In the interior of a triangle $A B C$ with area 1 , points $D, E$, and $F$ are chosen such that $D$ is the midpoint of $A E, E$ is the midpoint of $B F$, and $F$ is the midpoint of $C D$. Find the area of triangle $D E F$.
Let $x$ be the area of $\triangle D E F$. Comparing triangles $A B E$ and $D E F$, we find that base $A E$ is twice base $D E$ but, since $E$ bisects $B F$, the heights to these bases are equal. Thus $\triangle A B E$ has area $2 x$. Symmetrically, triangles $B C F$ and $C A D$ have area $2 x$. Since these four triangles fill up $\triangle A B C$, we have $1=x+2 x+2 x+2 x=7 x$, so $x=1 / 7$.