Given an ellipse, if you choose 5 points on the ellipse such that they are equal arc length's apart, does the pentagon formed by these points have the same area regardless of the starting position?
If so how would one prove that this is or is not the case?
So the conjecture is proven wrong however there are still special cases of the 3:4 ellipse where the conjecture seems to hold.
This problem arose from this discussion on brilliant.org.
I think this is not true. For a $(3,4)$ pentagon starting at the end of the major axis, I get an area of 28.4032, and at the minor axis I get 28.4020.
Obviously, this is a very small difference; I will try to justify the claim that this is not round off error. I ran my innermost computations with WorkingPrecision set to 100 digits. As a check on my computations, I directly traversed each of the 5 arcs around the ellipse, rather than manually setting the final vertex equal to the initial one; the difference between the initial vertex and the computed position for the final one was of order $10^{-12}$.
In terms of the parametrization $(a \sin \theta, b \cos \theta)$, I get that the vertices are at $\theta$ values of
{0.000000000000, 1.292954985206, 2.443082328571, 3.840102978604, 4.990230321969, 6.283185307175}
{1.570796326795, 2.779128093468, 4.147204875501, 5.277573085264, 6.645649867296, 7.853981633971}
Note that the last entry in each row is almost $2 \pi$ and $5 \pi/2$ respectively, but differs in the last digit given; if the computation were exact, they would be precisely equal.
Let me also give an intuitive explanation of why we expect the variation to be small. Let the arclength parametrization be $(a \sin \theta(t), b \cos \theta(t))$. Then $\theta'(t)$ is proportional to $1/\sqrt{a^2 \cos^2 \theta(t) + b^2 \sin^2(t)}$, which only varies between $1/a$ and $1/b$. It is not that far off from just taking $\theta$ to vary linearly with $t$. For example, if $(a,b)=(3,4)$ and you drive around an elliptical track at a constant rate of $\theta'$, then your minimum and maximum speed are in a 3/4 ratio, like changing between 60 MPH and 80 MPH. Noticeable, but not giant.
If we drive at a rate of constant $\theta'$, then the area doesn't depend on the starting point. To see this, notice that the pentagon with vertices at $(a \cos(\theta_0 + 2 \pi k/5), b \sin(\theta_0 + 2 \pi k/5))$ is the image under the linear map $(x,y) \mapsto (ax,by)$ of a regular pentagon. The area of the regular pentagon is independent of $\theta_0$, and linear maps preserve ratios of areas.
When we drive at arc length parametrization instead of constant $\theta'$, we are no longer considering linear images of regular pentagons, but of pentagons that are close to regular. I was surprised by how close: Starting at the end of the minor axis, I found that I need to go to $\theta = 2 \pi \cdot 0.20578$ to get $1/5$ of the way around; starting on the major axis, I needed $\theta = 2 \pi \cdot 0.18794$, so only about a 10% difference. Moreover, the discrepancies from regularity push both ways, with some edges getting longer and some getting shorter, so the total area change is even smaller.
To be honest, I am still surprised as to how small the variation is. I'd love to see an answer which works out approximately how much variation to expect for an ellipse as a function of its eccentricity.
Mathematica code follows:
(* Minor and major axis, and the total perimeter computed once and for all. *)
a = 3; b = 4; perimeter = NIntegrate[Sqrt[a^2 Cos[t]^2 + b^2 Sin[t]^2], {t, 0, 2 Pi}, WorkingPrecision -> 100]
(* If you are at position theta, and want to move a fraction delta
around the ellipse, what angle should you go to?
Because of the branch cut in EllipticE, delta must be between 0 and 1/2. *)
forward[theta_, delta_] :=
Re[phi] /.
FindRoot[
Integrate[Sqrt[a^2 Cos[t]^2 + b^2 Sin[t]^2], {t, theta, phi},
Assumptions -> theta < phi < theta + Pi] == delta*perimeter,
{phi, 2 Pi*delta}, WorkingPrecision -> 100]
eParam[theta_] := {a Sin[theta], b Cos[theta]}
(* Find the vertices of an equal arc n-gon, starting at theta=t0. *)
ePolygon[t0_, n_] := eParam /@ NestList[forward[#, 1/n] &, t0, n]
(* Computes the area of a convex polygon, given a list of vertices. *)
area[L_] := (1/2) Abs[Sum[Det[{L[[i]], L[[i + 1]]}], {i, 1, Length[L] - 1}]]
