2005 AMC 10A Problem 12

The dashed lines show a large equilateral triangle of side $2$ split into four equilateral triangles of side $1$. Each of the four curved arcs is a $60^\circ$ arc of radius $1$ drawn on a side of one of the small triangles, so the trefoil equals the big triangle plus four circular segments.

Each segment is a $60^\circ$ sector of radius $1$ minus a side-$1$ equilateral triangle. Therefore
$$\text{trefoil} = [\text{big triangle}] + 4\,[\text{sector}] - 4\,[\text{small triangle}].$$
But the big triangle is made of exactly four small triangles, so the triangle terms cancel and only the sectors remain:
$$4 \cdot \frac{60}{360} \cdot \pi \cdot 1^2 = \frac{4\pi}{6} = \frac{2\pi}{3}.$$

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

Curvy regions on a triangular grid are best written as sectors plus or minus triangles, then simplified. Here the four bulges each add a segment (sector minus triangle), and the four subtracted triangles are exactly the pieces of the base triangle, so the whole area is four sectors. The cancellation is not luck: each sector's straight part is one grid triangle, and there are four of each.

Reject choices with a $\sqrt{3}$ term by estimation: the trefoil looks only slightly larger than a triangle of area $\sqrt{3} \approx 1.73$, and $\frac{2\pi}{3} \approx 2.09$ fits, while choices (C)-(E) exceed $2.5$ and (A) is about $1.9$, which would correspond to only two sectors plus a triangle.

• Adding the four circular segments to the full side-2 triangle without noticing that the four small triangles they subtract cancel it exactly, leaving a stray root-3 term.
• Using radius $2$ for the sectors (the base length) instead of radius $1$ (the small triangle's side).