If any triangle has area at most 1 , points can be covered by a rectangle of area 2.

Question

I am working on this problem for some time, and I am not able to finish the argument:

There is a finite number of points in the plane, such that every triangle has area at most 1. Prove that the points can be covered by a rectangle of area 2.

My progress so far:

Consider the two points furthest apart, A and B. Draw lines through A and B, perpendicular to AB. Then every point has to be between the two lines, since AB has maximal length among our pair of points. Now take the points C and D, furthest apart from the line AB, on each of the two halfplanes determined by AB. Every other point in the plane has to be between the two lies through C and D, parallel to AB .

The rectangle determined by the four lines gives us an area of at most 4, but I feel like this argument can be improved somehow to give us the required area of 2.

Thank you! (Edit: thank you for the advice on posting problems here )

Answer 1

If the theorem is true, a circle that is approximated by a polygon of sufficiently enough vertices verifies the theorem too.

The radius of the circle is $r$. The rectangle is a square and has area $4 r^2$.

The triangle of greatest area in a circle is equilateral, so has area $3/2 \times r \times \sqrt{3}/2 \times r$.

If the theorem is true, $4r^2 \leq 2 \times (3/2 \times r \times \sqrt{3}/2 \times r)$.

So $16 \leq 27 /2$. Contradiction.

The theorem seems to be false.

Answer 2

An even simpler counterexample would be the corner points of any non-rectangular parallelogram of area 2, e.g. $(0,0)$, $(0,1)$, $(2,2)$ and $(2,1)$. Any triangles formed from those points must have half the area of the parallelogram, i.e. 1, but as the parallelogram is not a rectangle, any rectangle that covers it must have an area greater than 2.

Answer 3

In the arrangement of four points below, each of the $4$ triangles has area $1$.

$\hspace{7mm}$

As $L\to\infty$, it seems that the smallest rectangle enclosing the $4$ points would have to be about $L\times4/L$, which would give it an area of $4$, not $2$.