Difference between revisions of "2024 SSMO Relay Round 2 Problems/Problem 1"

(Created page with "==Problem== In a regular hexagon <math>ABCDEF</math>, let <math>X</math> be a point inside the hexagon such that <math>XA=XB=3.</math> If the area of the hexagon is <math>6\s...")
 
 
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==Solution==
 
==Solution==
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Let <math>s</math> be the sidelength of the hexagon. We have <math>\frac{6s^2\sqrt{3}}{4} = 6\sqrt{3}\implies s = 2.</math> Let <math>h</math> be the distance from <math>X</math> to <math>AB.</math> Thus, we have <math>\sqrt{h^2+\left(\frac{s}{2}\right)^2} = 3\implies h^2 = 8\implies h = 2\sqrt{2}.</math> Since the distance from <math>AB</math> to <math>DE</math> is <math>2\sqrt{3},</math> <cmath>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}.</cmath>
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~SMO_Team

Latest revision as of 14:43, 10 September 2025

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