How to find a line that divides two arbitrary triangles (in $R^2$ space) in half (area) at the same time?

We want to find a line that cut through the two triangles at the same time and each triangle is divided into two parts with equal areas.

There is a theorem call Ham Sandwich theorem (in two dimension, also called pancake theorem) saying that given n measurable "objects" in n-dimensional Euclidean space, it is possible to divide all of them in half (with respect to their measure, e.g. volume) with a single (n − 1)-dimensional hyperplane. The theorems give the existence of the line.

I have no clue about how to find such a line even for two arbitrary triangles case. I am thinking about the line going through the center of each individual triangle should be the solution. However, one can easily prove that not every line going through the center (centroid) will divide the triangle in equal area. It can will prove by contradiction and using the continuity of area as the line rotates around centroid .

  • 1
    $\begingroup$ First, define the problem. What does it mean "divide in half"? What if triangles overlap? Then, tell us what you think. $\endgroup$
    – Vasili
    Mar 24, 2021 at 13:26
  • $\begingroup$ @Vasya I think it means area. We want to show that there exists a line that cut through the two triangles at the same time, such that both parts have the same area. $\endgroup$ Mar 24, 2021 at 13:30
  • $\begingroup$ You also need to tell people here what have you tried. Otherwise, people here won't be able to help. $\endgroup$ Mar 24, 2021 at 13:38
  • $\begingroup$ Why overlapping is an issue? $\endgroup$
    – Moti
    Mar 24, 2021 at 21:04
  • $\begingroup$ I believe that the line through the centroids of the triangle is the answer. $\endgroup$
    – Moti
    Mar 24, 2021 at 21:07

1 Answer 1


Every tangent to one of the black curves cuts the area of the triangle in half:

The curves are hyperbolae each determined by two medians which are tangent to them in the vertices of the triangle and another tangent shown in grey which is parallel to one side of the triangle. These are obtained in the following way: $\frac{bh}{2} = ax \quad \frac{a}{x} = \frac{b}{h} \; \to \frac{h^2}{2} = x^2 \quad x = \frac{h}{\sqrt{2}}$ .

For two triangles you have to find the common tangent of the hyperbolae.


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