2022 AMC 10 A Problem 21

Dihedral Angle and Projection Method

If we look at the bowl directly from above, the rim forms an octagon. Because the bowl is perfectly symmetric, all 8 top vertices are at the same height, meaning the octagon is parallel to the ground. Therefore, its true area is exactly equal to the area of its 2D shadow (projection) on the ground.

Step 1: Set up the coordinate system and orient the base

Place the $1\times 1$ square base in the $xy$-plane, centered at the origin $(0,0,0)$. The edges of the square lie along the lines $x=\pm \frac{1}{2}$ and $y=\pm \frac{1}{2}$. We will focus on the rightmost hexagon, which is attached to the edge $x=\frac{1}{2}$.

Step 2: Analyze the dimensions of the un-tilted regular hexagon

In a regular hexagon of side length 1, the distance from the bottom edge to the adjacent vertices (measured perpendicular to the bottom edge) is $\frac{\sqrt{3}}{2}$. The distance from the bottom edge to the top edge is $\sqrt{3}$.

Step 3: Determine the projected position of the shared vertex

Let the rightmost hexagon be tilted upward by an angle $\alpha$. Let $E$ be the vertex shared with the top hexagon. In the projection, the step from the corner of the square $(\frac{1}{2},\frac{1}{2})$ to $E$ goes $\frac{1}{2}$ unit parallel to the attached edge (the $y$-direction) and $\frac{\sqrt{3}}{2}\cos \alpha$ units perpendicular to it (the $x$-direction).

Step 4: Use symmetry to find the scaling factor

Because $E$ is shared with the adjacent hexagon attached to $y=\frac{1}{2}$, the entire bowl is symmetric across the diagonal plane $y=x$. Therefore, the projected vertex $E$ must lie on the line $y=x$. We equate its $x$ and $y$ coordinates.

Step 5: Find the projected coordinates of the top edge

The top edge of the hexagon is parallel to the attached edge and is at a perpendicular distance of $\sqrt{3}$ in the hexagon’s plane. In the $xy$-projection, this distance is scaled by $\cos \alpha$. Substituting our finding from the previous step, we determine the $x$-coordinate of the top edge.

Step 6: Determine the vertices of the projected octagon

The top edge lies on the vertical line $x=\frac{3}{2}$, has length 1, and is centered at $y=0$. Its endpoints are $(\frac{3}{2},\frac{1}{2})$ and $(\frac{3}{2},-\frac{1}{2})$. By applying the same symmetry to all four hexagons, the 8 vertices of the projected octagon are $(\pm \frac{3}{2},\pm \frac{1}{2})$ and $(\pm \frac{1}{2},\pm \frac{3}{2})$.

Step 7: Calculate the area of the projected octagon

To find the area, draw a bounding square around the octagon. The $x$ and $y$ coordinates range from $-\frac{3}{2}$ to $\frac{3}{2}$, so the square has side length 3 and area 9. The octagon is formed by removing four right isosceles triangles from the corners of this square. Each triangle has legs of length $\frac{3}{2}-\frac{1}{2}=1$.

Proportional Projection Method — Shortcut Version of Method 1

The top-down projection compresses the hexagon perpendicularly. Since the middle diagonal of a regular hexagon is exactly halfway between the top and bottom edges, its projection will also be exactly halfway between them. This allows us to locate the top edge instantly without calculating any angles.

Step 1: Locate the middle diagonal using symmetry

Place the square base centered at the origin. Focus on the rightmost hexagon attached at $x=\frac{1}{2}$. The middle diagonal of this hexagon connects the two vertices shared with adjacent hexagons. Due to symmetry across $y=x$, the top shared vertex $E$ must have equal $x$ and $y$ coordinates. Since $E$ is $\frac{1}{2}$ unit along the $y$-axis from the corner $(\frac{1}{2},\frac{1}{2})$, its $y$-coordinate is $1$. Thus, its $x$-coordinate must also be 1. The projected middle diagonal therefore lies on the vertical line $x=1$.

Step 2: Use equal spacing to locate the top edge

The horizontal distance from the attached edge at $x=\frac{1}{2}$ to the middle diagonal at $x=1$ is $\frac{1}{2}$. Because the middle diagonal is exactly halfway to the top edge in the regular hexagon, this equal spacing is preserved in the projection. The top edge must be another $\frac{1}{2}$ unit to the right.

Step 3: Calculate the area of the octagon

The top edge has length 1, giving endpoints $(\frac{3}{2},\pm \frac{1}{2})$. Symmetry yields an octagon bounded by a $3\times 3$ square, minus four $1\times 1$ right corner triangles.

3D Coordinate and Vector Geometry Method

We can construct the bowl in a 3D coordinate system. By using the given side lengths and the $120^{\circ }$ interior angle of the regular hexagon, we can use the vector dot product to locate the shared vertices. From there, we find the top vertices and calculate the area of the resulting flat octagon.

Step 1: Define the base and the shared vertex

Place the square base in the $xy$-plane, centered at the origin. Let $A=(\frac{1}{2},-\frac{1}{2},0)$ and $B=(\frac{1}{2},\frac{1}{2},0)$. Let $E$ be the vertex shared by the hexagons on $AB$ and the top edge. By symmetry, $E$ lies on the plane $y=x$, so we can write $E=(x,x,z)$.

Step 2: Use the dot product to find the coordinates of $E$

The interior angle $\angle ABE$ is $120^{\circ }$. We create vectors $\overrightarrow{BA}$ and $\overrightarrow{BE}$. Because they are edges of a regular hexagon, their lengths are both 1. We compute their dot product to solve for $x$.

Step 3: Determine the full 3D coordinates of the shared vertices

Since the distance $BE=1$, we substitute $x=1$ into the 3D distance formula to find $z$.

Step 4: Find the top vertices using the hexagon’s center

The hexagon on $AB$ is symmetric. The vertex $H$ shared with the bottom hexagon is $H=(1,-1,\frac{\sqrt{2}}{2})$. The center $O$ of the hexagon is the midpoint of the diagonal $EH$. In a regular hexagon, opposite vertices reflect across the center, so the top vertex $F$ (opposite $A$) and $G$ (opposite $B$) can be found directly.

Step 5: Calculate the area of the resulting octagon

Rotating these points around the $z$-axis gives 8 vertices, all sharing the same height $z=\sqrt{2}$. Since they are coplanar and horizontal, the area of the 3D octagon is exactly the area of its projection onto the $xy$-plane. The bounding box in the $xy$-plane is $[-\frac{3}{2},\frac{3}{2}]\times [-\frac{3}{2},\frac{3}{2}]$, and removing the four corner triangles yields the final area.