TITLE: Why can't three unit regular triangles cover a unit square?
QUESTION [42 upvotes]: 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.

REPLY [2 votes]: (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.