AMC 10 · 2020 · #14
Grade 7 geometry-2dProblem
As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region ---- inside the hexagon but outside all of the semicircles?
Pick an answer.
AMC 10 2020 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 regular hexagon has side length $2$. Six semicircles lie inside it, one on each side, with each diameter equal to a side of the hexagon (so each semicircle has radius $1$ and bulges into the hexagon). Find the area of the region that is inside the hexagon but outside all six semicircles.
Givens: Regular hexagon with side length $2$; Six semicircles, each of diameter $2$ (radius $1$), one on each side; Each semicircle lies inside the hexagon; Answer choices: $6\sqrt{3} - 3\pi,\; \tfrac{9\sqrt{3}}{2} - 2\pi,\; \tfrac{3\sqrt{3}}{2} - \tfrac{\pi}{3},\; 3\sqrt{3} - \pi,\; \tfrac{9\sqrt{3}}{2} - \pi$
Unknowns: Area of the shaded region (inside hexagon, outside every semicircle)
Understand
Restated: A regular hexagon has side length $2$. Six semicircles lie inside it, one on each side, with each diameter equal to a side of the hexagon (so each semicircle has radius $1$ and bulges into the hexagon). Find the area of the region that is inside the hexagon but outside all six semicircles.
Givens: Regular hexagon with side length $2$; Six semicircles, each of diameter $2$ (radius $1$), one on each side; Each semicircle lies inside the hexagon; Answer choices: $6\sqrt{3} - 3\pi,\; \tfrac{9\sqrt{3}}{2} - 2\pi,\; \tfrac{3\sqrt{3}}{2} - \tfrac{\pi}{3},\; 3\sqrt{3} - \pi,\; \tfrac{9\sqrt{3}}{2} - \pi$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
Tool #1 (Draw): subdivide the hexagon into a grid of small equilateral triangles of side $1$ — $24$ of them — so every region (shaded or white) is a clean union of these tiles or circular sectors. Tool #9 (Easier Problem): by 6-fold rotational symmetry, the shaded region is $6$ congruent pieces; find the area of one piece and multiply. Tool #7 (Subproblems): each piece is (rhombus of $2$ small triangles) minus (one $60^\circ$ sector of radius $1$). Tool #3 (Eliminate): the simplified expression $3\sqrt{3} - \pi$ matches exactly one answer choice.
Execute — Answer: D
6.G.A.1 Step 1 - Tile the side-$2$ hexagon with $24$ small equilateral triangles of side $1$.
- Draw the three long diagonals and the three lines connecting midpoints of opposite sides — these cut the hexagon into $24$ congruent triangles of side $1$.
💡 Grade 6: regular hexagons tile cleanly into small equilateral triangles.
4.G.A.3 Step 2 - By the $6$-fold rotational symmetry of the figure (hexagon plus the $6$ congruent semicircles), the shaded region splits into $6$ congruent pieces — one around each vertex of the hexagon.
- Find one piece and multiply by $6$.
💡 Grade 4 symmetry: $6$ rotations map the figure to itself, so $6$ identical pieces.
7.G.B.6 Step 3 - Look at one such piece.
- It sits at a hexagon vertex and is bounded by parts of the two adjacent semicircles.
- Within the small-triangle grid, this piece is a rhombus made of $2$ unit equilateral triangles, minus a $60^\circ$ sector of radius $1$ centered at the hexagon vertex.
- (The two adjacent semicircles meet exactly along the long diagonals of the small grid, and inside the rhombus they together form one $60^\circ$ wedge of radius $1$.)
💡 Grade 7 area: subtract the circular wedge from the rhombus to isolate the shaded sliver.
7.G.B.4 Step 4 - Compute the two ingredients.
- The rhombus has area $2 \cdot \tfrac{\sqrt{3}}{4} = \tfrac{\sqrt{3}}{2}$.
- The $60^\circ$ sector has area $\tfrac{60^\circ}{360^\circ} \cdot \pi \cdot 1^2 = \tfrac{\pi}{6}$.
💡 Grade 7 circle: $60^\circ$ is one-sixth of a circle, so $\tfrac{1}{6}\pi r^2$.
7.EE.A.1 Step 5 Subtract to get one piece, then multiply by $6$ for the total shaded area.
