It is impossible for a regular polygon to have a focus, vertex, and point of a parabola

Question

A conjecture I made about four years ago about parabolas and vertex and focus regular polygons

I think the problem is related to algebra and number theory and not to geometry despite the geometric way of phrasing the problem

The conjecture states that it is impossible for a regular polygon whose sidelength is FV where F is the focus and V the vertex of a parabola to have a vertex belonging to the parabola that is different from vertex V. Although it is possible to get an infinite number of good approximations, it is impossible to get a completely accurate case

These are, for example, some numbers that give polygons that have a vertex that roughly belongs to a parabola:

$14,25,31,38,44,45,52,60,68,77,85,94,...$

I couldn't find a way to deal with the issue, whoever can help please be so kind

This conjecture can also be strengthened in several ways, for example we can accept regular stellar polygons

Answer 1

This is not a solution, just some thoughts on the question.

Let's consider WLOG a regular polygon of unit radius with $n$ sides, center $C$ and vertices $V, F, P_1, P_2,\dots P_{n-2}$.

The distance $P_kF$ from vertex $P_k$ to focus is given by: $$ P_kF=2\sin{k\pi\over n}. $$ The distance $P_kH$ from vertex $P_k$ to directrix is: $$ P_kH=\sin{(2k+1)\pi\over n}+3\sin{\pi\over n}. $$

$P_k$ lies on the parabola only if those distances are equal among them, that is if $$ 2\sin{k\pi\over n}=\sin{(2k+1)\pi\over n}+3\sin{\pi\over n}. $$ Hence we should prove that the above equality cannot hold for any integer $k$ between $1$ and $n-2$.

Answer 2

Here is a complement to the previous answer by @Intelligenci Pauca, providing an unexpected visualization of the issue.

Let us give the name $R(n,k)$ to relationship

$$R(n,k) : \ \ \ \sin{(2k+1)\pi\over n}-2\sin{k\pi\over n}+3\sin{\pi\over n}=0\tag{1}$$

considered in this answer.

The objective is to establish that

$$ \forall \ \text{(fixed)} \ n>2 , \ \forall k, \ 1 \le k \le n-2, \ R(n,k) \ \text{cannot hold} \tag{2}$$

The logic is now as follows. The negation of (2) is equivalent to :

$$z_{k,n}=w^{2k+1}-2w^k+3w=w(w^{2k}-2w^{k-1}+3) \ \text{where} \ w:=e^{i \pi/n} \ \text{is a real number}\tag{4}$$

(many thanks to @ronno who was the first to point a reasoning error in a first draft of this "answer").

We just have to express that this negation of (2) under the form (4) leads to a contradiction.

For a fixed $n$, when $k$ varies, the set of points $z_{k,n}$ belongs in fact to a cardioid (as shown on the figure). Why that ? They are images of the unit disk by the transformation

$$f(z)=w\left((z-\tfrac{1}{w})^2+3-\tfrac{1}{w^2}\right)\tag{5}$$

(just plug $z=w^k$ in (5) and expand to see the connection with (4)).

In this way, constraint (2) has been "visualized" : just look at the points of intersection of the astroid with the real axis : this point musn't be of the form $w^k$... which is the case in all examples I have taken.

Fig. 1 : Cardioids in the cases $n=10$ and $n=30$ with the relevant points $w_{k,n}$.

Remark : In order to understand why the cardioids can be obtained from circles by squaring in the complex domain, see for example here.

Answer 3

Again, not a full answer, but just to provide a different possible approach: say the parabola has equation $4y=x^2$, so that its vertex is $V(0,0)$ and focus $F(0,1)$. (There's no need to be more general about the position of the focus since it just scales the whole system.)

We can construct two regular $n$-gons with $FV$ as a side; let's take the one with non-positive $x$-coordinates to match the earlier diagrams.

This polygon has sidelength $1$, circumradius $\frac{1}{2}\csc \frac{\pi}{n}$, and centre $\left(-\frac{1}{2}\cot \frac{\pi}{n}, \frac{1}{2}\right)$. All points of the polygon lie on its circumcircle, $$\left(x+\frac{1}{2}\cot \frac{\pi}{n}\right)^2+\left(y-\frac{1}{2}\right)^2=\frac{1}{4}\csc^2 \frac{\pi}{n}$$

After a bit of tidying up, this is $$x^2+x\cot \frac{\pi}{n}+y(y-1)=0$$

This intersects the parabola at the origin, and at $$x=2\sqrt[3]{\tan\frac{\pi}{2n}}-2\sqrt[3]{\cot\frac{\pi}{2n}},\quad y=\sqrt[3]{\tan^2\frac{\pi}{2n}}+\sqrt[3]{\cot^2\frac{\pi}{2n}}-2$$

I haven't found these forms very useful to work with yet, but perhaps someone else might have an idea.

Two possible lines of thought are

1. Equate either the $x$ or $y$ coordinate above with one of the points on the polygon, and show no solution exists

2. Switch to the complex plane and use properties of roots of unity (this switch may be useful earlier on).

Any thoughts on a way to proceed?