Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Find the answer

The area of the largest regular hexagon that can fit inside of a rectangle with side lengths 20 and 22 can be expressed as $a \sqrt{b}-c$ , for positive integers $a, b$ , and $c$ , where $b$ is squarefree. Compute $100 a+10 b+c$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let $s$ be the sidelength of the hexagon. We can view this problem as finding the maximal rectangle of with sides $s$ and $s \sqrt{3}$ that can fit inside this rectangle. Let $A B C D$ be a rectangle with $A B=20$ and $B C=22$ and let $X Y Z W$ be an inscribed rectangle with $X$ on $A B$ and $Y$ on $B C$ with $X Y=s$ and $Y Z=s \sqrt{3}$ . Let $B X=a$ and $B Y=b$ . Then, by similar triangles, we have $A X=b \sqrt{3}$ and $C Y=a \sqrt{3}$ . Thus, we have $a+b \sqrt{3}=20$ and $a \sqrt{3}+b=22$ . Solving gives us $a=11 \sqrt{3}-10$ and $b=10 \sqrt{3}-11$ , so $s^{2}=a^{2}+b^{2}=884-440 \sqrt{3}$ . Thus, the area of the hexagon is $\frac{s^{2} \cdot 3 \sqrt{3}}{2}=1326 \sqrt{3}-1980$ .

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