2005 AMC 10A Problem 20

An equiangular octagon has interior angles $\frac{(8-2)\cdot 180^\circ}{8} = 135^\circ$, so each exterior angle is $45^\circ$. Orient the octagon so that the four sides of length $1$ are horizontal or vertical; the four sides of length $\frac{\sqrt{2}}{2}$ then run at $45^\circ$ and connect them.

Extend the four unit sides until they meet. Because consecutive extended sides are perpendicular, they form a square, and the octagon is that square with its four corners cut off. Each cut-off corner is an isosceles right triangle whose hypotenuse is a $\frac{\sqrt{2}}{2}$ side, so its legs are $\frac{\sqrt{2}/2}{\sqrt{2}} = \frac{1}{2}$.

The square's side is a unit side plus two legs: $1 + \frac{1}{2} + \frac{1}{2} = 2$, so its area is $4$. Each corner triangle has area $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$. Hence
$$\text{area} = 4 - 4 \cdot \frac{1}{8} = \frac{7}{2}.$$

The answer is $\boxed{\textbf{(A)}\ \frac{7}{2}}$.

Equiangular octagons are squares with corners clipped at $45^\circ$; the only question is which sides lie on the square and which are the clipped hypotenuses. Choosing the longer sides as the square's edges keeps the arithmetic rational. "Bounding shape minus corners" is the standard way to get areas of $135^\circ$-angled polygons.

Estimate: a regular octagon of side $1$ has area about $4.83$, and this octagon is clearly smaller since half its sides are shorter. Only $\frac{7}{2}$ is below that; choices (C), (D), (E) are all at least $5.3$ and (B) is about $4.95$.

• Treating the octagon as regular (a formula with side root 2 over 2 or 1), or clipping triangles with legs root 2 over 2 instead of 1/2.
• Extending the short sides instead, which also works but yields a square of side $\frac{\sqrt{2}}{2} + \sqrt{2}$ and messier subtraction, inviting errors.