원과 두 직선 거리 최솟값

[출제의도] 삼각함수의 합성을 활용하여 문제해결하기 점 $\mathrm{O}$에서 두 직선에 내린 수선의 발을 각각 $\mathrm{P}$, $\mathrm{Q}$라 하자. $\angle {\mathrm{OAP}} = \theta$$\displaystyle \left( 0 < \theta < \frac{\pi}{6} \right)$라 하면 $\displaystyle \angle {\mathrm{OAQ}} = \frac{\pi}{6} - \theta$ $l = 2 \sin \theta$, $\displaystyle m = 2 \sin \left( \frac{\pi}{6} - \theta \right) = \cos \theta - \sqrt{3} \sin \theta$

$\displaystyle 2 l ^{2} + m ^{2} = 8 \sin ^{2} \theta + \left( \cos \theta - \sqrt{3} \sin \theta \right) ^{2}$ $\displaystyle = 8 \sin ^{2} \theta + \cos ^{2} \theta - 2 \sqrt{3} \sin \theta \cos \theta + 3 \sin ^{2} \theta$ $\displaystyle = 1 + 10 \sin ^{2} \theta - \sqrt{3} \sin 2 \theta$ $\displaystyle = 6 - 5 \cos 2 \theta - \sqrt{3} \sin 2 \theta$ $\displaystyle = 6 - 2 \sqrt{7} \sin ( 2 \theta + \alpha )$ $\displaystyle \left( \begin{array}{l} \begin{matrix} \\ \end{matrix} \end{array} \right.$단, $\displaystyle \sin \alpha = \frac{5 \sqrt{7}}{14}$, $\displaystyle \cos \alpha = \frac{\sqrt{21}}{14}$$\displaystyle \begin{matrix} \\ \end{matrix} )$ $\displaystyle \tan \alpha = \frac{5 \sqrt{3}}{3}$ $\therefore$ $\displaystyle \frac{\pi}{3} < \alpha < \frac{\pi}{2}$ $\displaystyle 0 < \theta < \frac{\pi}{6}$이고 $\displaystyle \alpha < 2 \theta + \alpha < \frac{\pi}{3} + \alpha$이므로 $\displaystyle 2 \theta + \alpha = \frac{\pi}{2}$일 때 $\sin \left( 2 \theta + \alpha \right) = 1$ $2 l ^{2} + m ^{2}$의 최솟값은 $\displaystyle 6 - 2 \sqrt{7}$이므로 $p = 6$, $q = - 2$ 따라서 $30 \left( p + q \right) = 120$