Competition problem — Geometry — Maths Olympiad Prep

Library / /340 of 348

Geometry Difficulty 5.3 AIME, harder Find the answer

Let $A_{1} A_{2} \ldots A_{6}$ be a regular hexagon with side length $11 \sqrt{3}$ , and let $B_{1} B_{2} \ldots B_{6}$ be another regular hexagon completely inside $A_{1} A_{2} \ldots A_{6}$ such that for all $i \in\{1,2, \ldots, 5\}, A_{i} A_{i+1}$ is parallel to $B_{i} B_{i+1}$ . Suppose that the distance between lines $A_{1} A_{2}$ and $B_{1} B_{2}$ is 7 , the distance between lines $A_{2} A_{3}$ and $B_{2} B_{3}$ is 3 , and the distance between lines $A_{3} A_{4}$ and $B_{3} B_{4}$ is 8 . Compute the side length of $B_{1} B_{2} \ldots B_{6}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let $X=A_{1} A_{2} \cap A_{3} A_{4}$ , and let $O$ be the center of $B_{1} B_{2} \ldots B_{6}$ . Let $p$ be the apothem of hexagon $B$ . Since $O A_{2} X A_{3}$ is a convex quadrilateral, we have $\begin{aligned} {\left[A_{2} A_{3} X\right] } & =\left[A_{2} X O\right]+\left[A_{3} X O\right]-\left[A_{2} A_{3} O\right] \\ & =\frac{11 \sqrt{3}(7+p)}{2}+\frac{11 \sqrt{3}(8+p)}{2}-\frac{11 \sqrt{3}(3+p)}{2} \\ & =\frac{11 \sqrt{3}(12+p)}{2} \end{aligned}$ Since $\left[A_{2} A_{3} X\right]=(11 \sqrt{3})^{2} \frac{\sqrt{3}}{4}$ , we get that $\frac{12+p}{2}=(11 \sqrt{3}) \frac{\sqrt{3}}{4}=\frac{33}{4} \Longrightarrow p=\frac{9}{2}$ Thus, the side length of hexagon $B$ is $p \cdot \frac{2}{\sqrt{3}}=3 \sqrt{3}$ .

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