AMC 10 · 2005 · #12
Grade 7 geometry-2dThe figure shown is called a trefoil and is constructed by drawing circular sectors about sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length 2?
Pick an answer.
AMC 10 2005 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 trefoil is drawn by attaching circular sectors to the sides of congruent equilateral triangles, sitting on a horizontal base of length $2$. Find the total area of the trefoil.
Givens: The shape is made of circular sectors drawn about the sides of congruent equilateral triangles; The horizontal base has length $2$; The base is split into two equal parts by the center, so each equal triangle side (and each sector radius) is $1$; Answer choices: (A) $\frac{1}{3}\pi+\frac{\sqrt{3}}{2}$, (B) $\frac{2}{3}\pi$, (C) $\frac{2}{3}\pi+\frac{\sqrt{3}}{4}$, (D) $\frac{2}{3}\pi+\frac{\sqrt{3}}{3}$, (E) $\frac{2}{3}\pi+\frac{\sqrt{3}}{2}$
Unknowns: The total area of the trefoil
Understand
Restated: A trefoil is drawn by attaching circular sectors to the sides of congruent equilateral triangles, sitting on a horizontal base of length $2$. Find the total area of the trefoil.
Givens: The shape is made of circular sectors drawn about the sides of congruent equilateral triangles; The horizontal base has length $2$; The base is split into two equal parts by the center, so each equal triangle side (and each sector radius) is $1$; Answer choices: (A) $\frac{1}{3}\pi+\frac{\sqrt{3}}{2}$, (B) $\frac{2}{3}\pi$, (C) $\frac{2}{3}\pi+\frac{\sqrt{3}}{4}$, (D) $\frac{2}{3}\pi+\frac{\sqrt{3}}{3}$, (E) $\frac{2}{3}\pi+\frac{\sqrt{3}}{2}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #16 Change Focus / Count the Complement, #3 Eliminate Possibilities
The trefoil is a compound shape, so Tool #7 (Identify Subproblems) is the spine: cut it into pieces whose areas are easy. The cleanest cut sits the trefoil on a big equilateral triangle of side $2$ (vertices at the two base ends and the top point), then adds four circular "bumps." Each bump is a $60^\circ$-sector minus a small equilateral triangle of side $1$. Tool #1 (Draw a Diagram) locates those pieces. The payoff comes from Tool #16 (Change Focus): instead of grinding out the messy triangle areas, notice that the four small triangles rebuild exactly the big triangle, so every triangle term cancels and only the sectors survive. Tool #3 (Eliminate Possibilities) then matches the clean result to the choices.
Execute — Answer: B
6.G.A.1 Step 1 Fix the radius and the building blocks
- The center of the figure lies on the base, and the base runs from one side of the circle to the other, so the base is a diameter of length $2$.
- That makes every sector radius $r=1$.
- Each shaded arc turns through $60^\circ$, and joining the center to the two ends of such an arc gives two radii of length $1$ with a $60^\circ$ angle between them: a small equilateral triangle of side $1$.
- So the whole figure is built from radius-$1$, $60^\circ$ sectors sitting on side-$1$ equilateral triangles.
💡 A $60^\circ$ wedge with two equal radii always closes up into an equilateral triangle.
7.G.A.1 Step 2 Sit the trefoil on a big triangle
- Connect the two ends of the base to the top point of the trefoil.
- Those three points form one large equilateral triangle whose side is the full base, length $2$.
- The trefoil is exactly this big triangle with four rounded bumps pushed outward along its two slanted sides (each slanted side is split at its midpoint, giving two bumps per side).
- Doubling the side of an equilateral triangle multiplies its area by $2^2=4$, so the big triangle's area equals $4$ small side-$1$ triangles.
💡 Scaling a shape by a factor of $2$ stretches its area by $2\times 2=4$.
7.G.B.4 Step 3 Measure one bump
- Each bump is a circular segment: take the $60^\circ$ sector of radius $1$ and remove the equilateral triangle of side $1$ underneath its chord.
- A $60^\circ$ sector is one sixth of a full circle, so its area is $\frac{60}{360}\pi(1)^2=\frac{\pi}{6}$.
