# How to partition area of an ellipse into odd number of regions?

Is it possible to divide an ellipse into 3,5 or 7 etc. parts of equal area? If yes then how?

Describe a circle around the ellipse and the circle of an equilateral triangle we construct. Projection of points on the circle is an ellipse with the ellipse on the surface distribution of three equal parts. To divide into 5 the same procedure to construct a pentagon in a circle.

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Nice question! +1 –  Arjang Aug 6 '11 at 15:50
You should explicitly mention the rules of your game. It is clearly possible to partition the area of an ellipse into $N$ regions of equal area for any natural $N$, but you probably mean "by explicit construction", in the same way that it is possible to divide it into $2$ and $4$ regions. You should also mention how the areas are supposed to look like (e.g., are they contiguous? congruent?). –  Yuval Filmus Aug 6 '11 at 16:04
@Yuval : I would be intrested in the solution satisfying least or no requirements, I guess as long as the process is finite should be still be intresting. –  Arjang Aug 6 '11 at 16:11
I suppose a tag related to constructive mathematics is in order. Otherwise the answer is simply "Yes. You can.". –  Asaf Karagila Aug 6 '11 at 18:51
You posted an answer as part of the question -- it would be preferable to post it as an answer (which it is). –  joriki Aug 9 '11 at 14:21

Place an electric charge at one focus of the ellipse and fixate it. Place a second charge of opposite polarity on the ellipse with frictionless support and impart the initial velocity that will cause its trajectory to trace the ellipse periodically. Measure the period, divide it by the desired number of regions, mark the positions of the second charge at intervals of the resulting time, and connect them to the focus. By Kepler's second law, the resulting areas will be equal.

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@Paj You should perhaps post it as an answer to your own question? –  mixedmath Aug 6 '11 at 18:17
@Рајко: Yes, the FAQ explicitly allows answering your own question if you have a solution: math.stackexchange.com/faq#questions –  joriki Aug 6 '11 at 18:45
Very nice answer! –  barf Aug 6 '11 at 21:43
@Рајко: I'm very sorry, but what you're saying just isn't making any sense to me. If you have a solution to this question, please post it as an answer as mixedmath suggested. –  joriki Aug 7 '11 at 8:00
This is a cute idea but I can't go and actually get some charges and do this, so it just tells me it's possible. But the intermediate value theorem would easily tell me that. –  Ben Millwood Jun 30 '12 at 1:44

Scale in the $y$-direction until we get a circle, divide, scale back. Given the ellipse, if the division of the circle can be done with Euclidean tools, so can the division of the ellipse. If the construction must be done with straightedge and compass, this construction unfortunately excludes almost all odd numbers, though $3$ and $5$ are fine, since the equilateral triangle and the regular pentagon are Euclidean constructible.

Here is some detail. Suppose, for example, that the ellipse has equation $x^2/a^2+y^2/b^2=1$. Draw the circle with center the origin and radius $a$. Let $P$ be one of the division points of the circle. Drop a line from $P$ parallel to the $y$-axis until it meets the ellipse.

Whatever methods of construction are allowed, as long as they work for the circle and include the Euclidean tools, they can be extended to the ellipse.

Comment:: The general idea is fairly powerful. For example, suppose that we want the tangent lines to a standard ellipse that pass through a given point on the $x$-axis. Transform the ellipse to a circle by scaling in the $y$-direction, solve the problem for the circle using geometric ideas, scale back. No calculus needed. Or a simpler problem, find the area of the ellipse with major and minor axes of length $2a$ and $2b$. Using the transform technique, as long as we think we know what scaling in the $y$-direction does to areas, we can conclude that the area of the ellipse is $\pi ab$.

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Here are some easy solutions, satisfying Arjang's requirements:

1. Put the ellipse in some orientation, and pass a horizontal line from top to bottom. Continuously count the area covered "so far", and put a segment whenever it is exactly $1/N$ of the area.
2. Joriki's solution: select some point inside the ellipse, and do the same with a circular "sweep radius".
3. Pick some line orientation, and like in (1), mark off an area of size $1/N$. Continue with some other orientation, etc. Should work with any convex curve.
4. Mark off some rectangle which touches the ellipse in all its vertices. We get a rectangle and four convex shapes. Divide each of the convex shapes into $N$ parts using one of the methods above. Divide the rectangle into $N$ parts using a straightedge and compass.

And so on.

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Just to throw that out there, but the article i also linked in this question Ellipse 3-partition: same area and perimeter shows that it is impossible to divide a convex body in the plane into 7 equal parts of equal area using three lines. The article is from 2010 by Steiger, Szegedy and Zhao Six-way Equipartitioning by Three Lines in the Plane and is freely availiable from Steigers university website.

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