# How to fit the curve that this lines made? The length of lines are all fixed  but how to deduce it ? Thank you .

• For the record, this is the forum for the software Mathematica, you were probably looking for math.stackexchange.com Aug 4, 2016 at 9:49
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• Aug 4, 2016 at 11:03
• I don't think your lines on above illustrations are actually equal-length, since they seem to be from $(0, a)$ to $(1-a, 0)$. Aug 10, 2016 at 7:43

It's an astroid and not a circle. In any event:

With[{r = 1, n = 30},
Graphics[{AbsoluteThickness,
Table[Line[r {{Cos[θ], 0}, {0, Sin[θ]}}], {θ, 0, π/2, π/(2 n)}]}]] A short proof that the astroid is the correct envelope:

{x, InterpolatingPolynomial[{{Cos[θ], 0}, {0, Sin[θ]}}, x]} /.
First[Solve[D[InterpolatingPolynomial[{{Cos[θ], 0}, {0, Sin[θ]}}, x], θ] == 0, x]]
// FullSimplify

• Fair enough, I'm too trusting with what people claim to know. Aug 4, 2016 at 9:48
• great! Thank you! Aug 4, 2016 at 11:30

Based on your image, but not your statement on lengths of lines:

With[{eqn = a - a x / (1 - a)},
Show[
Quiet@Plot[Table[eqn, {a, 0, 1, 1 / 20}], {x, 0, 1},
Evaluated -> True, AspectRatio -> Automatic, PlotRange -> {0, 1},
PlotStyle -> Gray],
Plot[MaxValue[{eqn, 0 <= a <= 1}, a], {x, 0, 1},
Evaluated -> True, PlotStyle -> Directive[Red, Thick]]]] For the intended plot your can replace eqn with the following:

eqn = Sin[Pi a / 2] - Tan[Pi a / 2] x • Solve[{#, D[#, a]} &[x/a + y/(1 - a) == 1], {x, y}, {a}] also yields the red parabola (1 + (x - y)^2 - 2 (x + y)). Aug 10, 2016 at 19:00

$$\sqrt{1-x^{2/3}}-\sqrt{1-x^{2/3}}x^{2/3}$$

Plot[Sqrt[1 - x^(2/3)] - Sqrt[1 - x^(2/3)] x^(2/3), {x, 0, 1},
AspectRatio -> 1, PlotStyle -> {Thick, Red, Dashed} 