2022 AMC 10A Problem 21

Step 1: look from above. Put the square flat, with corners $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$. The four hexagons are placed the same way on the four sides, so they all lean by the same amount. Each hexagon's top edge is parallel to its bottom edge, which is level, so all eight rim points are at the same height. The octagon is therefore level, and its area equals the area of its shadow straight down. So we only need where the rim points are when seen from above.

Step 2: facts about one hexagon, laid flat. Put a regular hexagon of side $1$ with its bottom edge from $(0,0)$ to $(1,0)$. Its other corners are $\left(\frac32, \frac{\sqrt3}{2}\right)$, $(1, \sqrt3)$, $(0, \sqrt3)$, $\left(-\frac12, \frac{\sqrt3}{2}\right)$. So the corner next to the bottom edge sits $\frac12$ past the end of the edge and $\frac{\sqrt3}{2}$ "up the hexagon", and the top edge is directly above the bottom edge, $\sqrt3$ up the hexagon.

Step 3: find one corner that two hexagons share. Call the hexagon on side $(0,0)$ to $(1,0)$ the front one, and the hexagon on side $(1,0)$ to $(1,1)$ the right one. They share the edge that rises from the square's corner $(1,0)$ to a point $B$.
- The front hexagon's bottom edge runs in the $x$ direction, so $B$ is $\frac12$ past $x = 1$ in that direction: from above, $B$ has $x = \frac32$.
- The right hexagon's bottom edge runs in the $y$ direction from $y = 1$ down to $y = 0$, so $B$ is $\frac12$ past $y = 0$: from above, $B$ has $y = -\frac12$.

So from above, $B$ is at $\left(\frac32, -\frac12\right)$.

Step 4: how much the lean shrinks things. In the front hexagon, $B$ is $\frac{\sqrt3}{2}$ up the hexagon, and from above that shows up as moving outward from $y = 0$ to $y = -\frac12$, that is, by $\frac12$. The hexagon is a flat tilted sheet, so every distance "up the hexagon" shrinks by the same factor. The top edge is $\sqrt3$ up, twice as far as $B$, so from above it is $1$ outward. The front top edge appears as the segment from $(0, -1)$ to $(1, -1)$.

Step 5: all four top edges. By symmetry, from above the top edges are
$$(0,-1)\text{ to }(1,-1), \quad (2,0)\text{ to }(2,1), \quad (1,2)\text{ to }(0,2), \quad (-1,1)\text{ to }(-1,0).$$
The other four sides of the octagon join neighboring ones, such as $(1,-1)$ to $(2,0)$.

Step 6: area. The shadow is the $3 \times 3$ square with corners $(-1,-1)$ and $(2,2)$, with a right triangle with legs $1$ and $1$ cut off at each corner:
$$9 - 4 \times \tfrac12 = 7.$$

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

A symmetric 3D shape is often easiest to measure from a well-chosen view. The rim is level, so the top view shows its true area. The only unknown was how much the hexagons lean, and the edge shared by two neighboring hexagons pins that down with no trigonometry: seen from one hexagon it gives the $x$-coordinate, from the other the $y$-coordinate.

A check that the picture makes sense: $B$ is $1$ unit from the corner $(1,0)$, and from above it has moved $\frac12$ in $x$ and $\frac12$ in $y$. By the Pythagorean theorem its height is $\sqrt{1 - \frac14 - \frac14} = \sqrt{\frac12} \approx 0.71$, so the hexagons really do lean outward, as a bowl should.