2024 SSMO Relay Round 2 Problems/Problem 1

Problem

In a regular hexagon $ABCDEF$ , let $X$ be a point inside the hexagon such that $XA=XB=3.$ If the area of the hexagon is $6\sqrt{3}$ , then $XE^2 = a+b\sqrt{c},$ for squarefree $c$ . Find $a+b+c$ .

Solution

Let $s$ be the sidelength of the hexagon. We have $\frac{6s^2\sqrt{3}}{4} = 6\sqrt{3}\implies s = 2.$ Let $h$ be the distance from $X$ to $AB.$ Thus, we have $\sqrt{h^2+\left(\frac{s}{2}\right)^2} = 3\implies h^2 = 8\implies h = 2\sqrt{2}.$ Since the distance from $AB$ to $DE$ is $2\sqrt{3},$ $XE = \sqrt{(2\sqrt{3}-2\sqrt{2})^2+\left(\frac{s}{2}\right)^2}\implies XE^2 = 21-8\sqrt{6}\implies 21-8+6 = \boxed{19}.$

~SMO_Team

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2024 SSMO Relay Round 2 Problems/Problem 1

Problem

Solution