AMC 10 · 2022 · #21
Grade 8 geometry-2dProblem
A bowl is formed by attaching four regular hexagons of side 1 to a square of side 1. The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl?
Pick an answer.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A bowl is built by gluing four regular hexagons (side $1$) onto the four sides of a square (side $1$); adjacent hexagons meet along a shared slanted edge. The eight top vertices of the hexagons lie in a horizontal plane and form an octagon (the rim). Find the area of that octagon.
Givens: Central square has side $1$; Four regular hexagons of side $1$ are attached, one to each square edge; Adjacent hexagons share an edge (a slanted side that goes from the square corner up to the rim); The rim is the octagon joining the $8$ top hexagon vertices; Answer choices: (A) $6$, (B) $7$, (C) $5+2\sqrt{2}$, (D) $8$, (E) $9$
Unknowns: The area of the rim octagon
Understand
Restated: A bowl is built by gluing four regular hexagons (side $1$) onto the four sides of a square (side $1$); adjacent hexagons meet along a shared slanted edge. The eight top vertices of the hexagons lie in a horizontal plane and form an octagon (the rim). Find the area of that octagon.
Givens: Central square has side $1$; Four regular hexagons of side $1$ are attached, one to each square edge; Adjacent hexagons share an edge (a slanted side that goes from the square corner up to the rim); The rim is the octagon joining the $8$ top hexagon vertices; Answer choices: (A) $6$, (B) $7$, (C) $5+2\sqrt{2}$, (D) $8$, (E) $9$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #10 Create a Physical Representation, #1 Draw a Diagram, #3 Eliminate Possibilities
The octagon's area is the answer, but its shape is awkward. Tool #7 (Identify Subproblems) breaks the question into three pieces: (a) what are the side lengths? (b) what is the shape exactly? (c) how do we compute its area? Tool #10 (Physical Representation) — fold paper hexagons onto a square — makes the 3D geometry concrete enough to see that adjacent hexagons share a slanted edge of length $1$ rising from each square corner, and by symmetry the two rising edges at one corner stay perpendicular. Tool #1 (Diagram) of the rim, viewed from above, shows an equiangular octagon with alternating sides $1$ and $\sqrt{2}$, which fits exactly inside a $3\times 3$ square with four unit right-triangle corners cut off. Tool #3 (Eliminate) confirms the choice against the answer list.
Execute — Answer: B
6.G.A.4 Step 1 - Build the bowl in your head (or with paper).
- The square sits flat.
- Each hexagon tilts upward, hinged on one square edge.
- Two hexagons that share a square corner also share a slanted edge that rises from that corner — the shared edge has length $1$ (it is a side of each hexagon).
- Call the two top endpoints of that shared edge...
- actually call $P$ the top of the slanted edge — the two hexagons each have one top vertex adjacent to $P$ along the rim.
💡 Grade 6 — picturing a 3D figure from its faces tells you which edges meet at each vertex of the rim.
8.G.B.7 Step 2 - Find the lengths of the rim's edges.
- The rim alternates between two kinds of segments: (i) the top edge of a hexagon (one per hexagon, length $1$, four of them); (ii) the segment joining the two top vertices of adjacent hexagons that meet over a square corner.
- For type (ii), the two rim points are each one hexagon-edge above a square corner, and by the $4$-fold symmetry of the bowl the two upward edges at one square corner are perpendicular to each other.
- So the type-(ii) segment is the hypotenuse of a right triangle with legs $1, 1$.
💡 Grade 8 Pythagorean theorem on a right triangle whose legs are two unit hexagon edges meeting perpendicularly above a square corner.
4.G.A.2 Step 3 - Identify the octagon's shape.
- Each interior angle of the rim octagon is $135^\circ$ (the rim turns the same amount at every vertex by symmetry; $8 \times 135^\circ = 1080^\circ$ matches the octagon angle sum), so the octagon is equiangular.
- Its sides alternate $1, \sqrt{2}, 1, \sqrt{2}, 1, \sqrt{2}, 1, \sqrt{2}$ — four unit sides parallel to the square below, and four $\sqrt{2}$-sides making $45^\circ$ corner cuts.
