2005 AMC 10A Problem 20

All eight angles are equal, and an octagon's angles add to $(8 - 2) \times 180^\circ = 1080^\circ$, so each one is $135^\circ$. An octagon with $135^\circ$ angles looks like a square with its corners snipped off. Find that square.

1. Place the long sides straight. Turn the octagon so the four sides of length $1$ are horizontal or vertical. Because the side lengths alternate, the four short sides of length $\frac{\sqrt{2}}{2}$ are the slanted ones, cutting across at $45^\circ$.
2. Extend the long sides. Stretch the four sides of length $1$ until they meet. They form a square, and the octagon is this square with four corners cut off.
3. Size of a cut-off corner. Each corner piece is a right triangle with two equal legs, and its slanted side is a short side of the octagon, $\frac{\sqrt{2}}{2}$. In such a triangle the slanted side is $\sqrt{2}$ times a leg, so each leg is
$$\frac{\sqrt{2}/2}{\sqrt{2}} = \frac{1}{2}.$$
4. Size of the square. Each side of the square is a leg, a side of length $1$, and another leg: $\frac{1}{2} + 1 + \frac{1}{2} = 2$. Its area is $2 \times 2 = 4$.
5. Area of the corners. Each corner triangle has area $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$. Four of them make $\frac{4}{8} = \frac{1}{2}$.
6. Area of the octagon. $4 - \frac{1}{2} = \frac{7}{2}$.

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

Any octagon with all angles $135^\circ$ is a rectangle or square with its corners cut off at $45^\circ$. The only choice is which sides lie on the square. Putting the length $1$ sides on the square makes the corner triangles have neat legs of $\frac{1}{2}$, so all the square roots disappear. "Big shape minus the pieces cut away" is a strong method for odd-shaped areas.

Estimate. A regular octagon with all sides $1$ has area about $4.83$. This octagon has four sides shorter than $1$, so it must be smaller than that. Now estimate the choices: (B) is about $4.95$, (C) about $5.3$, (D) about $5.5$, and (E) is $7$. Only (A), $3.5$, is small enough.

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