In a recent post, I mentioned a problem posed in 1917, which remains unsolved and which lends itself to use in K-12 education:
Consider an n by n lattice. Is it always possible to choose 2n points in it so that no three points are in a line?
Today, I present a related unsolved problem I learned about at the same meeting: the Heilbronn triangle. Here is the problem as stated in Wikipedia:
(This is related to the previous problem, since three collinear points can be thought of as a triangle of area 0.)
Here is an attempt at a classroom-friendly version of the problem, which ties in well with other secondary school work on lattices (as described for example in this post.)
Consider a 5 by 5 lattice. Choose six points on the lattice. They determine many triangles. Among them, find the triangle of least area. Call this area A. Can you rearrange the six points so that the triangle of least area has an area greater than A? What is the greatest area you can get for the triangle of least area?
For example, in the arrangement above, the triangle of least area is the one on the diagonal, which has area 0. It is surely possible to do better than 0.
In this arrangement, the triangle with smallest area is shown. It has area 1, which is greater than 0, so we are doing better. (You can find the area either by subtracting triangles from a rectangle, or by using Pick’s formula.)
But we can still improve on this!
In the above arrangement, a triangle with smallest area is shown. It has area 2. (It is not the only one with this area. Can you find the others?) According to Wikipedia, this is the best possible arrangement for six points in a square.
Of course, the problem could be explored with a different-sized array, or a different number of points.
Because the idea of the “largest smallest” area is fairly subtle, I recommend introducing the problem with an example, such as the one above.