💡 Grade 4 — classify the figure by its sides and angles; the rim is an equiangular octagon with two side lengths.
6.G.A.1 Step 4 - Compute the area by enclosing the octagon in a big square.
- Fit the octagon snugly inside a square whose sides are parallel to the four unit edges of the octagon.
- The four corners of the big square are cut off by the $\sqrt{2}$-sides.
- Each cut-off triangle is an isosceles right triangle whose hypotenuse is $\sqrt{2}$, so its legs are $1$.
- The big square has side $1 + 1 + 1 = 3$ (one unit edge in the middle plus a leg of $1$ on each side).
💡 Grade 6 — decompose a polygon into a big rectangle minus simple pieces.
3.MD.C.7 Step 5 - Finish the arithmetic.
- Big square area $= 3^2 = 9$.
- Each corner triangle area $= \tfrac{1}{2}\cdot 1 \cdot 1 = \tfrac{1}{2}$, and there are four corners.
- Octagon area $= 9 - 4 \cdot \tfrac{1}{2} = 9 - 2 = 7$.
- This is choice (B).
💡 Grade 3 — area of a rectangle (big square) minus areas of four equal triangles.
6.G.A.4 Build the bowl in your head (or with paper). The square sits flat. Each hexagon 8.G.B.7 Find the lengths of the rim's edges. The rim alternates between two kinds of seg 4.G.A.2 Identify the octagon's shape. Each interior angle of the rim octagon is $135^\ci 6.G.A.1 Compute the area by enclosing the octagon in a big square. Fit the octagon snugl 3.MD.C.7 Finish the arithmetic. Big square area $= 3^2 = 9$. Each corner triangle area $= Review
Reasonableness: Sanity. The rim octagon contains the inner unit square (area $1$) plus a wide rim, so the answer should be well over $1$. The big square that just encloses the octagon has area $9$, so the octagon's area should be a bit less than $9$ — losing exactly $2$ (the four corner triangles) to get $7$ is consistent. Choice (C) $5 + 2\sqrt{2} \approx 7.83$ is what you'd get if you computed an octagon area formula in $\sqrt{2}$ wrongly; choice (B) $7$ is the exact value because all cut-off pieces are unit right triangles, no $\sqrt{2}$ leaks into the area.
Alternative: Tool #1 (Diagram): project the octagon onto the horizontal plane and split it into the inner unit square plus four $1 \times 1$ rectangles on the four sides plus four corner unit right triangles ($1 + 4 + 4 \cdot \tfrac{1}{2} = 7$). Tool #3 (Eliminate) on the choices: any answer with a $\sqrt{2}$ in it (only (C)) would require the four corner cut-offs to have non-integer area, but their legs are exactly $1$ — so (C) is out and (B) $= 7$ is the clean integer.
CCSS standards used (min grade 8)
3.MD.C.7Relate area to the operations of multiplication and addition (Computing $3 \times 3 = 9$ and $4 \times \tfrac{1}{2} = 2$ for the big square area and the four corner triangles.)4.G.A.2Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size (Recognizing the rim as an equiangular octagon with two alternating side lengths.)6.G.A.1Find the area of right triangles, other triangles, and polygons by composing into rectangles or decomposing into triangles (Computing the octagon's area as big square minus four corner triangles.)6.G.A.4Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area (Visualizing the bowl from its hexagon-and-square faces to see which edges meet at each rim vertex.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the diagonal rim segment of length $\sqrt{2}$ as the hypotenuse of a right triangle with legs $1, 1$.)
⭐ This AMC 10 problem only needs Grade 8 Pythagorean theorem you already know — once you see the rim is an equiangular octagon with sides $1, \sqrt{2}, 1, \sqrt{2}, \ldots$ inside a $3 \times 3$ square missing four unit right-triangle corners, the area is just $9 - 4 \cdot \tfrac{1}{2} = 7$.
⭐ This AMC 10 problem only needs Grade 8 Pythagorean theorem you already know — once you see the rim is an equiangular octagon with sides $1, \sqrt{2}, 1, \sqrt{2}, \ldots$ inside a $3 \times 3$ square missing four unit right-triangle corners, the area is just $9 - 4 \cdot \tfrac{1}{2} = 7$.