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Can someone tell me how to construct an equilateral pentagon (all its sides equal, not its angles) inscribed in an ellipse (with semi-axes 2 and 1) without approximations, with the exact coordinates of its vertices? (and if possible with a vertex at the point (0,1))

And with an equilateral polygon of n sides?

something approximate of what I ask

tomascatuxo
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  • Are you considering a coordinate geometry approach? It would be advisable to show what you have tried. – David Quinn Jun 09 '23 at 08:12
  • I have nothing clear. I tried to divide the 360 ​​degrees from the center by 5, as it is done to build the regular pentagon inscribed in a circle, but obviously it is not regular in the ellipse (not even equilateral). Thanks in advance for answering. – tomascatuxo Jun 09 '23 at 08:28
  • Please refer to the case for equilateral triangle here. – Ng Chung Tak Jun 09 '23 at 09:28
  • A brute force approach would be to assign parameter values to points $B$ and $G$ and start writing down some equations (but I haven’t actually tried it myself) – David Quinn Jun 09 '23 at 11:37

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Let the vertices of the pentagon be: $$ P_1=(0,1),\ P_2=(x,y),\ P_3=(h,-k),\ P_4=(-h,-k),\ P_5=(-x,y) $$ with $x,y,h,k$ all positive real numbers. It is straightforward to write down four equations for them: $$ \cases{ \strut x^2 + (y - 1)^2 = 4 h^2\\ \strut (x - h)^2 + (y + k)^2 = 4 h^2\\ \strut x^2/4 + y^2 = 1\\ \strut h^2/4 + k^2 = 1 } $$ I solved this system with Mathematica but the exact output is too large to be written here. Here's a numerical approximation: $$ h = 0.84145,\ k = 0.907188,\ x = 1.62958,\ y = 0.579753. $$

Intelligenti pauca
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