AMC 10 · 2023 · #24

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Rectilinear Frame / Box Border

area-regular-hexagonarea-trianglesspatial-visualization identify-subproblemseasier-related-problemarea-difference ↑ Prerequisites: area-trianglespythagorean-theorem

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A regular hexagonal frame contains six regular hexagonal blocks (side 1), each sitting along one inside edge and lined up with two other blocks as shown. The distance from each corner of the frame to the nearest block vertex is

\frac{3}{7}

. Find the area inside the frame not covered by the blocks.

Pick an answer.

(A)

$\frac{13 \sqrt{3}}{3}$

(B)

$\frac{216 \sqrt{3}}{49}$

(C)

$\frac{9 \sqrt{3}}{2}$

(D)

$\frac{14 \sqrt{3}}{3}$

(E)

$\frac{243 \sqrt{3}}{49}$

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram) is mandatory — sketch the big hex, the six small hexes pinned along the inside edges, and the central gap they leave. Tool #7 (Identify Subproblems) splits the area into two clean pieces: (large hex area) and (six small hex areas), with the answer their difference. The trickiest piece is finding the side S of the large hex; here Tool #9 (Solve an Easier Related Problem) shines — temporarily ignore the $\frac{3}{7}$ distance and look at the simplest aligned configuration (small hexes sharing edges around a central hexagonal hole of side 1). The vertical distance from center to top edge of the frame gives S = 3, and that side length is independent of the $\frac{3}{7}$ parameter.

1STEP 1

Draw it: big hex centered, six small hexes pinned to the inside edges, ringing a central hexagonal hole with six-fold symmetry.

Big hex: side S, small hexes: side 1

2STEP 2

The uncovered area is the big hexagon minus the six small hexagons: A_big - 6·A_small.

Answer = A_big - 6 A_small

3STEP 3

A regular hexagon splits into 6 equilateral triangles (side 1, area $\frac{√(3)}{4}$ each), so one small hexagon has area $\frac{3√(3)}{2}$ ; six give 9√(3).

A_small =

\frac{3√(3)}{2}

, 6 A_small = 9√(3)

4STEP 4

Find S via an easier case: drop the $\frac{3}{7}$ gap; six hexes ring a central hexagonal hole of side 1, so measure center-to-top-edge distance.

Center-to-top-edge distance =

\frac{S√(3)}{2}

5STEP 5

Stack heights up the center: the hole's apothem plus the small hex's full height equals the big-hex apothem $\frac{S√(3)}{2}$ , giving S = 3.

\frac{S√(3)}{2}

\frac{√(3)}{2}

+ √(3) =

\frac{3√(3)}{2}

→ S = 3

6STEP 6

Compute the area of the large hex. With S = 3, A_big = $\frac{3√(3)}{2}$ S² = $\frac{3√(3)}{2}$ · 9 = $\frac{27√(3)}{2}$ .

A_big =

\frac{27√(3)}{2}

7STEP 7

Subtract to finish: A_big - 6 A_small = $\frac{27√(3)}{2}$ - 9√(3) = $\frac{27√(3)}{2}$ - $\frac{18√(3)}{2}$ = $\frac{9√(3)}{2}$ , which is choice (C).

\frac{27√(3)}{2}

- 9√(3) =

\frac{9√(3)}{2}

→ (C)

Answer

\frac{9 √(3)}{2}

Sanity. Large hex area

\frac{27√(3)}{2}

≈ 23.4, six small hexes total 9√(3) ≈ 15.6, leftover ≈ 7.8, and

\frac{9√(3)}{2}

≈ 7.79 — matches. The answer is exactly

\frac{1}{3}

of the big-hex area (since

\frac{9}{27}

\frac{1}{3}

), which is a clean ratio that survives any rescaling of the configuration — strong sign the

\frac{3}{7}

corner gap really is a red herring, exactly as the easier-problem move predicted. The non-matching choices

\frac{216√(3)}{49}

and

\frac{243√(3)}{49}

have 49 in the denominator (they would arise if the frame side S depended on

\frac{3}{7}

); choices

\frac{13√(3)}{3}

and

\frac{14√(3)}{3}

would arise from an arithmetic slip near the apothem stacking.

💡Key takeaway

This AMC 10 problem only needs Grade 8 Pythagorean theorem on an equilateral triangle plus area-by-decomposing — once you spot that the $\frac{3}{7}$ corner gap doesn't affect the frame size (do the simplest case first), the big hex has side S=3 and the leftover area is just $\frac{27√(3)}{2}$ - 9√(3) = $\frac{9√(3)}{2}$ .

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