원과 정삼각형 내적

원의 중심을 O라 하면 $\mathrm{\vec{AP}} = \vec{AO} + \vec{OP}$이므로 $\mathrm{\vec{AP}} \cdot \vec{AQ} = ( \vec{AO} + \vec{OP} ) \cdot \vec{AQ}$ $= \mathrm{\vec{AO}} \cdot \vec{AQ} + \vec{OP} \cdot \vec{AQ}$ 여기서 $\mathrm{\vec{AO}} \cdot \vec{AC} \leq \vec{AO} \cdot \vec{AQ} \leq \vec{AO} \cdot \vec{AB}$

그림에서 $\displaystyle \mathrm{\overline{BM}} = \frac{1}{\sqrt{3}}$, $\displaystyle \mathrm{\overline{OD}} = \frac{\sqrt{3} + 1}{\sqrt{3}}$, $\displaystyle \mathrm{\overline{AD}} = \sqrt{3} - 1$이므로 $\displaystyle \mathrm{\vec{AO}} = \left( - \frac{\sqrt{3} + 1}{\sqrt{3}} , 1 - \sqrt{3} \right)$ $\displaystyle \mathrm{\vec{AB}} = \left( - 1 , - \sqrt{3} \right)$, $\displaystyle \mathrm{\vec{AC}} = \left( 1 , - \sqrt{3} \right)$ $\displaystyle \mathrm{\vec{AO}} \cdot \vec{AC} = 2 - \frac{4}{3} \sqrt{3}$, $\displaystyle \mathrm{\vec{AO}} \cdot \vec{AB} = 4 - \frac{2}{3} \sqrt{3}$ 또, $\mathrm{\vec{OP}} \cdot \vec{AQ} = \left| \vec{AQ} \right| \cos \theta$ 이므로 $- 2 \leq \mathrm{\vec{OP}} \cdot \vec{AQ} \leq 2$ 따라서 최댓값은 $\displaystyle \left( 4 - \frac{2}{3} \sqrt{3} \right) + 2 = 6 - \frac{2}{3} \sqrt{3}$ 최솟값은 $\displaystyle \left( 2 - \frac{4}{3} \sqrt{3} \right) - 2 = - \frac{4}{3} \sqrt{3}$ 이므로 최댓값과 최솟값의 합은 $\displaystyle a + b \sqrt{3} =$$\displaystyle 6 - 2 \sqrt{3}$ $\therefore$ $a ^{2} + b ^{2} = 36 + 4 = 40$