6. A triangular piece of paper of area 1 is folded along a line parallel to one of the sides and pressed flat. What is the minimum possible area of the resulting figure?
Solution:
$2 / 3$
Let the triangle be denoted $A B C$, and suppose we fold parallel to $B C$. Let the distance from $A$ to $B C$ be $h$, and suppose we fold along a line at a distance of $c h$ from $A$. We will assume that neither angle $B$ nor $C$ is obtuse, for the area of overlap will only be smaller if either is obtuse. If $c \leq \frac{1}{2}$, then $A$ does not fold past the edge $B C$, so the overlap is a triangle similar to the original with height $c h$; the area of the figure is then $1-c^{2} \geq \frac{3}{4}$. Suppose $c>\frac{1}{2}$, so that $A$ does fold past $B C$. Then the overlap is a trapezoid formed by taking a triangle of height $c h$ similar to the original and removing a triangle of height $(2 c-1) h$ similar to the original. The area of the resulting figure is thus $1-c^{2}+(2 c-1)^{2}=3 c^{2}-4 c+2$. This is minimized when $c=\frac{2}{3}$, when the area is $\frac{2}{3}<\frac{3}{4}$; the minimum possible area is therefore $\frac{2}{3}$.