AMC 10 · 2012 · #18
Grade 8 geometry-2dThe closed curve in the figure is made up of 9 congruent circular arcs each of length 32π, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
Pick an answer.
AMC 10 2012 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 closed curve is built from $9$ congruent circular arcs, each of length $\frac{2\pi}{3}$. The center of every arc's circle is one of the vertices of a regular hexagon with side length $2$. Find the area the curve encloses.
Givens: $9$ congruent arcs, each of arc length $\frac{2\pi}{3}$; Each arc's center is a vertex of a regular hexagon of side $2$; The figure has three-fold symmetry: some arcs bulge outward, some cave inward; Answer choices: (A) $2\pi+6$, (B) $2\pi+4\sqrt{3}$, (C) $3\pi+4$, (D) $2\pi+3\sqrt{3}+2$, (E) $\pi+6\sqrt{3}$
Unknowns: The area enclosed by the closed curve
Understand
Restated: A closed curve is built from $9$ congruent circular arcs, each of length $\frac{2\pi}{3}$. The center of every arc's circle is one of the vertices of a regular hexagon with side length $2$. Find the area the curve encloses.
Givens: $9$ congruent arcs, each of arc length $\frac{2\pi}{3}$; Each arc's center is a vertex of a regular hexagon of side $2$; The figure has three-fold symmetry: some arcs bulge outward, some cave inward; Answer choices: (A) $2\pi+6$, (B) $2\pi+4\sqrt{3}$, (C) $3\pi+4$, (D) $2\pi+3\sqrt{3}+2$, (E) $\pi+6\sqrt{3}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #16 Change Focus / Count the Complement
The curve is irregular, but its pieces are all circular arcs centered at the hexagon's vertices, so the honest move is to build the area out of pieces I can compute (Tool #7). Draw the hexagon that connects the centers (Tool #1) and use it as a baseline area. Then, instead of chasing the wavy boundary directly, reframe the curve as the hexagon with three outward bulges added and three inward bites removed (Tool #16). Every added or removed piece is a circular sector, so the whole answer becomes hexagon area plus outer sectors minus inner sectors.
Execute — Answer: E
7.G.B.4 Step 1 Find the arc radius
- Each arc has length $\frac{2\pi}{3}$.
- Reading the figure, each arc spans a $120^\circ$ turn, which is $\frac{1}{3}$ of a full circle.
- So the full circumference is $3\cdot\frac{2\pi}{3}=2\pi$.
- From circumference $=2\pi r$, the radius is $r=1$.
💡 A $120^\circ$ arc is one-third of its circle, so three of them make a whole trip around, which pins down the radius as $1$.
8.G.B.7 Step 2 Area of the center hexagon
- Connect the six arc centers; they form the regular hexagon of side $2$.
- Split it into $6$ equilateral triangles of side $2$.
- Each such triangle has height $\sqrt{2^2-1^2}=\sqrt{3}$ (drop an altitude and use the Pythagorean theorem), so its area is $\frac{1}{2}\cdot 2\cdot\sqrt{3}=\sqrt{3}$.
- The hexagon area is $6\sqrt{3}$.
💡 A regular hexagon is just six equilateral triangles glued at the center, and each triangle's height comes straight from the Pythagorean theorem.
7.G.B.6 Step 3 Reframe curve vs. hexagon
- Compare the curve to the hexagon boundary.
- By the three-fold symmetry, three arcs bulge outward past the hexagon (adding area), and three arcs cave inward (removing area).
- Each inward bite is a $120^\circ$ sector of a radius-$1$ circle; each outward bulge is a $240^\circ$ sector of a radius-$1$ circle.
- So enclosed area $=$ hexagon $+$ (outer sectors) $-$ (inner sectors).
💡 Rather than integrate a wavy edge, treat the curve as the hexagon with bumps added and dents subtracted.
