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, 2023

Geometry Difficulty 5.7 AIME, harder Prove it United States

Problem:
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 $B_{1} B_{2} \ldots B_{6}$ .

Solution

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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