Little explorations with HP calculators (no Prime)

03272017, 06:01 PM
(This post was last modified: 03272017 07:33 PM by Dieter.)
Post: #33




RE: Little explorations with the HP calculators
(03272017 12:14 PM)pier4r Wrote: ...so if the side of the inner square is 2r so the sides of the triangle are "s+2r" and "s" from which one can say that "s^2+(s+2r)^2=1" . This plus the knowledge that 4 times the triangles plus the inner square add up to 1 as area. Still, those were not enough for a solution Right, in the end you realize that both formulas are the same. ;) The second constraint for s and r could be the formula of a circle inscribed in a triangle. This leads to two equations in two variables s and r. Or with d = 2r you'll end up with something like this: (d² + d)/2 + (sqrt((d² + d)/2) + d)² = 1 I did not try an analytic solution, but using a numeric solver returns d = 2r = (√3–1)/2 = 0,36603 and s = 1/2 = 0,5. Edit: finally this seems to be the correct solution. ;) Dieter 

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