Among all the triangles containing a square of side 1, which ones have the smaller area?

Question

I started assuming wlog that one verex of the square lays on a side of the triangle. Then I considered that side of the triangle and its length $l$, the distance $d$ of the vertex of the square from one of the two vertices of the triangle contained on the side and the angle $\alpha$ formed by the side of the square and the side of the triangle. My aim was to write the area of the triangle as a function of $l$, $d$ and $\alpha$, but I really think there is an easier way...

Answer 1

It can be easily proved that, in order to have the smallest possible area of the triangle, one side of the square need to lie on a side of the triangle. Let say that the square is $ABCD$, and that the triangle is $EFG$. Let us assume that the side $AB$ lies on $EF$. Finally, we name $h$ the distance between the vertex $G$ and $CD$. In order to have the smallest area, $D\in EG$ and $C\in GF$. The area of the triangle $GCD$ is $(1\cdot h)/2$, and since $EFG$ and $GCD$ are similar triangles, it follows that the area of $EFG$ is $$\frac{h}{2}\biggl(\frac{(h+1)^2}{h^2}\biggr).$$ By differentiating, you can find that the minimum value for the expression above is achieved when $h=1$, which proves that the smallest area is $2$.