반원 속 정삼각형 넓이 극한

[출제의도] 도형의 성질을 이용하여 삼각함수의 극한값을 구할 수 있는가? 선분 $\mathrm{A} B$의 중점을 $\mathrm{M}$이라 하면 $\angle \mathrm{AMQ}$$= 2 \times \angle \mathrm{ABQ}$$= 2 \times 2 \theta$$= 4 \theta$이므로, (부채꼴 $\mathrm{A} MQ$의 넓이)$\displaystyle = \frac{1}{2} \times 1 ^{2} \times 4 \theta$$= 2 \theta$ (삼각형 $\mathrm{M} BQ$의 넓이)$\displaystyle = \frac{1}{2} \times 1 ^{2} \times \sin ( \pi - 4 \theta )$$\displaystyle = \frac{1}{2} \sin 4 \theta$ 삼각형 $\mathrm{R} AB$에서 $\angle \mathrm{ARB} = \pi - 3 \theta$이므로, 사인법칙에 의하여 $\displaystyle \frac{2}{\sin ( \pi - 3 \theta )} = \frac{\overline{\mathrm{BR}}}{\sin \theta}$ 즉, $\displaystyle \overline{\mathrm{BR}} = \frac{2 \sin \theta}{\sin 3 \theta}$이므로, (삼각형 $\mathrm{R} AB$의 넓이) $\displaystyle = \frac{1}{2} \times \overline{\mathrm{AB}} \times \overline{\mathrm{BR}} \times \sin 2 \theta$ $\displaystyle = \frac{1}{2} \times 2 \times \frac{2 \sin \theta}{\sin 3 \theta} \times \sin 2 \theta$ $\displaystyle = \frac{2 \sin \theta \sin 2 \theta}{\sin 3 \theta}$ 그러므로, $f ( \theta )$$\displaystyle = 2 \theta + \frac{1}{2} \sin 4 \theta - \frac{2 \sin \theta \sin 2 \theta}{\sin 3 \theta}$이므로, $\displaystyle \lim\limits _{\theta \rightarrow 0 +} \frac{f ( \theta )}{\theta}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} \left\{ 2 + 2 \times \frac{\sin 4 \theta}{4 \theta} - \frac{4 \times \frac{\sin \theta}{\theta} \times \frac{\sin 2 \theta}{2 \theta}}{3 \times \frac{\sin 3 \theta}{3 \theta}} \right\}$ $\displaystyle = 2 + 2 - \frac{4}{3}$$\displaystyle = \frac{8}{3}$ $\cdots \cdots$ ㉠

정삼각형 $\mathrm{S} TU$의 한 변의 길이를 $a$라 하면 삼각형 $\mathrm{T} SB$에서 사인법칙에 의하여 $\displaystyle \frac{a}{\sin 2 \theta} = {\frac{\overline{\mathrm{BT}}}{\sin {\frac{\pi}{3}}}}$ 즉, $\displaystyle \overline{\mathrm{BT}}$$\displaystyle = \frac{\sqrt{3} a}{2 \sin 2 \theta}$ 두 삼각형 $\mathrm{R} UT , RAB$가 서로 닮음이므로, $\displaystyle \mathrm{\overline{RT}} : \overline{RB} = \overline{UT} : \overline{AB}$ $\displaystyle \frac{2 \sin \theta}{\sin 3 \theta} - \frac{\sqrt{3} a}{2 \sin 2 \theta} : \frac{2 \sin \theta}{\sin 3 \theta} = a : 2$ $\displaystyle \frac{2 \sin \theta}{\sin 3 \theta} a = \frac{4 \sin \theta}{3 \sin 3 \theta} - \frac{\sqrt{3}}{\sin 2 \theta} a$ $\displaystyle \left( \frac{2 \sin \theta}{\sin 3 \theta} + \frac{\sqrt{3}}{\sin 2 \theta} \right) a = \frac{4 \sin \theta}{\sin 3 \theta}$ $\displaystyle \frac{2 \sin \theta \sin 2 \theta + \sqrt{3} \sin 3 \theta}{\sin 3 \theta \sin 2 \theta} a = \frac{4 \sin \theta}{\sin 3 \theta}$ $a$$\displaystyle = \frac{4 \sin \theta}{\sin 3 \theta} \times \frac{\sin 3 \theta \sin 2 \theta}{2 \sin \theta \sin 2 \theta + \sqrt{3} \sin 3 \theta}$ 이 때, $\displaystyle g ( \theta ) = \frac{\sqrt{3}}{4} a ^{2}$이고, $\displaystyle \lim\limits _{\theta \rightarrow 0 +} \frac{a}{\theta}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} \left( \frac{4 \sin \theta}{\sin 3 \theta} \times \frac{\sin 3 \theta \sin 2 \theta}{2 \sin \theta \sin 2 \theta + \sqrt{3} \sin 3 \theta} \times \frac{1}{\theta} \right)$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} \left( \frac{4 \sin \theta}{\sin 3 \theta} \times \frac{\frac{\sin 3 \theta \sin 2 \theta}{\theta ^{2}}}{\frac{2 \sin \theta \sin 2 \theta + \sqrt{3} \sin 3 \theta}{\theta}} \right)$ $\displaystyle = \frac{4}{3} \times \frac{2 \times 3}{0 + 3 \sqrt{3}}$$\displaystyle = \frac{8}{3 \sqrt{3}}$ 이므로, $\displaystyle \lim\limits _{\theta \rightarrow 0 +} \frac{g ( \theta )}{\theta ^{2}}$$\displaystyle = \frac{\sqrt{3}}{4} \lim\limits _{\theta \rightarrow 0 +} \left( \frac{a}{\theta} \right) ^{2}$ $\displaystyle = \frac{\sqrt{3}}{4} \times \left( \frac{8}{3 \sqrt{3}} \right) ^{2}$ $\displaystyle = \frac{16 \sqrt{3}}{27}$ $\cdots \cdots$ ㉡ 따라서, ㉠, ㉡에서 $\displaystyle \lim\limits _{\theta \rightarrow 0 +} \frac{g ( \theta )}{\theta \times f ( \theta )}$$\displaystyle = \lim\limits _{\theta \rightarrow 0 +} \frac{\frac{g ( \theta )}{\theta ^{2}}}{\frac{f ( \theta )}{\theta}}$ $\displaystyle = \frac{\frac{16 \sqrt{3}}{27}}{\frac{8}{3}}$ $\displaystyle = \frac{2}{9} \sqrt{3}$ 이므로, $p + q$$= 9 + 2$$= 11$