부채꼴 내접원 극한

[출제의도] 삼각함수의 극한을 활용하여 문제해결하기 $\angle \mathrm{CBD} = \alpha$라 하면 $\displaystyle \cos \alpha = \frac{3}{5} = 1 - 2 \sin ^{2} \frac{\alpha}{2}$이므로 $\displaystyle \sin \frac{\alpha}{2} = \frac{\sqrt{5}}{5}$ $\displaystyle \mathrm{\overline{BP}} = \mathit{a}$라 하면 $\displaystyle \sin \frac{\alpha}{2} = \frac{r \left( \theta \right)}{a - r \left( \theta \right)} = \frac{\sqrt{5}}{5}$이므로 $\displaystyle r \left( \theta \right) = \frac{a}{1 + \sqrt{5}}$ 삼각형 $\mathrm{PAB}$에서 사인법칙에 의해 $\displaystyle \frac{a}{\sin \theta} = \frac{4}{\sin \left( \frac{\pi}{2} - \theta + \alpha \right)}$이므로 $\displaystyle a = \frac{4 \sin \theta}{\cos \left( \theta - \alpha \right)}$ $\therefore$$\displaystyle r \left( \theta \right) = \frac{4 \sin \theta}{\left( 1 + \sqrt{5} \right) \cos \left( \theta - \alpha \right)}$ $\displaystyle \lim\limits _{\theta \rightarrow + 0} {\frac{r \left( \theta \right)}{\theta}}$$\displaystyle = \lim\limits _{\theta \rightarrow + 0} {\frac{4}{1 + \sqrt{5}} \times \frac{\sin \theta}{\theta}} \times \frac{1}{\cos \left( \theta - \alpha \right)}$ $\displaystyle = \frac{5}{3} \left( \sqrt{5} - 1 \right)$ $\displaystyle \left( \because \cos \alpha = \frac{3}{5} \right)$ $\therefore$$\displaystyle p = - \frac{5}{3}$, $\displaystyle q = \frac{5}{3}$ 따라서 $36 \left( p ^{2} + q ^{2} \right) = 200$