(In reply to
Solution by Brian Smith)
'Then substitute into the first equation:
(R +/- sqrt(100-3R^2))/2 + (R +/- sqrt(576-3R^2))/2 = 13'
To be exact: ((576-3R^2)^(1/2)-R)/2+(R + (100-3R^2)^(1/2))/2 =13
Then ((100-3R^2)^(1/2)+(576-3R^2)^(1/2))^2 = 26^2, and there is a sort of quartic, which cancels: (300-9R^2)(192-R^2) = 9R^4
But if so, then 57600-2028R^2 = 0; R^2 = 4800/169: R = 40*3^(1/2)/13, as expected.
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Posted by broll
on 2016-06-12 05:13:37 |