UKMT Intermediate Challenge-II

• All answers are wrong. The correct answer is 2:3 .
• See, let's use your equations up there. If the perimeters are equal, we can set the regular sides length as $a = \frac{p}{3}$ , for the triangle; and $a = \frac{p}{6}$ for the hexagon.
• We have the area of the triangle is $\frac{\sqrt{3}}{4} \times (\frac{p}{3})^{2} = \frac{p^{2}\sqrt{3}}{36}$ .
• And we find the area of the hexagon as $\frac{3\sqrt{3}}{2} \times (\frac{p}{6})^{2} = \frac{p^{2}\sqrt{3}}{24}$ .
• So, the ratio is: $\frac{p^{2}\sqrt{3}}{36} \times \frac{24}{p^{2}\sqrt{3}} =\frac{2}{3}$ , which means the ratio is 2:3 .

Is there a problem with this solution? I think the ratio of the area of the triangle to the area of the hexagon is more like 1/6. How can you calculate and have 2/3 as result?