삼각형 선분 길이의 극한

$AH = a$ , $BH = b$ 라 두면 $a \tan \theta = CH , b \tan 2 \theta = CH$ $\displaystyle a = \frac{CH}{\tan \theta} , b = \frac{CH}{\tan 2 \theta}$ $\displaystyle a + b = \frac{CH}{\tan \theta} + \frac{CH}{\tan 2 \theta} = 1$ $\displaystyle \therefore CH = \frac{\tan \theta \tan 2 \theta}{\tan \theta + \tan 2 \theta}$ 또, $\angle ACB = \pi - 3 \theta$ 이므로 $\displaystyle \angle BCD = \frac{\pi}{3} - \theta , \angle CDA = \frac{\pi}{3} + \theta$ $\displaystyle \sin \left( \frac{\pi}{3} + \theta \right) = \frac{CH}{CD}$ $\displaystyle \therefore CD = \frac{\tan \theta \tan 2 \theta}{\sin ( \frac{\pi}{3} + \theta ) \times ( \tan \theta + \tan 2 \theta )}$ $\displaystyle \therefore \lim\limits _{\theta \rightarrow 0 +} {\frac{\mathrm{\overline{CD}}}{\theta}}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {\frac{\frac{\tan \theta}{\theta} \times \frac{\tan 2 \theta}{\theta}}{\sin \left( \frac{\pi}{3} + \theta \right) \times \left( \frac{\tan \theta}{\theta} + \frac{\tan 2 \theta}{\theta} \right)}}$ $\displaystyle = \frac{1 \times 2}{\frac{\sqrt{3}}{2} \times \left( 1 + 2 \right)}$ $\displaystyle = \frac{4}{3 \sqrt{3}} = a$ $\displaystyle \therefore 27 a ^{2} = 27 \times \frac{16}{27} = 16$