Equilateral triangle touching three sides of a square Consider a unit square. What is the largest, and smallest, equilateral triangle  with vertices touching the sides? 
Clearly the sides have to be larger than one, and it looks like the biggest would be with side $\frac{1}{\cos(\pi/12)}=\sqrt{\frac{4}{\sqrt{3}+2}}$, by placing one vertex in a corner of the square, but 


*

*Is it really the biggest such triangle ? 

*what is the smallest one?


You could call it an Ikea type problem, how to fit your box into the back of the van..
 A: The whole set of equilateral triangles inscribed in a unit square $ABCD$ can be generated through the following procedure:



*

*Pick a point $P\in AB$;

*Rotate $ABCD$ by $60^\circ$ counter/clockwise around $P$;

*Consider the suitable intersections between $ABCD$ and the rotated squares.


By symmetry it is pretty clear that the minimum area of the inscribed equilateral triangle is attained when $P$ is the midpoint of $AB$ and the maximum area is attained when a vertex of the inscribed equilateral triangle is a vertex of the original square. In the former case the mentioned triangle has unit side length, in the latter the side length equals $\frac{1}{\cos 15^\circ}=\sqrt{6}-\sqrt{2}$.

It is interesting to point out that if $P$ is a vertex of an inscribed equilateral triangle close to the midpoint of $AB$, the opposite sides goes through a fixed point $Q$, and the area of the inscribed equilateral triangle just equals $\frac{1}{\sqrt{3}} PQ^2$. Here it is an animation, too:

A: Your calculation for the largest triangle is correct. The expression for the side length can be simplified to $\sqrt{6}-\sqrt{2}$.
Since the equilateral triangle touches three sides of the square, at least one of the sides of the triangle must touch opposite sides of the square. From this it's easily seen that the smallest such equilateral triangle has side $1$.
Both of these are confirmed at the bottom of Mathworld's Equilateral Triangle page.
