2003 AMC 10B Problem 23

1. Let $O$ be the center of the octagon. Draw lines from $O$ to all $8$ corners. This cuts the octagon into $8$ identical triangles, so each has area $\frac{1}{8}$.
2. $A$ and $E$ are opposite corners, so the segment $AE$ goes straight through the center $O$, and $O$ is its midpoint. The same is true for $B$ and $F$.
3. So the two diagonals $AE$ and $BF$ of the rectangle cross at $O$ and cut the rectangle into four triangles: $OAB$, $OBE$, $OEF$, $OFA$.
4. Triangles $OAB$ and $OEF$ are two of the octagon's $8$ pieces. Each has area $\frac{1}{8}$.
5. Now compare $OBE$ with $OAB$. Both have the side $OB$ as a base. Their third corners, $A$ and $E$, are on opposite sides of $O$ along one straight line and the same distance from $O$, so they are the same distance from the line $OB$. Same base and same height means same area: $OBE$ also has area $\frac{1}{8}$.
6. In the same way, $OFA$ and $OEF$ share the base $OF$, and $A$, $E$ are equally far from line $OF$. So $OFA$ has area $\frac{1}{8}$ too.
7. The rectangle's area is
$$4 \times \frac{1}{8} = \frac{1}{2}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{1}{2}}$.

A regular polygon splits into identical triangles from its center, which gives a unit of area to count with. Triangles with the same base and the same height have the same area even if they look different, so the "wide" triangles inside the rectangle are worth exactly one unit each. Always try the center first with regular polygons.

Use side lengths. Let each side of the octagon be $1$. The octagon is a big square with its four corners cut off. Each corner cut is a right triangle with two equal legs and a long side of $1$, so each leg is $\frac{\sqrt{2}}{2}$ and each corner has area $\frac{1}{2} \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{4}$. The big square has side $1 + \sqrt{2}$, so the octagon's area is $(1 + \sqrt{2})^2 - 4 \cdot \frac{1}{4} = 3 + 2\sqrt{2} - 1 = 2 + 2\sqrt{2}$. The rectangle $ABEF$ is $1$ by $1 + \sqrt{2}$, area $1 + \sqrt{2}$, which is exactly half of $2 + 2\sqrt{2}$.

Let $O$ be the center of the octagon. Joining $O$ to all eight vertices cuts the octagon into $8$ congruent triangles, each of area $\frac{1}{8}$.

Vertices $A$ and $E$ are opposite, so $A$, $O$, $E$ are collinear with $OA = OE$; the same holds for $B$, $O$, $F$. Hence $O$ lies inside rectangle $ABEF$, and the diagonals $AE$, $BF$ split it into four triangles $OAB$, $OBE$, $OEF$, $OFA$.

• $OAB$ and $OEF$ are two of the eight congruent pieces: area $\frac{1}{8}$ each.
• $OAB$ and $OBE$ share the side $OB$. Since $A$ and $E$ are equally far from $O$ along the same line through $O$, their distances to line $OB$ are equal, so $[OBE] = [OAB] = \frac{1}{8}$.
• By the same argument with side $OF$, $[OFA] = [OEF] = \frac{1}{8}$.

Therefore
$$[ABEF] = 4 \cdot \frac{1}{8} = \frac{1}{2}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{1}{2}}$.