2013 AMC 10B Problem 15

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

Regular polygons with a given perimeter have area proportional to the square of the perimeter, with a shape-dependent constant. Comparing the two constants (triangle $\frac{\sqrt3}{36}$, hexagon $\frac{\sqrt3}{24}$) is the whole problem; the hexagon is the more "efficient" shape, so it needs less wire for the same area, and $a > b$.

Let the hexagon have side $1$ (so $b = 6$, area $\frac{3\sqrt3}{2}$). A triangle of equal area has side $s$ with $\frac{\sqrt3}{4}s^2 = \frac{3\sqrt3}{2}$, so $s^2 = 6$ and $a = 3\sqrt6$. Then $\frac{a}{b} = \frac{3\sqrt6}{6} = \frac{\sqrt6}{2}$.

• Using side length a and b instead of a/3 and b/6, or forgetting to take the square root at the end and answering 3/2.
• Misremembering the hexagon area as $\frac{3\sqrt3}{4}s^2$ or the triangle area as $\frac{\sqrt3}{2}s^2$, which changes the ratio to a different choice.