- Thus one bump has area $\frac{\pi}{6}-T_{\text{small}}$, and there are four identical bumps.
💡 A segment is just the pie slice with its straight-sided triangle carved off.
6.G.A.1 Step 4 Add the pieces and watch the triangles cancel
- The trefoil is the big triangle plus the four bumps: $4\,T_{\text{small}}+4\left(\frac{\pi}{6}-T_{\text{small}}\right)$.
- Distribute the $4$ over the bumps: the $+4\,T_{\text{small}}$ from the big triangle and the $-4\,T_{\text{small}}$ from the bumps are equal and opposite, so every triangle term disappears.
- No triangle area ever has to be computed, which is why no $\sqrt{3}$ can appear in the answer.
💡 The four small triangles you added are the very same four you subtract back out, so they wash away.
5.NF.B.4 Step 5 Compute the surviving sectors
- Only the four sectors remain: $4\cdot\frac{\pi}{6}=\frac{4\pi}{6}=\frac{2\pi}{3}$.
- Scanning the choices, this has no $\sqrt{3}$ term, which rules out (A), (C), (D), and (E) immediately, and it matches (B) exactly.
- The area of the trefoil is $\frac{2}{3}\pi$, choice (B).
💡 Four sixth-of-a-circle wedges glue into two thirds of a whole circle.
6.G.A.1 The center of the figure lies on the base, and the base runs from one side of th 7.G.A.1 Connect the two ends of the base to the top point of the trefoil. Those three po 7.G.B.4 Each bump is a circular segment: take the $60^\circ$ sector of radius $1$ and re 6.G.A.1 The trefoil is the big triangle plus the four bumps: $4\,T_{\text{small}}+4\left 5.NF.B.4 Only the four sectors remain: $4\cdot\frac{\pi}{6}=\frac{4\pi}{6}=\frac{2\pi}{3} Review
Reasonableness: A sanity check: $\frac{2}{3}\pi\approx 2.09$. The big equilateral triangle of side $2$ has area $\sqrt{3}\approx 1.73$, and the trefoil is that triangle plus four outward bumps, so its area should be a bit more than $1.73$ — and $2.09$ is indeed a little larger, which fits. Every other choice adds a $\sqrt{3}$ term, but the four added triangles and four subtracted triangles cancel exactly, so a lone $\pi$ answer is exactly what the geometry predicts; $\frac{2}{3}\pi$ is choice (B).
Alternative: Compute directly without the cancellation trick. Using the four bump segments each equal to $\frac{\pi}{6}-\frac{\sqrt{3}}{4}$ (sector minus a side-$1$ equilateral triangle of area $\frac{\sqrt{3}}{4}$) and the big triangle of area $\sqrt{3}$: total $=\sqrt{3}+4\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)=\sqrt{3}+\frac{2\pi}{3}-\sqrt{3}=\frac{2\pi}{3}$. The $\sqrt{3}$ terms cancel and the same answer $\frac{2}{3}\pi$ appears, confirming (B).
CCSS standards used (min grade 7)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Cutting the trefoil into a large equilateral triangle plus four circular segments, and adding the pieces back up.)7.G.A.1Solve problems involving scale drawings of geometric figures (Using that doubling a triangle's side quadruples its area, so the side-$2$ triangle equals four side-$1$ triangles.)7.G.B.4Know the formulas for area and circumference of a circle (Finding a $60^\circ$ sector of radius $1$ as one sixth of a circle: $\frac{1}{6}\pi(1)^2=\frac{\pi}{6}$.)5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Combining the four surviving sectors: $4\cdot\frac{\pi}{6}=\frac{2\pi}{3}$.)
⭐ Sit the trefoil on a big triangle and add four pie-slice bumps: the triangles you add and subtract cancel, leaving just $4\times\frac{1}{6}$ of a circle $=\frac{2}{3}\pi$.
⭐ Sit the trefoil on a big triangle and add four pie-slice bumps: the triangles you add and subtract cancel, leaving just $4\times\frac{1}{6}$ of a circle $=\frac{2}{3}\pi$.
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