Uniting Polygons And Parabolas

Relevant wiki: Regular Polygons - Problem Solving - Medium

Rotate the plane so that the parabola is facing up, and put a coordinate system so that the parabola is $y = cx^2$ for some $c > 0$ . Label the vertices of the polygon as $A_1, A_2, \ldots, A_n$ in order from left to right. (There is exactly one point on the parabola for each abscissa, so no two points of the polygon will have the same abscissa, thus we can label them from left to right.) Since the polygon is regular, it is convex, so it must be $A_1 A_2 \ldots A_n$ (in that order).

Without loss of generality, assume that $A_1$ is not lower than $A_n$ (otherwise reflect the plane through the y-axis). Note that because a parabola is convex, $A_1$ is now also the topmost point. (If for some $A_k$ , we have $A_k$ higher than both $A_1$ and $A_n$ , then it contradicts the convexity of the parabola.)

Let $A_1 = (x_1, y_1), A_2 = (x_2, y_2), A_n = (x_n, y_n)$ . Observe that $x_1 < x_2$ and $x_1 < x_n$ (because $A_1$ is the leftmost point), but $y_1 \ge y_2$ and $y_1 \ge y_n$ (because $A_1$ is the topmost point). Thus the vector $\overrightarrow{A_1 A_2}$ and $\overrightarrow{A_1 A_n}$ are both pointing to the fourth quadrant, and thus the angle between them cannot be greater than $90^\circ$ . In fact, it cannot be $90^\circ$ either, since neither of them is pointing to the negative y-axis.

But we know that the angles of a regular polygon is given by the formula $\frac{n-2}{n} \cdot 180^\circ$ . For $n \ge 4$ , we have $\frac{n-2}{n} \cdot 180^\circ \ge \frac{2}{4} \cdot 180^\circ = 90^\circ$ , contradicting the result that the angle $A_2 A_1 A_n$ is less than $90^\circ$ . This proves that $n \le 3$ .

To show that $n = 3$ is possible, just choose $c = \sqrt{3}, A_1 = (-1, \sqrt{3}), A_2 = (0, 0), A_3 = (1, \sqrt{3})$ . This proves that the maximum number of vertices is $\boxed{3}$ .

Easy way to do it is :

A regular polygon can be inscribed in a circle

A circle can intersect a parabola at a max of 4 times

Proof of above statement:

Let

$y = mx^2$

$(y-a)^2+(x-b)^2 = r^2$

Subbing first equation into second you get a polymonial of deg 4

hence there are only 4 possible solutions at the max

you are left with 3 and 4 vertices (4 is impossible due to the fact that is has to be a square and has 90 deg)

Therefore answer is $\boxed{3}$

But Ivan Koswara's solution is more interesting

Clearly if it is a regular polygon then at max. It can have 3 vertices each eqiuidistant from each other & lying on the same parabola .

Hence the polygon is equilateral triangle. And thus the ans. Is 3.!

I dont know much about the concept. So it would be nice if someone explains it in a better way. ! Thank you.