반원 속 내접원 반지름 극한

점 $\mathrm{O} _{1}$에서 선분 $\mathrm{AB}$에 내린 수선의 발을 $\mathrm{F}$, 직선 $\mathrm{O} _{1} O _{2}$와 현 $\mathrm{BC}$, 호 $\mathrm{BC}$의 교점을 각각 $\mathrm{D}$, $\mathrm{E}$라 하자.

$\displaystyle \mathrm{\overline{O _{1} F}}$$\displaystyle = \mathrm{\overline{O _{1} E}}$$= f ( \theta )$, $\mathrm{\angle} OO _{1} F = \theta$ $\displaystyle \mathrm{\overline{OE}} = \overline{OO _{1}} + \overline{O _{1} E}$$\displaystyle = \frac{f ( \theta )}{\cos \theta} + f ( \theta )$$= 1$ $\displaystyle f \left( \theta \right) = \frac{\cos \theta}{1 + \cos \theta}$ $\displaystyle \mathrm{\overline{OD}} = \sin \theta$, $\displaystyle \mathrm{\overline{ED}} = 1 - \sin \theta$이므로 $\displaystyle g ( \theta ) = \frac{1 - \sin \theta}{2}$ $\displaystyle \lim\limits _{\theta \rightarrow \frac{\pi}{2} - 0} {\frac{g ( \theta )}{\left\{ f ( \theta ) \right\} ^{2}}} = \lim\limits _{\theta \rightarrow \frac{\pi}{2} - 0} {\frac{\frac{1 - \sin \theta}{2}}{\left( \frac{\cos \theta}{1 + \cos \theta} \right) ^{2}}}$ $\displaystyle = \lim\limits _{\theta \rightarrow \frac{\pi}{2} - 0} {\frac{( 1 - \sin \theta ) ( 1 + \cos \theta ) ^{2}}{2 \cos ^{2} \theta}}$ $\displaystyle \frac{\pi}{2} - \theta = t$라 하면, $\displaystyle \theta = \frac{\pi}{2} - t$이고 $\displaystyle \sin \theta = \sin \left( \frac{\pi}{2} - t \right) = \cos t$, $\displaystyle \cos \theta = \cos \left( \frac{\pi}{2} - t \right) = \sin t$ $\displaystyle \theta \rightarrow \frac{\pi}{2} - 0$일때 $t \rightarrow + 0$이다. (주어진 식) $\displaystyle = \lim\limits _{t \rightarrow + 0} {} \frac{( 1 - \cos t ) ( 1 + \sin t ) ^{2}}{2 \sin ^{2} t}$ $\displaystyle = \lim\limits _{t \rightarrow + 0} {} \frac{( 1 - \cos t ) ( 1 + \sin t ) ^{2}}{2 \left( 1 - \cos ^{2} t \right)}$ $\displaystyle = \lim\limits _{t \rightarrow + 0} {} \frac{( 1 + \sin t ) ^{2}}{2 ( 1 + \cos t )}$ $\displaystyle = \frac{1}{4} = \frac{q}{p}$ 따라서 $p = 4$, $q = 1$이므로 $p ^{2} + q ^{2} = 17$이다.