삼각형 넓이 비 극한

[출제의도] 삼각형의 성질을 이용하여 삼각함수의 극한에 대한 문제를 해결한다. 삼각형 ${\mathrm{ABC}}$에서 코사인법칙에 의하여 $\displaystyle {\overline{\mathrm{AC}}} ^{2}$$= 1 ^{2} + 2 ^{2} - 2 \times 1 \times 2 \times \cos \theta$$= 5 - 4 \cos \theta$ 이므로 $\displaystyle \overline{\mathrm{AC}} = \sqrt{5 - 4 \cos \theta}$ 직선 ${\mathrm{AE}}$가 $\angle {\mathrm{BAC}}$의 이등분선이므로 $\displaystyle \overline{\mathrm{BE}} : \overline{\mathrm{CE}}$$\displaystyle = \overline{\mathrm{AB}} : \overline{\mathrm{AC}}$$\displaystyle = 1 : \sqrt{5 - 4 \cos \theta}$에서 $\displaystyle \overline{\mathrm{BE}}$$\displaystyle = \frac{1}{1 + \sqrt{5 - 4 \cos \theta}} \times \overline{\mathrm{BC}}$$\displaystyle = \frac{2}{1 + \sqrt{5 - 4 \cos \theta}}$ 그러므로 $f ( \theta )$$\displaystyle = \frac{1}{2} \times \overline{\mathrm{AB}} \times \overline{\mathrm{BE}} \times \sin ( \angle {\mathrm{CBA}} )$ $\displaystyle = \frac{1}{2} \times 1 \times \frac{2}{1 + \sqrt{5 - 4 \cos \theta}} \times \sin \theta$ $\displaystyle = \frac{\sin \theta}{1 + \sqrt{5 - 4 \cos \theta}}$ 두 직선 ${\mathrm{AB}}$, ${\mathrm{DM}}$이 서로 평행하므로 $\angle {\mathrm{CDM}} = \theta$, $\angle {\mathrm{BAE}} = \angle {\mathrm{DFE}}$ 이다. 이때 $\angle {\mathrm{BAE}} = \angle {\mathrm{FAC}}$이므로 삼각형 ${\mathrm{AMF}}$는 이등변삼각형이다. 점 ${\mathrm{M}}$은 선분 ${\mathrm{AC}}$의 중점이므로 $\displaystyle \overline{{\mathrm{FM}}}$ $\displaystyle = \frac{1}{2} \times \overline{\mathrm{AC}}$$\displaystyle = \frac{\sqrt{5 - 4 \cos \theta}}{2}$이고 $\displaystyle \overline{\mathrm{BD}}$$\displaystyle = \overline{\mathrm{CD}}$$= 1$, $\displaystyle \overline{{\mathrm{DM}}} = \frac{1}{2}$ 그러므로 $\displaystyle \overline{{\mathrm{DF}}}$$\displaystyle = \overline{{\mathrm{FM}}} - \overline{\mathrm{DM}}$$\displaystyle = \frac{\sqrt{5 - 4 \cos \theta} - 1}{2}$ $\angle {\mathrm{FDC}}$$= \pi - \angle {\mathrm{CDM}}$$= \pi - \theta$이므로 $g ( \theta )$$\displaystyle = \frac{1}{2} \times \overline{\mathrm{CD}} \times \overline{{\mathrm{DF}}} \times \sin ( \angle {\mathrm{FDC}} )$ $\displaystyle = \frac{1}{2} \times 1 \times \frac{\sqrt{5 - 4 \cos \theta} - 1}{2} \times \sin ( \pi - \theta )$ $\displaystyle = \frac{\sqrt{5 - 4 \cos \theta} - 1}{4} \times \sin \theta$ 따라서 $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {} \frac{g ( \theta )}{\theta ^{2} \times f ( \theta )}$$\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \frac{\frac{\sin \theta}{4} ( \sqrt{5 - 4 \cos \theta} - 1 )}{\theta ^{2} \times \frac{\sin \theta}{1 + \sqrt{5 - 4 \cos \theta}}}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \frac{( \sqrt{5 - 4 \cos \theta} - 1 ) ( \sqrt{5 - 4 \cos \theta} + 1 )}{4 \theta ^{2}}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \frac{( 5 - 4 \cos \theta ) - 1}{4 \theta ^{2}}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \frac{1 - \cos \theta}{\theta ^{2}}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \frac{( 1 - \cos \theta ) ( 1 + \cos \theta )}{\theta ^{2} ( 1 + \cos \theta )}$ $\displaystyle = \frac{1}{2} \times \lim\limits _{\theta \rightarrow 0 +} {} \frac{\sin ^{2} \theta}{\theta ^{2}}$ $\displaystyle = \frac{1}{2}$