2013 AMC 10B Problem 15

1. Find the side lengths. The triangle has $3$ equal sides made from length $a$, so each side is $\frac{a}{3}$. The hexagon has $6$ equal sides made from length $b$, so each side is $\frac{b}{6}$.
2. Recall that an equilateral triangle with side $s$ has area $\frac{\sqrt{3}}{4} s^2$.
3. Triangle area:
$$\frac{\sqrt{3}}{4} \cdot \left(\frac{a}{3}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{a^2}{9} = \frac{\sqrt{3}\, a^2}{36}.$$
4. Hexagon area: lines from the center to the six corners cut a regular hexagon into $6$ equilateral triangles, each with side $\frac{b}{6}$:
$$6 \cdot \frac{\sqrt{3}}{4} \cdot \left(\frac{b}{6}\right)^2 = 6 \cdot \frac{\sqrt{3}}{4} \cdot \frac{b^2}{36} = \frac{\sqrt{3}\, b^2}{24}.$$
5. The areas are equal. Both sides have $\sqrt{3}$, which cancels:
$$\frac{a^2}{36} = \frac{b^2}{24}.$$
Multiply both sides by $36$: $a^2 = \frac{36}{24} b^2 = \frac{3}{2} b^2$. So $\frac{a^2}{b^2} = \frac{3}{2}$.
6. The question asks for $\frac{a}{b}$, not its square, so take the square root:
$$\frac{a}{b} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{3} \cdot \sqrt{2}}{2} = \frac{\sqrt{6}}{2}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{\sqrt{6}}{2}}$.

Both shapes are built from equilateral triangles, so one area formula does all the work. It makes sense that $a > b$: a hexagon is closer to a circle than a triangle is, so it wraps more area with the same amount of wire and needs less wire for the same area.

Pick easy numbers. Let the hexagon have side $1$, so $b = 6$ and its area is $6 \cdot \frac{\sqrt{3}}{4} = \frac{3\sqrt{3}}{2}$. A triangle with side $s$ and the same area has $\frac{\sqrt{3}}{4} s^2 = \frac{3\sqrt{3}}{2}$, so $s^2 = 6$ and $s = \sqrt{6}$. Then $a = 3\sqrt{6}$, and $\frac{a}{b} = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2}$.

The triangle has side $\frac{a}{3}$, so its area is
$$\frac{\sqrt3}{4}\left(\frac{a}{3}\right)^2 = \frac{\sqrt3\, a^2}{36}.$$
The hexagon has side $\frac{b}{6}$ and consists of six equilateral triangles of that side, so its area is
$$6 \cdot \frac{\sqrt3}{4}\left(\frac{b}{6}\right)^2 = \frac{6\sqrt3\, b^2}{144} = \frac{\sqrt3\, b^2}{24}.$$
Setting the areas equal and cancelling $\sqrt3$:
$$\frac{a^2}{36} = \frac{b^2}{24} \quad\Longrightarrow\quad \frac{a^2}{b^2} = \frac{36}{24} = \frac{3}{2}.$$
Therefore $\frac{a}{b} = \sqrt{\frac32} = \frac{\sqrt6}{2}$.

The answer is $\boxed{\textbf{(B)}\ \frac{\sqrt{6}}{2}}$.