2005 AMC 10A Problem 12

The dashed lines split a big equilateral triangle with side $2$ into four small equilateral triangles with side $1$. Each curved edge of the trefoil is a $60^\circ$ arc of radius $1$ drawn across a side of one small triangle.

1. Break the shape up. The trefoil is the big triangle plus four curved bulges, one outside each arc's straight edge.
2. What is one bulge? Take a $60^\circ$ slice of a circle of radius $1$ (a sector). Its straight part is a small equilateral triangle with side $1$. The bulge is the sector with that triangle removed:
$$\text{bulge} = \text{sector} - \text{small triangle}.$$
3. Add everything up.
$$\text{trefoil} = \text{big triangle} + 4 \times \text{sector} - 4 \times \text{small triangle}.$$
4. Watch the triangles cancel. The big triangle is made of exactly four small triangles. So "big triangle" and "$4 \times$ small triangle" are equal and cancel out:
$$\text{trefoil} = 4 \times \text{sector}.$$
5. Area of one sector. $60^\circ$ is $\frac{1}{6}$ of a full turn, so one sector has area $\frac{1}{6} \cdot \pi \cdot 1^2 = \frac{\pi}{6}$.
6. Total. $4 \times \frac{\pi}{6} = \frac{2\pi}{3}$.

The answer is $\boxed{\textbf{(B)}\ \frac{2}{3}\pi}$.

Curved shapes built on a triangle grid are best written as sectors plus or minus triangles. Here each sector covers one small grid triangle plus one bulge, and there are exactly as many sectors as small triangles, so the whole trefoil is just four sectors. Before computing $\sqrt{3}$ terms, check whether they cancel.

Estimate. The big triangle has area $\sqrt{3} \approx 1.73$, and the trefoil is that plus four thin bulges, so it should be a bit bigger than $1.73$. Now estimate the choices: (B) is about $2.09$; (C), (D) and (E) are all above $2.5$; (A) is about $1.91$. Each bulge is $\frac{\pi}{6} - \frac{\sqrt{3}}{4} \approx 0.09$, so four of them add about $0.36$, giving about $2.09$. That matches (B).