Shapes that can be cut in half by any line/plane going through its center

387 Views Asked by At

I am aware that a line passing through the center of a square cuts the area in half. Same is true for a circle and a rectangle. What other shapes is this true for? More interestingly, can one guide me to theorems like this that is true in N-dimensional shapes?

Thanks

2

There are 2 best solutions below

1
On BEST ANSWER

First, let me suggest a characterization of contractible shapes with your property in the plane. I‘ll assume that the center is the origin. Then, a shape should have the required property if and only if it is point-symmetric with respect to the origin.

Clearly, if it is point-symmetric, then every line through the origin will split the shape in half. On the other hand, assume that every line through the origin splits your shape in half, and rotate the line for an arbitrarily small angle. This gives you a double wedge, where both sides need to have the same area, as the rotated line again splits the shape in half. In the limit, these wedges are segments of the same length that lie on the same line through the origin, meaning that on this line, the shape is point-symmetric. As this is true for every line through the origin, we can conclude that the whole shape is point-symmetric.

Of course, one needs to check the details, and it could be that there are some subtelties that I have overlooked.

I also think that similar arguments should work in higher dimensions, showing that the intersection of a shape with a hyperplane must have the same property one dimension lower, and then using induction to give the same characterization as in the plane.

As for non-contractible shapes, like an annulus, more things might happen.

In general, this problem is related to the Centerpoint Theorem, which states that there is always a point such that each half-space containing this point contains at least a 1/(d+1)-fraction of the area of the shape (or more generally, of a mass, here d is the dimension).

1
On

Take any point in a plane and consider the set of oriented lines (having a normal vector defining the "up" direction relative to the line) through that point.

For any region in the plane with finite, non-zero area, the area of the subregions above and below a line sum to the area of the region. If we let $\theta$ be the angle of the normal of the line with respect to the $x$-axis, then the function $f$ which identifies the difference of the area above and below the line with angle $\theta$ is a continuous function of $\theta$.

Further, $f(\pi) = - f(0)$ because the angles $0$ and $\pi$ refer to the same line with opposite orientation. By intermediate value theorem, this implies that there is some $\theta \in [0, \pi)$ such that $f(\theta) = 0$, and so the line with that angle will bisect the area of the region.

Notice that this is true for any point in the plane, and so is true for any possible definition of "center" for the region, provided such a point exists.

This is a special case of the ham sandwich theorem (as noted by Eric Towers in the comments). This holds for up to $n$ distinct regions in $n$-dimensional Euclidean space via the Borsuk-Ulam theorem, a kind of variation/generalization of the intermediate value theorem that fits these kinds of "choice of orientation" problems well.