Three balls of radius 2 are all tangent to one another as well as to a ball of radius 6. A fifth, smaller ball is tangent to all four balls. What is its radius?
I think this is right, but not 100% sure.
Consider the three balls sitting on a plane with the large ball on top. The fifth ball must be directly underneath the large one.
Call A, B, C the centers of the small ball, the large ball, and any of the three of radius 2 respectively.
The line containing AB must pass through the center of the equilateral triangle containing the centers of the three 2 balls. Call this point D. CD=4/sqrt(3)
r is the missing radius. AB=r+6, AC=r+2, BC=8. After simplifying, cos<BAC=(r^2+8r+12)/(r^2+8r-12)
<CAD=180-<BAC
sin<CAD=sin<BAC=CD/(r+2)
turning the cosine into the sine of <BAC eventually leads to the quadratic equation 2r^2+15r-9=0 from which r=(-15+3sqrt(33))/4 or about 0.5584
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Posted by Jer
on 2019-08-26 19:11:58 |
Solution (I'd like confirmation)
