# Does there exist a point in every closed shape such that all lines that pass through the point divide the shape in two equal areas?

I was refreshing myself on some basic geometry for fun and thought it was neat that any line through the midpoint of a parallelogram divides its area in two. I started thinking about whether or not such a point exists for all shapes. My guess is no, since there are some pretty crazy shapes out there, but the more I've been considering this the more it's driving me nuts not knowing.

I'm really curious to know what kind of math would go into an answer like this. It seems like a higher level topic, but maybe there's a simple counterexample out there? What do you think?

• Not true even for the equilateral triangle.
– lhf
Jul 29, 2021 at 13:43
• I'd guess that it is true iff the shape is center-symmetric.
– lhf
Jul 29, 2021 at 13:44
• The statement is false for every non-degenerate triangle. Hint: look at those 3 lines that parallel to the sides of triangle and bisect the triangle into regions of equal areas. Jul 29, 2021 at 14:19

First of all, it is crucial to understand the definition of the "center" of a polygon in mathematical terms. It refers to a point in the polygon that is equidistant from all of its vertices.

Your statement is not true all the time. Sure, it is true for all parallelograms, but not for all closed shapes.

Take an equilateral triangle as an example to disprove your statement. It only works when you draw a perpendicular bisector to one of the sides. Other than that, it doesn't give same area.

Now, if we consider all closed polygons, the line through the center of the polygon divides the figure into equal areas if and only if the figure is symmetric by its center. However, not all polygons have this property. In fact, very few of them do. This can be seen through experimentation. Try taking examples of quadrilaterals that are not parallelograms, since it is easier to experiment with them.

Also, the funny thing is, some irregular polygons might not even have a center, as not necessarily there is a point in the wacky shape that is equally distanced from all of the polygon's vertices.

Remark:

By the way, I have an interesting topic for you to think about, which is somewhat related to this: Does every closed polygon have a center? If so, prove it. If you are interested in this posed question, see below for my answer to it.

-------------------------------------------------------------------------------------------------------------------------------

If the polygon is regular, then we can divide it up into triangles by drawing all the diagonals of it. For example, we can draw all three diagonals of a hexagon to form its 6 triangles. Since all of these triangles have their base , which is the side of the polygon, in common, they all have one side congruent. Also note that in a regular polygon, all interior angles are congruent, and the diagonals are the angle bisectors of those interior angles (by symmetry), so these triangles have two angles in common as well. Therefore, using the two facts above, we know that these triangles are all congruent by ASA (angle-side-angle congruence).

And as said before, if it is irregular, it doesn't necessarily have to have a center. For example, the figure below doesn't have a center: .

$$\blacksquare$$

• Your definition of "center" implies that no polygon other than a cyclic polygon can have a center. Most parallelograms do not have a center according to this definition. Jul 31, 2021 at 15:13

Not a higher level topic I suppose, it is about symmetry and anti-symmetry or a semi full-rotation about central pole for certain special shapes with anti-symmetry.

Find a midpoint A of two given points C,B, draw a perpendicular bisector,

Double reflection on these two mirrors any continuous line between C,B gives this property across the arbitrary red line through A.

This equivalent to a single rotation of region through angle $$180^{\circ}$$.

Another example with smooth continuous boundary with red line area bisector.