자연계열 1번

(1)

선분 $\displaystyle \mathrm{B}{}_{0}\mathrm{C}_{0}$와 $\displaystyle \mathrm{B}{}_{1}\mathrm{C}_{1}$가 서로 평행이고, 선분 $\displaystyle \mathrm{OA}$와 수직이므로 삼각형 $\displaystyle \mathrm{OB}{}_{0}\mathrm{C}_{0}$와 $\displaystyle \mathrm{OB}{}_{1}\mathrm{C}_{1}$도 서로 닮은 이등변삼각형이다.따라서 $\displaystyle \overline{\mathrm{B}{}_{0}\mathrm{C}_{0}} : \overline{\mathrm{B}{}_{1}\mathrm{C}_{1}} = {\overline{\mathrm{OA}}_{1} : \overline{\mathrm{OA}}_{2}}$.$\displaystyle \angle \mathrm{BOC} = 60 ^{\circ}$이므로 삼각형 $\displaystyle \mathrm{OB}{}_{0}\mathrm{C}_{0}$와 $\displaystyle \mathrm{OB}{}_{1}\mathrm{C}_{1}$ 모두 정삼각형이다.따라서 삼각형 $\displaystyle \mathrm{A}_{1} \mathrm{B}{}_{1}\mathrm{C}_{1}$과 $\displaystyle \mathrm{OB}{}_{1}\mathrm{C}_{1}$은 합동인 정삼각형이다.따라서 $\displaystyle \overline{\mathrm{OA}}_{1} = 2 \overline{\mathrm{OA}}_{2}$.그러므로 $\displaystyle \overline{\mathrm{B}{}_{1}\mathrm{C}_{1}} = \frac{1}{2} \overline{\mathrm{B}{}_{0}\mathrm{C}_{0}}$.같은 방법을 이용하면 $\displaystyle \overline{\mathrm{B}{}_{k + 1}\mathrm{C}_{k + 1}} = \frac{1}{2} \overline{\mathrm{B}{}_{k}\mathrm{C}_{k}}$이다.$\displaystyle \overline{\mathrm{B}{}_{0}\mathrm{C}_{0}} = a$이므로 $\displaystyle \overline{\mathrm{B}{}_{k}\mathrm{C}_{k}} = a \left( \frac{1}{2} \right)^{k} = \frac{a}{2^{k}}$.한 변의 길이가 $\displaystyle x$인 정삼각형의 넓이는 $\displaystyle \frac{{\sqrt{3} x^{2}}}{4}$이고, 둘레의 길이는 $\displaystyle 3 x$이므로,$\displaystyle S_4=\frac{\sqrt3}4\left(\frac a{2^4}\right)^2=\frac{\sqrt3a^2}{1024}$이고, $\displaystyle L_4=3\times\frac a{2^4}=\frac{3a}{16}$.

(2)

