In the diagram, the two regular octagons have side lengths of 1 and 2. The smaller octagon is completely contained within the larger octagon. What is the area of the region inside the larger octagon and outside the smaller octagon?

![](https://cdn.mathpix.com/cropped/2024_04_17_5042e00a68bea6f5e8e9g-2.jpg?height=307&width=315&top_left_y=123&top_left_x=1493)
Let $A$ be the area of a regular octagon with side length 1 .

Then the area of a regular octagon with side length 2 equals $2^{2} A=4 A$.

Thus, the area between the octagons is $4 A-A=3 A$.

This means that it is sufficient to find the area of a regular octagon with side length 1.

Label the octagon $P Q R S T U V W$. Join $P U, Q T, W R, V S$.

![](https://cdn.mathpix.com/cropped/2024_04_30_dfe99ade8da3ee5d11fdg-03.jpg?height=385&width=390&top_left_y=1369&top_left_x=930)

We note that the sum of the angles in an octagon is $6 \cdot 180^{\circ}=1080^{\circ}$ and so in a regular octagon, each interior angle equals $\frac{1}{8}\left(1080^{\circ}\right)=135^{\circ}$.

Therefore, $\angle W P Q=\angle P Q R=135^{\circ}$.

By symmetry, $\angle P W R=\angle Q R W$. Looking at quadrilateral $P Q R W$, each of these angles equals $\frac{1}{2}\left(360^{\circ}-2 \cdot 135^{\circ}\right)=45^{\circ}$.

Since $\angle P W R+\angle W P Q=45^{\circ}+135^{\circ}=180^{\circ}$, then $P Q$ is parallel to $W R$. Using a similar argument, $P Q, W R, V S$, and $U T$ are all parallel, as are $W V, P U, Q T$, and $R S$.

Since $\angle P W R=45^{\circ}$ and $\angle P W V=135^{\circ}$, then $\angle R W V=90^{\circ}$, which means that $W R$ is perpendicular to $W V$.

This means that $P Q, W R, V S$, and $U T$ are all perpendicular to $W V, P U, Q T$, and $R S$.

Therefore, the innermost quadrilateral is a square with side length 1. (Note that its sides are parallel and equal in length to $P Q, R S, U T, W V$, each of which has length 1.) Its area is 1.

The four right-angled triangles with hypotenuses $Q R, S T, U V$, and $W P$ are all isosceles rightangled triangles with hypotenuse of length 1 .

These can be pieced together to form a square with side length 1, as shown.

![](https://cdn.mathpix.com/cropped/2024_04_30_dfe99ade8da3ee5d11fdg-04.jpg?height=209&width=220&top_left_y=259&top_left_x=1012)

Therefore, their combined area is 1.

The four remaining rectangles are identical. Each has one side of length 1 (one of the sides of the octagon) and one side equal to the shorter side length of an isosceles right-angled triangle with hypotenuse 1 , which equals $\frac{1}{\sqrt{2}}$.

Therefore, the combined area of these rectangles is $4 \cdot 1 \cdot \frac{1}{\sqrt{2}}=2 \sqrt{2}$.

Putting these pieces together, we obtain $A=2+2 \sqrt{2}$, and so the area between the octagons is $3 A=6+6 \sqrt{2}$.

ANSwER: $6+6 \sqrt{2}$