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Posted by markr on 24/7/2009, 6:27 am, in reply to "Circles ahoy"
If I didn't screw up the calculations, I get six solutions depending on whether the circles are inside or outside of the trapezoid. There is one pair (out of four) of horizontal tangents that cannot be used together. So, that gives three pairs of horizontal tangents with two solutions each (circles with y=9 inside and outside of the trapezoid).
2*(12-17/sqrt(3)), 2*(12+5/sqrt(3)), 44/sqrt(3), 44/sqrt(3)
2*(12+17/sqrt(3)), 2*(12-5/sqrt(3)), 44/sqrt(3), 44/sqrt(3)
2*(12+1/sqrt(3)), 2*(12-13/sqrt(3)), 28/sqrt(3), 28/sqrt(3)
2*(12-1/sqrt(3)), 2*(12+13/sqrt(3)), 28/sqrt(3), 28/sqrt(3)
2*(12-3*sqrt(3)), 2*(12-sqrt(3)), 12/sqrt(3), 12/sqrt(3)
2*(12+3*sqrt(3)), 2*(12+sqrt(3)), 12/sqrt(3), 12/sqrt(3)
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