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

Six regular hexagonal blocks of side length 1 unit are arranged inside a regular hexagonal frame. Each block lies along an inside edge of the frame and is aligned with two other blocks, as shown in the figure below. The distance from any corner of the frame to the nearest vertex of a block is $\frac{3}{7}$ unit. What is the area of the region inside the frame not occupied 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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: Each small block is a regular hexagon with side $1$; Six blocks; each lies along a different inside edge of the frame; Each block is aligned (touching) two other blocks; Corner-to-nearest-vertex distance = $\frac{3}{7}$; Answer choices: (A) $\tfrac{13\sqrt{3}}{3}$, (B) $\tfrac{216\sqrt{3}}{49}$, (C) $\tfrac{9\sqrt{3}}{2}$, (D) $\tfrac{14\sqrt{3}}{3}$, (E) $\tfrac{243\sqrt{3}}{49}$

Unknowns: Area inside the frame not covered by the six blocks

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem

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 $\tfrac{3}{7}$ 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 $\tfrac{3}{7}$ parameter.

Execute — Answer: C

#1 Draw a Diagram 4.G.A.2 Step 1

Draw the setup.
Place the large hex centered at the origin with two horizontal sides (top and bottom).
The six small hexes sit one against each inside edge and touch two neighbors.
By symmetry, the six small hexes ring a central hexagonal hole.
The whole figure has the same six-fold symmetry as the frame.

$$\text{Big hex: side } S,\;\text{small hexes: side } 1$$

💡 Grade 4 "classify by parallel/perpendicular sides" — set the frame up so opposite sides are parallel and the symmetry is visible.

#7 Identify Subproblems 6.G.A.1 Step 2

Split the target area.
Let $A_{\text{big}}$ be the area of the large hexagon and $A_{\text{small}}$ the area of one small hexagon.
The uncovered area is $A_{\text{big}} - 6 A_{\text{small}}$.

$$\text{Answer} = A_{\text{big}} - 6\,A_{\text{small}}$$

💡 Grade 6 "area by composing/decomposing" — split a complicated region into two pieces whose areas are easy.

#7 Identify Subproblems 8.G.B.7 Step 3

Compute the area of one small hexagon.
A regular hexagon with side $s$ decomposes into $6$ equilateral triangles of side $s$.
Each equilateral triangle has height $\tfrac{s\sqrt{3}}{2}$ (Pythagoras on the half-triangle: $1^2 = (\tfrac{1}{2})^2 + h^2$ gives $h = \tfrac{\sqrt{3}}{2}$ for $s = 1$), so its area is $\tfrac{1}{2}\cdot s\cdot \tfrac{s\sqrt{3}}{2} = \tfrac{s^2\sqrt{3}}{4}$.
For $s = 1$, hexagon area $= 6 \cdot \tfrac{\sqrt{3}}{4} = \tfrac{3\sqrt{3}}{2}$.

$$A_{\text{small}} = \tfrac{3\sqrt{3}}{2}, \quad 6\,A_{\text{small}} = 9\sqrt{3}$$

💡 Grade 8 Pythagorean theorem gives the equilateral height $\tfrac{\sqrt{3}}{2}$; then six congruent triangles fill the hexagon.

#9 Solve an Easier Related Problem 8.G.B.7 Step 4

Find the side $S$ of the big hexagon by looking at an easier version.
Drop the $\tfrac{3}{7}$ distance temporarily; the six small hexes ring a central hexagonal hole of side $1$ (each pair of small hexes shares a side, and the inner sides of the ring form the central hole).
Measure the vertical distance from the center of the figure straight up to the top of the frame.

$$\text{Center-to-top-edge distance} = \tfrac{S\sqrt{3}}{2}$$

💡 Grade 9-easier-problem move — the actual size of the frame should not depend on the $\tfrac{3}{7}$ corner gap, so solve the cleanest case first.

#1 Draw a Diagram 8.NS.A.2 Step 5

Add the dimensions stacked vertically through the center.
From center to the top of the central hole: apothem of the central hexagon $= \tfrac{\sqrt{3}}{2}$.
From there to the top of the top small hex: full height of the small hex $= 2 \cdot \tfrac{\sqrt{3}}{2} = \sqrt{3}$.
Total $= \tfrac{\sqrt{3}}{2} + \sqrt{3} = \tfrac{3\sqrt{3}}{2}$, which equals the apothem of the big hex $\tfrac{S\sqrt{3}}{2}$.
So $S = 3$.

$$\tfrac{S\sqrt{3}}{2} = \tfrac{\sqrt{3}}{2} + \sqrt{3} = \tfrac{3\sqrt{3}}{2} \;\Rightarrow\; S = 3$$

💡 Grade 8 "use rational approximations of irrationals" — the $\sqrt{3}$ factors cancel cleanly because every length is a rational multiple of $\sqrt{3}$.

