내접원 반지름 극한

[출제의도] 삼각형과 내접원의 성질을 이용하여 삼각함수의 극한 문제를 해결한다. 삼각형 $\mathrm{OAP}$가 이등변삼각형이므로 $\displaystyle \mathrm{\angle} OAP = \angle OPA = \frac{\pi}{2} - \frac{\theta}{2}$이고 삼각형 $\mathrm{APH}$에서 $\displaystyle \mathrm{\angle} APH + \angle PAH = \frac{\pi}{2}$이므로 $\displaystyle \mathrm{\angle} APH = \frac{\theta}{2}$이다. 내접원의 중심을 $\mathrm{Q}$라 하고, 내접원과 선분 $\mathrm{PH}$의 교점을 $\mathrm{T}$라 하면 $\displaystyle \mathrm{\angle} QPT = \frac{\theta}{4}$이다.

$\displaystyle \mathrm{\overline{PH}} = \sin \theta$이므로 삼각형 $\mathrm{QPT}$에서 $\displaystyle \tan \frac{\theta}{4} = \frac{\mathrm{\overline{QT}}}{\overline{PT}} = \frac{\mathit{r ( \theta )}}{\sin \theta - r ( \theta )}$ $\displaystyle \left( 1 + \tan \frac{\theta}{4} \right) r ( \theta ) = \sin \theta \tan \frac{\theta}{4}$ 이므로 $\displaystyle r ( \theta ) = \frac{\sin \theta \tan \frac{\theta}{4}}{1 + \tan \frac{\theta}{4}}$ 따라서 $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {} \frac{r ( \theta )}{\theta ^{2}}$$\displaystyle = \lim\limits _{\theta \rightarrow 0 +} \frac{\sin \theta \tan \frac{\theta}{4}}{\theta ^{2} \left( 1 + \tan \frac{\theta}{4} \right)}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} \left\{ \frac{\sin \theta}{\theta} \times \frac{\tan \frac{\theta}{4}}{\frac{\theta}{4}} \times \frac{1}{4} \times \frac{1}{\left( {1 + \tan \frac{\theta}{4} )} \right.} \right.$ $\displaystyle = 1 \times 1 \times \frac{1}{4} \times 1 = \frac{1}{4}$ [다른풀이] 내접원의 반지름의 길이를 $r ( \theta )$라 하면 $\triangle \mathrm{A} PH$$\displaystyle = \frac{1}{2} \times$($\triangle \mathrm{A} PH$의 둘레의 길이)$\times r ( \theta )$ $\triangle \mathrm{A} PH$$\displaystyle = \frac{1}{2} r ( \theta ) \left( \overline{\mathrm{P}} H + \overline{AH} + \overline{AP} \right)$ $\displaystyle \mathrm{=} \frac{1}{2} \times \overline{AH} \times \overline{PH}$ 따라서 $\displaystyle \frac{\sin \theta + ( 1 - \cos \theta ) + 2 \sin \frac{\theta}{2}}{2} \times r ( \theta )$ $\displaystyle = \frac{1}{2} \sin \theta \times ( 1 - \cos \theta )$ $\displaystyle \frac{1}{2} \left( 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} + 2 \sin ^{2} \frac{\theta}{2} + 2 \sin \frac{\theta}{2} \right) r ( \theta )$ $\displaystyle = \frac{1}{2} \times 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} ( 1 - \cos \theta )$ 따라서 $\displaystyle \left( {\cos \frac{\theta}{2} + \sin \frac{\theta}{2} + 1} \right) r ( \theta ) = {( 1 - \cos \theta ) \cos \frac{\theta}{2}}$ $\displaystyle r ( \theta ) = \frac{( 1 - \cos \theta ) \cos \frac{\theta}{2}}{1 + \cos \frac{\theta}{2} + \sin \frac{\theta}{2}}$ 이므로 $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {} \frac{r ( \theta )}{\theta ^{2}}$$\displaystyle = \lim\limits _{\theta \rightarrow 0 +} \frac{( 1 - \cos \theta ) \cos \frac{\theta}{2}}{\theta ^{2} \left( 1 + \cos \frac{\theta}{2} + \sin \frac{\theta}{2} \right)}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} \left\{ \frac{\cos \frac{\theta}{2}}{1 + \cos \frac{\theta}{2} + \sin \frac{\theta}{2}} \times \frac{\sin ^{2} \theta}{\theta ^{2}} \times \frac{1}{1 + \cos \theta} \right\}$ $\displaystyle = \frac{1}{1 + 1 + 0} \times 1 ^{2} \times \frac{1}{1 + 1}$$\displaystyle = \frac{1}{4}$