Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 2.7 Junior Find the answer

A wire is cut into two pieces, one of length $a$ and the other of length $b$ . The piece of length $a$ is bent to form an equilateral triangle, and the piece of length $b$ is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is $\frac{a}{b}$ ?

Pick one

Solution

Using the area formulas for an equilateral triangle $\left(\frac{{s}^{2}\sqrt{3}}{4}\right)$ and regular hexagon $\left(\frac{3{s}^{2}\sqrt{3}}{2}\right)$ with side length $s$ , plugging $\frac{a}{3}$ and $\frac{b}{6}$ into each equation, we find that $\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}$ . Simplifying this, we get $\frac{a}{b}=\boxed{\textbf{(B)} \frac{\sqrt{6}}{2}}$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.