1. Octagon $A B C D E F G H$ is equiangular. Given that $A B=1, B C=2, C D=3$, $D E=4$, and $E F=F G=2$, compute the perimeter of the octagon.
Answer: $20+\sqrt{2}$
Solution: Extend sides $A B, C D, E F, G H$ to form a rectangle: let $X$ be the intersection of lines $G H$ and $A B ; Y$ that of $A B$ and $C D ; Z$ that of $C D$ and $E F$; and $W$ that of $E F$ and $G H$.
As $B C=2$, we have $B Y=Y C=\sqrt{2}$. As $D E=4$, we have $D Z=Z E=2 \sqrt{2}$. As $F G=2$, we have $F W=W G=\sqrt{2}$.
We can compute the dimensions of the rectangle: $W X=Y Z=Y C+C D+D Z=$ $3+3 \sqrt{2}$, and $X Y=Z W=Z E+E F+F W=2+3 \sqrt{2}$. Thus, $H X=X A=X Y-$ $A B-B Y=1+2 \sqrt{2}$, and so $A H=\sqrt{2} H X=4+\sqrt{2}$, and $G H=W X-W G-H X=2$. The perimeter of the octagon can now be computed by adding up all its sides.