2022 AMC 10A Problem 21

Put the square in the horizontal plane with corners $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$. By the square's symmetry all four hexagons tilt equally, so the eight rim vertices are at one common height: the octagon is horizontal, and its area equals the area of its top view.

Look at hexagon $H_1$ on the side from $(0,0)$ to $(1,0)$ and hexagon $H_2$ on the side from $(1,0)$ to $(1,1)$. They share the edge rising from the corner $(1,0)$ to a vertex $B$. In a regular hexagon of side $1$, the vertex adjacent to one end of a side lies $\frac12$ farther along that side's direction and $\frac{\sqrt3}{2}$ away perpendicular to it, within the hexagon's plane. The plane of $H_1$ contains the $x$-direction, so $B$ has $x$-coordinate $1 + \frac12 = \frac32$; the plane of $H_2$ contains the $y$-direction, so $B$ has $y$-coordinate $0 - \frac12 = -\frac12$. Thus the in-plane perpendicular offset $\frac{\sqrt3}{2}$ in $H_1$ projects to a horizontal offset of exactly $\frac12$ outward (from $y = 0$ to $y = -\frac12$).

The top edge of $H_1$ is at in-plane distance $\sqrt3 = 2 \cdot \frac{\sqrt3}{2}$ from its base edge, so it projects twice as far: seen from above it runs from $(0,-1)$ to $(1,-1)$. By symmetry the four top edges project to $(0,-1)$-$(1,-1)$, $(2,0)$-$(2,1)$, $(1,2)$-$(0,2)$, $(-1,1)$-$(-1,0)$, and the other four rim edges join consecutive ones, such as $(1,-1)$-$(2,0)$.

The octagon is therefore the $3 \times 3$ square with corners $(-1,-1)$, $(2,-1)$, $(2,2)$, $(-1,2)$ with four isosceles right triangles of legs $1$ cut off:
$$9 - 4 \cdot \tfrac12 = 7.$$

The answer is $\boxed{\textbf{(B)}\ 7}$.

A symmetric three-dimensional figure is best measured through a well-chosen projection: the rim is horizontal, so its area is visible from above. The only unknown was the tilt of the hexagons, and a shared edge, viewed from two hexagons at once, pins it down with no trigonometry. Once the tilt is known, everything about the rim follows by symmetry.

Sanity check the heights: $B$ is $1$ unit from $(1,0,0)$ with horizontal displacement $(\frac12, -\frac12)$, so its height is $\sqrt{1 - \frac14 - \frac14} = \frac{1}{\sqrt2}$, and the top edge, at in-plane distance $\sqrt3$ with horizontal offset $1$, is at height $\sqrt{3 - 1} = \sqrt2$, consistent with a rising bowl.

• Guessing the octagon is regular or that the hexagons stand vertically; the rim's sides alternate 1 and sqrt(2), and the tilt is fixed by the shared edges.
• Computing the area of a regular octagon of side $1$ ($2 + 2\sqrt2$) or misreading the top view as a $3 \times 3$ square without the cut corners ($9$).