Given lengths R and r with R > 2r. Construct a triangle with side lengths in arithmetic progression and R and r its circumradius and inradius respectively.
(In reply to
re: solution by Bractals)
My mistake: thanks for correcting. It should be as you noted:
a=√(4r(2R-r))
x=√(2r(R-2r))
The right answer for triangle sides are:
√(4r(2R-r))-√(2r(R-2r))
√(4r(2R-r))
√(4r(2R-r))+√(2r(R-2r))
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Posted by Art M
on 2007-01-31 20:09:58 |
re(2): solution
