AMC 10 · 2022 · #21

Grade 8 geometry-2d
spatial-visualizationarea-regular-hexagonpythagorean-theorem physical-representationidentify-subproblems ↑ Prerequisites: spatial-visualization
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
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.

Pick an answer.

(A)
6
(B)
7
(C)
$5+2\sqrt{2}$
(D)
8
(E)
9

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

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 √(2), which fits exactly inside a 3× 3 square with four unit right-triangle corners cut off. Tool #3 (Eliminate) confirms the choice against the answer list.

1STEP 1

Picture the bowl: a flat square with four hexagons tilting up. Two hexagons at each corner share a slanted unit edge rising to the rim.

rim = 8 vertices, alternating type
2STEP 2

The rim alternates: a hexagon top edge (length 1), then a segment where two unit edges meet at 90° — a diagonal of √(2).

type (ii) length = √(1² + 1²) = √(2)
3STEP 3

By symmetry every interior angle is 135°, so the rim is an equiangular octagon whose sides alternate 1, √(2), 1, √(2), …

sides: 1,√(2),1,√(2),1,√(2),1,√(2)
4STEP 4

Enclose the octagon in a square parallel to its unit edges. The four cut corners are unit right triangles, so the big square has side 3.

big square side = 1 + 2 · 1 = 3
5STEP 5

Big square area 3² = 9; subtract four corner triangles of 12\frac{1}{2} each, so octagon area = 9 - 4 · 12\frac{1}{2} = 7, choice (B).

9 - 4 · 12\frac{1}{2} = 7 → (B)
Answer
7
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√(2) ≈ 7.83 is what you'd get if you computed an octagon area formula in √(2) wrongly; choice (B) 7 is the exact value because all cut-off pieces are unit right triangles, no √(2) leaks into the area.
💡Key takeaway

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, √(2), 1, √(2), … inside a 3 × 3 square missing four unit right-triangle corners, the area is just 9 - 4 · 12\frac{1}{2} = 7.