2003 AMC 10B Problem 23

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

Two triangles with a common side and third vertices symmetric through a point on that side's line have equal areas; here that turns the "wide" triangles $OBE$ and $OFA$ into copies of the octagon's basic triangle. In general, for a regular $2n$-gon the rectangle on two opposite sides is built from four of the $2n$ central triangles, so its area is $\frac{4}{2n}$ of the whole; for the octagon that is $\frac{1}{2}$.

With side length $s$: the octagon is a square of side $s(1 + \sqrt{2})$ minus four corner right triangles with legs $\frac{s}{\sqrt2}$, so its area is $s^2(1+\sqrt2)^2 - s^2 = 2s^2(1 + \sqrt2)$. Rectangle $ABEF$ has sides $s$ and $s(1 + \sqrt2)$, area $s^2(1 + \sqrt2)$, exactly half.

• Assuming the rectangle is "a bit less than half" from the picture and guessing a radical answer, or setting up side lengths and mishandling the $45$-$45$-$90$ corner triangles.
• Believing $ABEF$ is a square; its sides are $s$ and $s(1 + \sqrt{2})$, and the equal-area argument is what makes the answer clean, not equal sides.