = sqrt[(15+R).R.7.8] ~ sqrt(840R)

Sum of areas of 3 smaller Ts = area of large T

=> sqrt(546R) + sqrt(672R) + sqrt(840R) ~ 84

=> sqrt(R) ~ 84/[sqrt(546)+sqrt(672)+sqrt(840)]

=> R ~ 1.15

Edited on August 1, 2004, 7:37 pm

Edited on August 1, 2004, 7:39 pm
 Posted by David on 2004-08-01 19:29:33

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