부채꼴 속 삼각형 넓이 극한

[출제의도] 도형에 활용된 삼각함수의 극한값을 구할 수 있는가? 직각삼각형 $\mathrm{AHP}$에서 $\angle {\mathrm{APH}} = \theta$이므로 $\displaystyle \angle {\mathrm{HAP}} = \frac{\pi}{2} - \theta$ 한편, 삼각형 $\mathrm{OPA}$는 $\displaystyle \overline{\mathrm{OP}} = \overline{\mathrm{OA}} = 1$인 이등변삼각형이므로 $\angle {\mathrm{AOP}}$$= \pi - 2 \times \angle {\mathrm{HAP}}$$\displaystyle = \pi - 2 \times \left( \frac{\pi}{2} - \theta \right)$$= 2 \theta$ 그러므로 $\displaystyle \overline{\mathrm{AH}}$$\displaystyle = 1 - \overline{\mathrm{OH}}$$\displaystyle = 1 - \overline{\mathrm{OP}} \cos 2 \theta$$= 1 - \cos 2 \theta$ $\cdots \cdots$ ㉠ 또, $\angle {\mathrm{HAQ}}$$\displaystyle = \frac{\pi}{4} - \frac{\theta}{2}$이므로 $f ( \theta )$$\displaystyle = \frac{1}{2} {\overline{\mathrm{AH}}} ^{2} \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right)$ $\displaystyle = \frac{1}{2} \times \left( 1 - \cos 2 \theta \right) ^{2} \times \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right)$ $\displaystyle = \frac{1}{2} \times \frac{\sin ^{4} 2 \theta}{\left( 1 + \cos 2 \theta \right) ^{2}} \times \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right)$ 그러므로 $\displaystyle \lim\limits _{\theta \rightarrow 0 -} {\frac{f \left( \theta \right)}{\theta ^{4}}}$ $\displaystyle = \frac{1}{2} \times 16 \lim\limits _{\theta \rightarrow 0 +} {\left( \frac{\sin 2 \theta}{2 \theta} \right)} ^{4} \times \lim\limits _{\theta \rightarrow 0 +} {\frac{1}{\left( 1 + \cos 2 \theta \right) ^{2}}}$$\displaystyle \times \lim\limits _{\theta \rightarrow 0 +} {\tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right)}$ $\displaystyle = \frac{1}{2} \times 16 \times 1 ^{2} \times \frac{1}{4} \times 1$ $= 2$ $\cdots \cdots$ ㉢ 한편, 이등변삼각형 $\mathrm{OPA}$에서 점 $\mathrm{O}$에서 선분 $\mathrm{PA}$에 내린 수선의 발을 $\mathrm{H}'$이라 하면 ㉠에서 $\angle {\mathrm{H}' OP} = \theta$이므로 $\displaystyle \overline{\mathrm{AP}}$$\displaystyle = 2 \overline{\mathrm{PH}'}$$\displaystyle = 2 \times \overline{\mathrm{OP}} \times \sin \theta$$= 2 \sin \theta$ 삼각형 $\mathrm{AOP}$에서 각의 이등분선이 선분 $\mathrm{OP}$와 만나는 점이 $\mathrm{R}$이므로 $\displaystyle \overline{\mathrm{AO}} : \overline{\mathrm{AP}} = \overline{\mathrm{OR}} : \overline{\mathrm{RP}}$ $\displaystyle 1 : 2 \sin \theta = \overline{\mathrm{OR}} : 1 - \overline{\mathrm{OR}}$ $\displaystyle 2 \sin \theta \times \overline{\mathrm{OR}} = 1 - \overline{\mathrm{OR}}$ $\displaystyle \overline{\mathrm{OR}} = \frac{1}{1 + 2 \sin \theta}$ $\cdots \cdots$ ㉣ 또, $\displaystyle \overline{\mathrm{OS}}$$\displaystyle = \overline{\mathrm{OA}} \tan \left( \angle {\mathrm{SAO}} \right)$$\displaystyle = 1 \times \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right)$$\displaystyle = \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right)$ $\cdots \cdots$ ㉤ ㉣과 ㉤에서 $g \left( \theta \right)$$= \triangle {\mathrm{OSP}} - \triangle {\mathrm{OSR}}$ $\displaystyle = \frac{1}{2} \times \overline{\mathrm{OS}} \times \overline{\mathrm{OP}} \sin \left( \angle {\mathrm{POS}} \right)$ $\displaystyle = \frac{1}{2} \times \overline{\mathrm{OS}} \times \overline{\mathrm{OR}} \times \sin \left( \angle {\mathrm{POS}} \right)$ $\displaystyle = \frac{1}{2} \times \overline{\mathrm{OS}} \times \sin \left( \angle {\mathrm{POS}} \right) \times \left( \overline{\mathrm{OP}} - \overline{\mathrm{OR}} \right)$ $\displaystyle = \frac{1}{2} \times \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \times \sin \left( \frac{\pi}{2} - 2 \theta \right)$$\displaystyle \times \left( 1 - \frac{1}{2 \sin \theta + 1} \right)$ $\displaystyle = \frac{1}{2} \times \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \times \sin \left( \frac{\pi}{2} - 2 \theta \right) \times \frac{2 \sin \theta}{2 \sin \theta + 1}$ 그러므로 $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {\frac{g \left( \theta \right)}{\theta}}$$\displaystyle = \frac{1}{2} \times \lim\limits _{\theta \rightarrow 0 +} {\tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \times \lim\limits _{\theta \rightarrow 0 +} {\sin \left( \frac{\pi}{2} - 2 \theta \right)}}$ $\displaystyle \times 2 \lim\limits _{\theta \rightarrow 0 +} {\frac{\sin \theta}{\theta}} \times \lim\limits _{\theta \rightarrow 0 +} {\frac{1}{2 \sin \theta + 1}}$ $\displaystyle = \frac{1}{2} \times 1 \times 1 \times 2 \times 1 \times 1$$= 1$ $\cdots \cdots$ ㉥ 따라서, ㉢과 ㉥을 이용하면 $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {\frac{\theta ^{3} \times g \left( \theta \right)}{f \left( \theta \right)}}$$\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {\frac{\frac{g \left( \theta \right)}{\theta}}{\frac{f \left( \theta \right)}{\theta ^{4}}}}$$\displaystyle = \frac{1}{2}$ 이므로 $\displaystyle 100 k = 100 \times \frac{1}{2} = 50$