2023 AMC 10 A Problem 24

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To help you master 2023 AMC 10 A Problem 24, this page includes:

• The original AMC problem to try, plus the correct answer

• Clear, step-by-step solutions using different methods

• Key ideas, useful formulas, and takeaways you can apply to similar problems

• Related problems from past AMC exams, plus new practice problems that build the same skills

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2023 AMC10 A Problem 24

Solution

Solution 1

Alignment and Area-Subtraction Method

Use the aligned groups of three unit hexagons to determine the side length of the outer regular hexagon. Once both the frame and the blocks have known side lengths, calculate their regular-hexagon areas and subtract.

A regular hexagon of side length $s$ can be divided into six equilateral triangles of side length $s$, so its area is $A=\frac{3\sqrt{3}}{2}{s}^2$. Also, equal endpoint offsets cancel when measuring the total displacement between adjacent frame corners.

Step 1: Determine the side length of the frame

Choose a direction parallel to one side of the outer frame and trace the aligned group of three blocks indicated in the diagram. Each unit hexagon contributes one unit-length side vector in this direction. At the two ends, the corner-to-block distances are equal and point in opposite endpoint directions, so one offset is added and the other is subtracted. Thus the offsets do not affect the total distance between adjacent frame corners.

$$d=\frac{3}{7}$$

$$S=d+(1+1+1)-d$$

$$S=3$$

Step 2: Find the area of the outer frame

The frame is a regular hexagon with the side length found above. Apply the regular-hexagon area formula.

$$A_{\text{frame}}=\frac{3\sqrt{3}}{2}{S}^2$$

$$A_{\text{frame}}=\frac{3\sqrt{3}}{2}(3{)}^2$$

$$A_{\text{frame}}=\frac{27\sqrt{3}}{2}$$

Step 3: Find the total area occupied by the blocks

Each block is a regular hexagon of side length $1$. The six blocks have disjoint interiors, so their areas can be added.

$$A_{\text{one block}}=\frac{3\sqrt{3}}{2}(1{)}^2$$

$$A_{\text{one block}}=\frac{3\sqrt{3}}{2}$$

$$A_{\text{six blocks}}=6\left(\frac{3\sqrt{3}}{2}\right)$$

$$A_{\text{six blocks}}=9\sqrt{3}$$

Step 4: Subtract the occupied area

The blocks are entirely inside the frame, so the uncovered region is exactly the part of the frame remaining after the six block areas are removed.

$$A_{\text{uncovered}}=A_{\text{frame}}-A_{\text{six blocks}}$$

$$A_{\text{uncovered}}=\frac{27\sqrt{3}}{2}-9\sqrt{3}$$

$$A_{\text{uncovered}}=\frac{27\sqrt{3}-18\sqrt{3}}{2}$$

$$A_{\text{uncovered}}=\frac{9\sqrt{3}}{2}$$

Correct answer: C. $\frac{9\sqrt{3}}{2}$
The uncovered area is $\boxed{\frac{9\sqrt{3}}{2}}$ square units, corresponding to choice $\boxed{\text{C}}$. For a check, the covered and uncovered areas add to $9\sqrt{3}+\frac{9\sqrt{3}}{2}=\frac{27\sqrt{3}}{2}$, which equals the area of the side-$3$ frame.

Similar Problems in Real Exams

The following five AMC 10 A problems are closely related to AMC 10 A 2023 Question 24. These real AMC 10 A problems test the same key concept—Area. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2020 AMC 10 A Problem 20

2025 AMC 10 A Problem 10

2025 AMC 10 A Problem 15

2024 AMC 10 A Problem 14

2024 AMC 10 A Problem 16

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Area of a regular hexagon:

Here $s$ is the side length of a regular hexagon. The formula applies because joining the center to the six vertices divides the hexagon into six equilateral triangles of side length $s$. Both the outer frame and each of the six blocks are regular hexagons.

• Area subtraction:

The occupied pieces must lie entirely inside the whole region and must not overlap in their interiors. The six blocks lie inside the frame and have disjoint interiors, so removing their total area leaves exactly the requested region.

• Cancellation of equal endpoint offsets:

The two offsets must have equal length and contribute in opposite endpoint directions along the measured span. The distance from every frame corner to the nearest block vertex is the same, so the two $\frac{3}{7}$ endpoint contributions cancel when the aligned span is measured.

Key Takeaway: Remember This Strategy for Next Time

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