7.G.B.4 Step 4 Compute the sector areas
- A full radius-$1$ circle has area $\pi\cdot 1^2=\pi$.
- Inner bites: three $120^\circ$ sectors, each $\frac{1}{3}\pi$, together $3\cdot\frac{\pi}{3}=\pi$ (one whole circle).
- Outer bulges: three $240^\circ$ sectors, each $\frac{240}{360}\pi=\frac{2\pi}{3}$, together $3\cdot\frac{2\pi}{3}=2\pi$ (two whole circles).
💡 Each sector is a plain fraction of $\pi r^2$, and the fractions line up so the three inner sectors make one circle and the three outer sectors make two.
7.G.B.6 Step 5 Add it all up
- Combine the pieces: start from the hexagon $6\sqrt{3}$, subtract the inner bites $\pi$, and add the outer bulges $2\pi$.
- The $\pi$ terms give $-\pi+2\pi=\pi$, leaving $6\sqrt{3}+\pi$.
💡 The bulges outweigh the bites by exactly one circle's worth, so a single $+\pi$ survives on top of the hexagon.
7.G.B.4 Each arc has length $\frac{2\pi}{3}$. Reading the figure, each arc spans a $120^ 8.G.B.7 Connect the six arc centers; they form the regular hexagon of side $2$. Split it 7.G.B.6 Compare the curve to the hexagon boundary. By the three-fold symmetry, three arc 7.G.B.4 A full radius-$1$ circle has area $\pi\cdot 1^2=\pi$. Inner bites: three $120^\c 7.G.B.6 Combine the pieces: start from the hexagon $6\sqrt{3}$, subtract the inner bites Review
Reasonableness: Count the arcs the framing uses: three inward $120^\circ$ arcs ($3$ arcs) plus three outward $240^\circ$ bulges, each of which is two $120^\circ$ arcs ($6$ arcs), totals $3+6=9$ arcs of length $\frac{2\pi}{3}$ each — exactly the $9$ congruent arcs the problem names. The area $\pi+6\sqrt{3}\approx 3.14+10.39=13.53$, a bit larger than the hexagon's $6\sqrt{3}\approx 10.39$, which fits a shape whose outward bulges beat its inward dents. Among the choices, (E) is the only one carrying a lone $\pi$ with the hexagon's $6\sqrt{3}$, matching our net of one extra circle.
Alternative: Shortcut by tracking only what changes. Every arc is a piece of a radius-$1$ circle, so the whole answer is $6\sqrt{3}$ (the fixed hexagon) plus some whole number of circle-areas $\pi$ from the arcs. The straight-sided hexagon already gives the $\sqrt{3}$ part, so the answer must read $(\text{integer})\pi+6\sqrt{3}$. Only choice (E) has this exact form, so (E) follows even before computing that the integer is $1$.
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Turning arc length $\frac{2\pi}{3}$ into radius $r=1$, and computing each sector as a fraction of the circle area $\pi r^2$.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding the height $\sqrt{2^2-1^2}=\sqrt{3}$ of the side-$2$ equilateral triangle to get the hexagon area $6\sqrt{3}$.)7.G.B.6Solve real-world and mathematical problems involving area of two-dimensional objects composed of triangles, quadrilaterals, and polygons (Assembling the enclosed area as hexagon $+$ outer sectors $-$ inner sectors to reach $\pi+6\sqrt{3}$.)
⭐ Draw the hexagon through the arc centers ($6\sqrt{3}$), then treat the curve as that hexagon with three outward bulges added ($2\pi$) and three inward bites removed ($\pi$); the bulges win by one circle, giving $\textbf{(E)}\ \pi+6\sqrt{3}$.
⭐ Draw the hexagon through the arc centers ($6\sqrt{3}$), then treat the curve as that hexagon with three outward bulges added ($2\pi$) and three inward bites removed ($\pi$); the bulges win by one circle, giving $\textbf{(E)}\ \pi+6\sqrt{3}$.
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