# Why can't three unit regular triangles cover a unit square?

A square with edge length $1$ has area $1$. An equilateral triangle with edge length $1$ has area $\sqrt{3}/4 \approx 0.433$. So three such triangles have area $\approx 1.3$, but it requires four such triangles to cover the unit square, e.g.:

Q. How can it be proved that three unit triangles cannot cover a unit square?

I am not seeing a straightforward route to proving this.

• Consider eight points that are on the sides of the square, a small fixed distance from the corners. Can you show that no unit triangle can cover more than two such points? – mjqxxxx Apr 7 '15 at 0:26
• @mjqxxxx: Nice idea! Unfortunately, I think it is false... – Joseph O'Rourke Apr 7 '15 at 0:28
• I posted a natural follow-up to this question, asking for the smallest equilateral triangle such that three copies may cover a square. – Milo Brandt Apr 7 '15 at 2:43

Suppose we define the corner-power of a triangle as follows:

• A triangle gets $1$ corner-power point for each corner of the square that it contains excluding the vertices of the triangle.

• A triangle gets $\frac{1}2$ of a corner-power point for each corner of the square that is also a vertex of the triangle.

It should be clear that, in order for a neighborhood of the square's corner to be covered by a set of triangles, it must either:

• Be contained within a triangle, but not on the vertex.

• Be contained as the vertex of two triangles.

Hence, a cover of the square must have at least $4$ corner-power points. However, each triangle can either have a corner in its interior (1 point) or two vertices on corners (1 point) for a maximum of $1$ point. Three triangles gets at most 3 corner-power points, which is not enough to cover a square.

• Very nice! A triangle can contain a corner on its boundary but still cover a neighborhood of that corner. See the bottommost triangle in my example. Shouldn't that have corner-power point $1$? – Joseph O'Rourke Apr 7 '15 at 0:35
• How about change your $1$-point triangles to: If the triangle covers a neighborhood of a corner. Then $\frac{1}{2}$-point triangles cover the corner itself and at least one point in a neighborhood of the corner. – Joseph O'Rourke Apr 7 '15 at 0:47
• @JosephO'Rourke Good catch; the key is that the first condition implies that the triangle is contained within an open ball of radius $1$ around that corner (excluding all other corners), which is true of any point in the triangle except the vertices. – Milo Brandt Apr 7 '15 at 0:54
• @JosephO'Rourke Your definition is clearly correct (and defines a tighter measure of how useful a triangle is), but, unless I'm missing something, it seems like the difference from mine is that it transfers cases of $1$-point triangles to $\frac{1}2$ point triangles. It's true that a $1$-point triangle, with my definition, might not cover a neighborhood of a corner, but my argument only relies on the fact that there are no triangles with more than $1$ corner-power point - not on the fact that $1$-point triangles are actually doing more good than $\frac{1}2$ point triangles. – Milo Brandt Apr 7 '15 at 1:31
• A constructive version of the same idea, without the notion of "corner power": It is impossible for two vertices of the square to be inside the boundary of one triangle, so to cover 4 vertices with 3 triangles, at least one triangle must share two vertices with the square. The only way to do this is to share an edge of the triangle, E say, with the square. The only way to completely cover the two edges adjacent to E is with two more triangles, and that leaves the edge opposite E uncovered. So three triangles are not enough. – alephzero Apr 7 '15 at 20:22

(This was intended to be a comment but was too long.)

Here is another approach, but a bit messy. We can calculate how much of the perimeter of the square an equilateral triangle can hold. There are two cases, one where a side of the triangle is a side of the square which gives that one equilateral triangle can hold only length 1. The other case is where a corner of the square is inside the triangle. WLOG, we can assume the corner touches a side of the triangle since moving the corner inside until it does only increases the length of the perimeter inside the triangle. After some calculations, we get that the length of the perimeter inside the triangle is at most $\frac{\sqrt{3}}2(1-x)+x$ where $0 < x < 1$ which has a maximum of $1$ also.

• If a vertex of the square is placed at the centre of one of the edges of the triangle, such that the two edges of the square incident to this vertex pass through two vertices of the triangle, then ${\sqrt 3 + 1 \over 2} \approx 1.366$ of the perimeter of the square is contained in the triangle; this is greater than $4 \over 3$ so it does not rule out the whole perimeter being contained in three triangles. – Gareth Rees Apr 7 '15 at 10:14