💡 Grade 7 expressions: distribute $6$ across the two terms.
7.EE.A.2 Step 6 - Match $3\sqrt{3} - \pi$ to the choices: option (D).
- The other choices have wrong coefficients on either $\sqrt{3}$ or $\pi$, corresponding to common over- or under-counting mistakes.
💡 Grade 7: match coefficients of $\sqrt{3}$ and $\pi$ to a single option.
6.G.A.1 Tile the side-$2$ hexagon with $24$ small equilateral triangles of side $1$. Dra 4.G.A.3 By the $6$-fold rotational symmetry of the figure (hexagon plus the $6$ congruen 7.G.B.6 Look at one such piece. It sits at a hexagon vertex and is bounded by parts of t 7.G.B.4 Compute the two ingredients. The rhombus has area $2 \cdot \tfrac{\sqrt{3}}{4} = 7.EE.A.1 Subtract to get one piece, then multiply by $6$ for the total shaded area. 7.EE.A.2 Match $3\sqrt{3} - \pi$ to the choices: option (D). The other choices have wrong Review
Reasonableness: The hexagon's area is $\tfrac{3\sqrt{3}}{2} \cdot 4 = 6\sqrt{3} \approx 10.39$. The total semicircle area (with overlap) is $6 \cdot \tfrac{\pi}{2} = 3\pi \approx 9.42$ — so the white region cannot be that large; some semicircle area is double-counted. The shaded answer $3\sqrt{3} - \pi \approx 5.196 - 3.14 \approx 2.06$ is positive and well below the hexagon area, exactly as expected for the leftover slivers near the vertices.
Alternative: Tool #5 (Pattern) via the leaf-counting identity (Solution #3 of the reference). Let $S$ be the shaded area and $L$ be the total area of the six lens-shaped overlaps where adjacent semicircles meet. Then $S = (\text{hexagon}) - (\text{semicircles}) + L = 6\sqrt{3} - 3\pi + L$, and the six leaves rearrange into one full unit circle minus a unit hexagon (an algebraic identity from the figure). Solving gives $S = 3\sqrt{3} - \pi$, confirming (D).
CCSS standards used (min grade 7)
4.G.A.3Recognize a line of symmetry for a two-dimensional figure (Using the hexagon's $6$-fold rotational symmetry to reduce to one congruent piece.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Tiling the side-$2$ hexagon into $24$ unit equilateral triangles.)7.G.B.4Know the formulas for area and circumference of a circle (Computing the $60^\circ$ sector area $\tfrac{\pi}{6}$ from $\pi r^2$.)7.G.B.6Solve real-world problems involving area, surface area, and volume (Subtracting the circular sector from the rhombus to find one shaded piece.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Multiplying $6 \left( \tfrac{\sqrt{3}}{2} - \tfrac{\pi}{6} \right) = 3\sqrt{3} - \pi$.)7.EE.A.2Rewrite an expression in different forms to shed light on the problem (Matching the simplified form to a single answer choice by its coefficients.)
⭐ This AMC 10 problem only needs Grade 7 area formulas you already know! Cut the side-$2$ hexagon into $24$ small triangles of side $1$. By 6-fold symmetry, the shaded region is $6$ identical pieces near the vertices — each piece is a $2$-triangle rhombus (area $\tfrac{\sqrt{3}}{2}$) minus a $60^\circ$ sector of radius $1$ (area $\tfrac{\pi}{6}$). Multiply by $6$: $6\left(\tfrac{\sqrt{3}}{2} - \tfrac{\pi}{6}\right) = \mathbf{3\sqrt{3} - \pi}$, answer (D).
⭐ This AMC 10 problem only needs Grade 7 area formulas you already know! Cut the side-$2$ hexagon into $24$ small triangles of side $1$. By 6-fold symmetry, the shaded region is $6$ identical pieces near the vertices — each piece is a $2$-triangle rhombus (area $\tfrac{\sqrt{3}}{2}$) minus a $60^\circ$ sector of radius $1$ (area $\tfrac{\pi}{6}$). Multiply by $6$: $6\left(\tfrac{\sqrt{3}}{2} - \tfrac{\pi}{6}\right) = \mathbf{3\sqrt{3} - \pi}$, answer (D).
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