 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  The Five Touching Spheres 2 (Posted on 2019-09-15) Two balls of radius 2 and two balls of radius 6 are all externally tangent to each other. A fifth, smaller ball is tangent to all four balls. What is its radius?

 Submitted by Brian Smith Rating: 4.0000 (1 votes) Solution: (Hide) Let points A and B be the centers of the radius 6 balls. Let C be the midpoint of AB; C is also where the 6 radius balls are tangent. AC=CB=6 Let points X and Y be the centers of the radius 2 balls. Let Z be the midpoint of XY; Z is also where the 2 radius balls are tangent. XZ=ZY=2 AY, AZ, BY, and BZ connect the centers of a 6 radius ball and a 2 radius ball. Then AY=AZ=BY=BZ=8. Let M center of the fifth ball. M lies on the line segment between C and Z. Let r be the radius of the fifth ball. AM=BM=6+r and XM=YM=2+r. AB is orthogonal to CZ; XY is orthogonal to CZ; and the vectors of AB and XY are orthogonal. Then CM^2 + 6^2 = (6+r)^2 and MZ^2 + 2^2 = (2+r)^2 and CZ^2 + 6^2 + 2^2 = 8^2 Combining these into one equation yields sqrt[r^2+12r] + sqrt[r^2+4r] = sqrt. After squaring, rearranging, squaring again, and simplifying yields a quadratic r^2+24r-18 = 0. The positive root is r = -12+9*sqrt = 0.727922. Generalizing to general radii a and b gives an equation sqrt[r^2+2ar] + sqrt[r^2+2br] = sqrt[2ab]. Simplifying yields a quadratic in r: (-a^2+4ab-b^2)r^2 + (2a^2b+2ab^2)r - a^2b^2 = 0. The positive root of this eventually yields r = ab/(a+b+sqrt[6ab]). Plugging in a=6 and b=2 yields the same answer. Comments: ( You must be logged in to post comments.)
 Subject Author Date re: The Five Touching Spheres 2 kexegiho 2019-09-30 13:19:23 No Subject kexegiho 2019-09-30 13:17:57 exact solution Steven Lord 2019-09-18 02:50:14 iterative soln Steven Lord 2019-09-16 02:06:06 Please log in:
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