#7 Identify Subproblems 6.G.A.1 Step 6

Compute the area of the large hex.
With $S = 3$, $A_{\text{big}} = \tfrac{3\sqrt{3}}{2}\, S^2 = \tfrac{3\sqrt{3}}{2} \cdot 9 = \tfrac{27\sqrt{3}}{2}$.

$$A_{\text{big}} = \tfrac{27\sqrt{3}}{2}$$

💡 Plug $S = 3$ into the hexagon area formula derived in step 3 (scaled up by factor $9$).

#7 Identify Subproblems 7.NS.A.1 Step 7

Subtract to finish: $A_{\text{big}} - 6 A_{\text{small}} = \tfrac{27\sqrt{3}}{2} - 9\sqrt{3} = \tfrac{27\sqrt{3}}{2} - \tfrac{18\sqrt{3}}{2} = \tfrac{9\sqrt{3}}{2}$, which is choice (C).

$$\tfrac{27\sqrt{3}}{2} - 9\sqrt{3} = \tfrac{9\sqrt{3}}{2} \;\Rightarrow\; \textbf{(C)}$$

💡 Grade 7 "add and subtract rationals" — common denominator of $2$ and subtract.

[1] #1 4.G.A.2 Draw the setup. Place the large hex centered at the origin with two horizontal s

[2] #7 6.G.A.1 Split the target area. Let $A_{\text{big}}$ be the area of the large hexagon and

[3] #7 8.G.B.7 Compute the area of one small hexagon. A regular hexagon with side $s$ decompose

[4] #9 8.G.B.7 Find the side $S$ of the big hexagon by looking at an easier version. Drop the $

[5] #1 8.NS.A.2 Add the dimensions stacked vertically through the center. From center to the top

[6] #7 6.G.A.1 Compute the area of the large hex. With $S = 3$, $A_{\text{big}} = \tfrac{3\sqrt

[7] #7 7.NS.A.1 Subtract to finish: $A_{\text{big}} - 6 A_{\text{small}} = \tfrac{27\sqrt{3}}{2}

Review

Reasonableness: Sanity. Large hex area $\tfrac{27\sqrt{3}}{2} \approx 23.4$, six small hexes total $9\sqrt{3} \approx 15.6$, leftover $\approx 7.8$, and $\tfrac{9\sqrt{3}}{2} \approx 7.79$ — matches. The answer is exactly $\tfrac{1}{3}$ of the big-hex area (since $\tfrac{9}{27} = \tfrac{1}{3}$), which is a clean ratio that survives any rescaling of the configuration — strong sign the $\tfrac{3}{7}$ corner gap really is a red herring, exactly as the easier-problem move predicted. The non-matching choices $\tfrac{216\sqrt{3}}{49}$ and $\tfrac{243\sqrt{3}}{49}$ have $49$ in the denominator (they would arise if the frame side $S$ depended on $\tfrac{3}{7}$); choices $\tfrac{13\sqrt{3}}{3}$ and $\tfrac{14\sqrt{3}}{3}$ would arise from an arithmetic slip near the apothem stacking.

Alternative: Tool #3 (Eliminate Possibilities): the answer must be a rational multiple of $\sqrt{3}$ (every length is a rational multiple of $\sqrt{3}$ or $1$, and area is linear in $\sqrt{3}$ when squared) and should equal the uncovered fraction times a clean hex area. The fraction $\tfrac{1}{3}$ of the big-hex area is the simplest combinatorial answer (six hexes ring a hole, leaving a $7$-hex shape's worth of area — the hole hexagon contributes the $\tfrac{1}{3}$ surplus). That points to (C) immediately.

CCSS standards used (min grade 8)

4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines, or specified angles (Identifying the configuration as a regular hexagon containing six congruent regular hexagons arranged around a central hexagonal hole.)
6.G.A.1 Find the area of triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles (Decomposing each regular hexagon into 6 congruent equilateral triangles and computing the area as the difference of two hexagon areas.)
7.NS.A.1 Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers (Subtracting $\tfrac{27\sqrt{3}}{2} - 9\sqrt{3}$ via a common denominator to get $\tfrac{9\sqrt{3}}{2}$.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Deriving the height $\tfrac{s\sqrt{3}}{2}$ of an equilateral triangle with side $s$, then the apothem $\tfrac{S\sqrt{3}}{2}$ of the big hex.)
8.NS.A.2 Use rational approximations of irrational numbers to compare their size and locate them on the number line (Comparing apothem lengths as multiples of $\sqrt{3}$ to solve $\tfrac{S\sqrt{3}}{2} = \tfrac{3\sqrt{3}}{2}$ and conclude $S = 3$.)

⭐ This AMC 10 problem only needs Grade 8 Pythagorean theorem on an equilateral triangle plus area-by-decomposing — once you spot that the $\tfrac{3}{7}$ 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 $\tfrac{27\sqrt{3}}{2} - 9\sqrt{3} = \tfrac{9\sqrt{3}}{2}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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