삼각형 $\displaystyle \mathrm{A}_{0} \mathrm{B}{}_{0}\mathrm{C}_{0}$와 $\displaystyle \mathrm{A}_{1} \mathrm{B}{}_{1}\mathrm{C}_{1}$가 서로 닮은 삼각형이므로 $\displaystyle \overline{\mathrm{B}{}_{0}\mathrm{C}_{0}} : \overline{\mathrm{B}{}_{1}\mathrm{C}_{1}} = \overline{\mathrm{A}_{0} \mathrm{A}_{1}} : \overline{\mathrm{A}_{1} \mathrm{A}_{2}}$.$\displaystyle \angle \mathrm{BOA} = 45 ^{\circ}$이므로 $\displaystyle \overline{\mathrm{B}_{0} \mathrm{A}_{1}} = \overline{\mathrm{OA}}_{1} {}$, $\displaystyle \overline{\mathrm{B}_{1} \mathrm{A}_{2}} = \overline{\mathrm{OA}}_{2}$.따라서 $\displaystyle \overline{\mathrm{B}_{1} \mathrm{A}_{2}} = x$라고 하면 $\displaystyle \overline{\mathrm{A}_{1} \mathrm{A}_{2}} = 3 \frac{x}{2}$이므로$\displaystyle \overline{\mathrm{OA}}_{2} = x$, $\displaystyle \overline{\mathrm{OA}}_{1} = \overline{\mathrm{OA}}_{2} + \overline{\mathrm{A}_{1} \mathrm{A}_{2}}$$\displaystyle = x + 3 \frac{x}{2} = 5 \frac{x}{2}$.(1)에서와 마찬가지로 삼각형 $\displaystyle \mathrm{OB}{}_{0}\mathrm{C}_{0}$와 $\displaystyle \mathrm{OB}{}_{1}\mathrm{C}_{1}$가 서로 닮은 이등변삼각형이다.$\displaystyle \overline{\mathrm{B}{}_{0}\mathrm{C}_{0}} : \overline{\mathrm{B}{}_{1}\mathrm{C}_{1}} = \overline{\mathrm{OA}_{1}} : \overline{\mathrm{OA}_{2}}$$\displaystyle = 5 \frac{x}{2} : x = 5 : 2$이므로 $\displaystyle \overline{\mathrm{B}{}_{1}\mathrm{C}_{1}} = \frac{2}{5} \overline{\mathrm{B}{}_{0}\mathrm{C}_{0}}$.같은 방법을 이용하면 $\displaystyle \overline{\mathrm{B}{}_{k + 1}\mathrm{C}_{k + 1}} = \frac{2}{5} \overline{\mathrm{B}{}_{k}\mathrm{C}_{k}}$이다.$\displaystyle \overline{\mathrm{B}{}_{0}\mathrm{C}_{0}} = 4 b$이므로 $\displaystyle \overline{\mathrm{B}{}_{k}\mathrm{C}_{k}}$$\displaystyle = 4 b \left( \frac{2}{5} \right)^{k}$.따라서 $\displaystyle S_{k} = \frac{1}{2} \times 4 b \left( \frac{2}{5} \right)^{k} \times 3 b \left( \frac{2}{5} \right)^{k} = 6 b^{2} \left( \frac{4}{25} \right)^{k}$이므로 $\displaystyle \sum_{k = 0}^{\infty} S_{k} = 6 \frac{b^{2}}{1 - \frac{4}{25}} = 50 \frac{b^{2}}{7}$.

(3)

$\displaystyle \overline{B_0A_1}=2$이고 $\displaystyle \angle BOA=\theta$이므로 $\displaystyle \overline{OA_1}=\frac2{\tan\theta}$.$\displaystyle \overline{B_1A_2}=x$라고 하면 $\displaystyle \overline{OA_2}=\frac x{\tan\theta}$이고 $\displaystyle \overline{A_1A_2}=\frac{3x}2$.따라서 $\displaystyle \frac2{\tan\theta}=\frac x{\tan\theta}+\frac{3x}2$이므로 $\displaystyle x=\frac4{2+3\tan\theta}$.(1), (2)에서와 마찬가지로 삼각형 $\displaystyle \mathrm{OB}{}_{0}\mathrm{C}_{0}$와 $\displaystyle \mathrm{OB}{}_{1}\mathrm{C}_{1}$가 서로 닮은 이등변삼각형이다.$\displaystyle \overline{\mathrm{B}{}_{0}\mathrm{C}_{0}} : \overline{\mathrm{B}{}_{1}\mathrm{C}_{1}} = \overline{\mathrm{B}_{0} \mathrm{A}_{1}} : \overline{\mathrm{B}_{1} \mathrm{A}_{2}}$$\displaystyle = 2 : x = 2 : \frac{4}{2 + 3 \tan \theta}$이므로 $\displaystyle \overline{\mathrm{B}{}_{1}\mathrm{C}_{1}} = \frac{2}{2 + 3 \tan \theta} \overline{\mathrm{B}{}_{0}\mathrm{C}_{0}}$.같은 방법을 이용하면 $\displaystyle \overline{\mathrm{B}{}_{k + 1}\mathrm{C}_{k + 1}} = \frac{2}{2 + 3 \tan \theta} \overline{\mathrm{B}{}_{k}\mathrm{C}_{k}}$이고 $\displaystyle \overline{\mathrm{B}{}_{0}\mathrm{C}_{0}} = 4$이므로$\displaystyle \overline{\mathrm{B}{}_{k}\mathrm{C}_{k}} = 4 \left( \frac{2}{2 + 3 \tan \theta} \right)^{k}$.따라서 $\displaystyle L_k=\left(1+\frac{\sqrt{13}}4+\frac{\sqrt{13}}4\right)\overline{B_kC_k}=(4+2\sqrt{13})\left(\frac2{2+3\tan\theta}\right)^k$이므로$\displaystyle \sum_{k = 0}^{\infty} L_{k} = \frac{{4 + 2 \sqrt{13}}}{1 - \frac{2}{2 + 3 \tan \theta}} = \frac{{( 4 + 2 \sqrt{13} ) ( 2 + 3 \tan \theta )}}{3 \tan \theta}$.

