내접원 넓이 극한

삼각형 $\mathrm{PAO}$에 내접하는 원의 중심을 $\mathrm{Q}$, 반지름의 길이를 $r$라 하자.

$\triangle \mathrm{AOP}$$= \triangle \mathrm{AOQ} + \triangle OPQ + \triangle PAQ$이므로 $\displaystyle \frac{1}{2} \cdot \overline{\mathrm{AO}} \cdot \mathrm{\overline{OP}} \cdot \sin \left( \pi - \theta \right)$$\displaystyle = \frac{r}{2} \cdot \overline{\mathrm{AO}} + \frac{r}{2} \cdot \overline{\mathrm{OP}} + \frac{r}{2} \cdot \overline{\mathrm{PA}}$ $\therefore$ $\displaystyle \sin \theta = r \left( 2 + \overline{\mathrm{PA}} \right)$ … ㉠ 직각삼각형 $\mathrm{A} BP$에서 $\displaystyle \overline{\mathrm{PA}} = \overline{\mathrm{AB}} \cos \frac{\theta}{2} = 2 \cos \frac{\theta}{2}$ … ㉡이므로 ㉠, ㉡에 의하여 $\displaystyle r = \frac{\sin \theta}{2 + 2 \cos \frac{\theta}{2}}$이다. 따라서 $\displaystyle f ( \theta ) = \pi r ^{2} = \frac{\pi \sin ^{2} \theta}{\left( 2 + 2 \cos \frac{\theta}{2} \right. ) ^{2}}$이므로 $\displaystyle \lim\limits _{\theta \rightarrow + 0} {\frac{f ( \theta )}{\theta ^{2}}} = \lim\limits _{\theta \rightarrow + 0} {\frac{\sin ^{2} \theta}{\theta ^{2}}} \cdot \frac{\pi}{\left( 2 + 2 \cos \frac{\theta}{2} \right) ^{2}}$ $\displaystyle = \lim\limits _{\theta \rightarrow + 0} {\frac{\sin ^{2} \theta}{\theta ^{2}}} \cdot \lim\limits _{\theta \rightarrow + 0} {\frac{\pi}{\left( 2 + 2 \cos \frac{\theta}{2} \right. ) ^{2}}}$ $\displaystyle = 1 ^{2} \cdot \frac{\pi}{16}$$\displaystyle = \frac{\pi}{16}$ [다른풀이] 삼각형 $\mathrm{PAO}$에 내접하는 원의 중심을 $\mathrm{Q}$라 하고 원 $\mathrm{Q}$와 변 $\mathrm{AP}$의 접점을 $\mathrm{R}$라 하면 $\displaystyle \overline{\mathrm{OR}} \perp \overline{\mathrm{AP}}$, $\displaystyle \overline{\mathrm{QR}} \perp \overline{\mathrm{AP}}$이다. 삼각형 $\mathrm{AOR}$에서 $\displaystyle \angle \mathrm{RAO} = \frac{\theta}{2}$이므로 $\displaystyle \overline{\mathrm{AR}} = \overline{\mathrm{OA}} \cos \frac{\theta}{2} = \cos \frac{\theta}{2}$ 삼각형 $\mathrm{AQR}$에서 $\displaystyle \angle \mathrm{RAQ} = \frac{\theta}{4}$이므로 $\displaystyle \overline{\mathrm{QR}} = \overline{\mathrm{AR}} \tan \frac{\theta}{4} = \cos \frac{\theta}{2} \tan \frac{\theta}{4}$ 따라서 $\displaystyle f ( \theta ) = \pi \overline{\mathrm{QR}} ^{2} = \pi \cos ^{2} \frac{\pi}{2} \tan ^{2} \frac{\theta}{4}$이다. $\therefore$ $\displaystyle \lim\limits _{\theta \rightarrow + 0} {\frac{f ( \theta )}{\theta ^{2}}} = \lim\limits _{\theta \rightarrow + 0} {\frac{\pi \cos ^{2} \frac{\theta}{2} \tan ^{2} \frac{\theta}{4}}{\theta ^{2}}}$ $\displaystyle = \lim\limits _{\theta \rightarrow + 0} {\frac{\pi \cos ^{2} \frac{\theta}{2} \tan ^{2} \frac{\theta}{4}}{\left( \frac{\theta}{4} \right) ^{2} \cdot 16}}$$\displaystyle = \frac{\pi}{16} \lim\limits _{\theta \rightarrow + 0} {} \frac{\tan ^{2} \frac{\theta}{4}}{\left( \frac{\theta}{4} \right) ^{2}} \lim\limits _{\theta \rightarrow + 0} {\cos ^{2} \frac{\theta}{2}}$ $\displaystyle = \frac{\pi}{16} \cdot 1 ^{2} \cdot 1 ^{2} = \frac{\pi}{16}$