(4)

$\displaystyle f ( \theta ) = \frac{{6 \sin \theta}}{2 + \sqrt{13}} \frac{{( 4 + 2 \sqrt{13} ) ( 2 + 3 \tan \theta )}}{3 \tan \theta} = ( 8 + 12 \tan \theta ) \cos \theta$$\displaystyle = 8 \cos \theta + 12 \sin \theta$$\displaystyle f^{\prime} ( \theta ) = - 8 \sin \theta + 12 \cos \theta = 4 \cos \theta ( 3 - 2 \tan \theta )$.$\displaystyle 30 ^{\circ} \le \theta \le 60 ^{\circ}$일 때, $\displaystyle \frac{1}{\sqrt{3}} \le \tan \theta \le \sqrt{3}$이므로, 이 구간에 $\displaystyle \tan \theta = \frac{3}{2}$인 $\displaystyle \theta$가 존재한다.$\displaystyle \frac{1}{\sqrt{3}} \le \tan \theta < \frac{3}{2}$일 때 $\displaystyle f^{\prime} ( \theta ) > 0$이고, $\displaystyle \frac{3}{2} < \tan \theta \le \sqrt{3}$일 때 $\displaystyle f^{\prime} ( \theta ) < 0$이므로,$\displaystyle \tan \theta = \frac{3}{2}$인 $\displaystyle \theta$에서 $\displaystyle f ( \theta )$는 최대가 된다.$\displaystyle \tan \theta = \frac{3}{2}$이고 $\displaystyle 30 ^{\circ} \le \theta \le 60 ^{\circ}$이므로, $\displaystyle \sin \theta = \frac{3}{\sqrt{13}}$, $\displaystyle \cos \theta = \frac{2}{\sqrt{13}}$.따라서 $\displaystyle f ( \theta )$는 $\displaystyle \tan \theta = \frac{3}{2}$일 때 최댓값 $\displaystyle \frac{52}{\sqrt{13}} = 4 \sqrt{13}$을 가진다.$\displaystyle f ( \theta )$는 $\displaystyle \theta = 30 ^{\circ}$일 때 $\displaystyle 4 \left( \sqrt{3} + \frac{3}{2} \right)$이고, $\displaystyle \theta = 60 ^{\circ}$일 때 $\displaystyle 4 \left( 1 + \frac{{3 \sqrt{3}}}{2} \right)$인데,$\displaystyle \sqrt{3} + \frac{3}{2} < \frac{{3 \sqrt{3}}}{2} + 1$이므로,$\displaystyle f ( \theta )$는 $\displaystyle \theta = 30 ^{\circ}$일 때 최솟값 $\displaystyle 4 \left( \sqrt{3} + \frac{3}{2} \right) = 4 \sqrt{3} + 6$을 가진다.