자연계열 2번

[문제 2-1]$\displaystyle r_1=1$, $\displaystyle r_2=2$, $\displaystyle r_3=4$이고 세 원 $\displaystyle C_1$, $\displaystyle C_2$, $\displaystyle C_3$가 제시문 〈그림 1〉과 같이 서로 외접하므로$\displaystyle \begin{aligned}\overline{\mathrm{O}_1\mathrm{O}_2}&=r_1+r_2=1+2=3,\\\overline{\mathrm{O}_2\mathrm{O}_3}&=r_2+r_3=2+4=6,\\\overline{\mathrm{O}_1\mathrm{O}_3}&=r_1+r_3=1+4=5.\end{aligned}\qquad(1)$

삼각형 $\displaystyle \mathrm{O}_1\mathrm{O}_2\mathrm{O}_3$에 코사인법칙을 적용하면$\displaystyle \cos(\angle\mathrm{O}_1\mathrm{O}_2\mathrm{O}_3)=\frac{\overline{\mathrm{O}_1\mathrm{O}_2}^2+\overline{\mathrm{O}_2\mathrm{O}_3}^2-\overline{\mathrm{O}_1\mathrm{O}_3}^2}{2\cdot\overline{\mathrm{O}_1\mathrm{O}_2}\cdot\overline{\mathrm{O}_2\mathrm{O}_3}}=\frac{3^2+6^2-5^2}{2\cdot3\cdot6}=\frac{20}{2\cdot3\cdot6}=\frac59.\qquad(2)$

[문제 2-2]조건 (나)에 의하여 수열 $\displaystyle \{r_n\}$이 첫째항 1, 공비 2인 등비수열이므로$\displaystyle r_n=1\cdot2^{n-1}=2^{n-1}$이다. 따라서 원 $\displaystyle C_n$의 넓이 $\displaystyle S_n$은$\displaystyle S_n=\pi(2^{n-1})^2=2^{2n-2}\pi\qquad(3)$

이고, 로그의 정의에 의하여$\displaystyle \log_2\frac{S_n}{\pi}=\log_22^{2n-2}=2n-2.\qquad(4)$

(5)에 의하여Correction. 이곳의 “(5)”는 앞에서 구한 로그값인 “(4)”의 오기이다. 그러나 원칙에 따라 원문 표기를 그대로 실었다.$\displaystyle \begin{aligned}\sum_{n=1}^N\log_2\frac{S_n}{\pi}&=\sum_{n=1}^N(2n-2)=2\sum_{n=1}^Nn-\sum_{n=1}^N2=2\cdot\frac12N(N+1)-2N\\&=N(N-1).\end{aligned}\qquad(5)$

$\displaystyle N(N-1)$이 자연수 $\displaystyle N$에 대한 증가함수이고 $\displaystyle 10(10-1)=90<100$, $\displaystyle 11(11-1)=110>100$이므로, 구하는 $\displaystyle N$의 최솟값은 11이다.

제시문의 조건 (가), (나)에 의하여$\displaystyle \begin{aligned}\overline{\mathrm{O}_n\mathrm{O}_{n+1}}&=r_n+r_{n+1}=2^{n-1}+2^n,\\\overline{\mathrm{O}_n\mathrm{O}_{n+2}}&=r_n+r_{n+2}=2^{n-1}+2^{n+1},\\\overline{\mathrm{O}_{n+1}\mathrm{O}_{n+2}}&=r_{n+1}+r_{n+2}=2^n+2^{n+1},\\\overline{\mathrm{O}_{n+1}\mathrm{O}_{n+3}}&=r_{n+1}+r_{n+3}=2^n+2^{n+2},\\\overline{\mathrm{O}_{n+2}\mathrm{O}_{n+3}}&=r_{n+2}+r_{n+3}=2^{n+1}+2^{n+2}\end{aligned}\qquad(6)$

이므로, 코사인법칙을 이용하면$\displaystyle \begin{aligned}\cos(\angle\mathrm{O}_n\mathrm{O}_{n+1}\mathrm{O}_{n+2})&=\frac{\overline{\mathrm{O}_n\mathrm{O}_{n+1}}^2+\overline{\mathrm{O}_{n+1}\mathrm{O}_{n+2}}^2-\overline{\mathrm{O}_n\mathrm{O}_{n+2}}^2}{2\cdot\overline{\mathrm{O}_n\mathrm{O}_{n+1}}\cdot\overline{\mathrm{O}_{n+1}\mathrm{O}_{n+2}}}\\&=\frac{(2^{n-1}+2^n)^2+(2^n+2^{n+1})^2-(2^{n-1}+2^{n+1})^2}{2(2^{n-1}+2^n)(2^n+2^{n+1})}\\&=\frac{(3^2+6^2-5^2)(2^{n-1})^2}{2\cdot3\cdot6\cdot(2^{n-1})^2}=\frac{3^2+6^2-5^2}{2\cdot3\cdot6}\quad\left(=\frac59\right)\\\cos(\angle\mathrm{O}_{n+1}\mathrm{O}_{n+2}\mathrm{O}_{n+3})&=\frac{\overline{\mathrm{O}_{n+1}\mathrm{O}_{n+2}}^2+\overline{\mathrm{O}_{n+2}\mathrm{O}_{n+3}}^2-\overline{\mathrm{O}_{n+1}\mathrm{O}_{n+3}}^2}{2\cdot\overline{\mathrm{O}_{n+1}\mathrm{O}_{n+2}}\cdot\overline{\mathrm{O}_{n+2}\mathrm{O}_{n+3}}}\\&=\frac{(2^n+2^{n+1})^2+(2^{n+1}+2^{n+2})^2-(2^n+2^{n+2})^2}{2(2^n+2^{n+1})(2^{n+1}+2^{n+2})}\\&=\frac{(3^2+6^2-5^2)(2^n)^2}{2\cdot3\cdot6\cdot(2^n)^2}=\frac{3^2+6^2-5^2}{2\cdot3\cdot6}\quad\left(=\frac59\right)\end{aligned}$따라서$\displaystyle \cos(\angle\mathrm{O}_n\mathrm{O}_{n+1}\mathrm{O}_{n+2})=\cos(\angle\mathrm{O}_{n+1}\mathrm{O}_{n+2}\mathrm{O}_{n+3})\qquad(7)$

을 얻는다. 각 $\displaystyle \mathrm{O}_n\mathrm{O}_{n+1}\mathrm{O}_{n+2}$와 각 $\displaystyle \mathrm{O}_{n+1}\mathrm{O}_{n+2}\mathrm{O}_{n+3}$이 삼각형의 내각들이므로$\displaystyle 0<\angle\mathrm{O}_n\mathrm{O}_{n+1}\mathrm{O}_{n+2}<\pi,\ 0<\angle\mathrm{O}_{n+1}\mathrm{O}_{n+2}\mathrm{O}_{n+3}<\pi$이고, 따라서 (7)에 의하여$\displaystyle \angle\mathrm{O}_n\mathrm{O}_{n+1}\mathrm{O}_{n+2}=\angle\mathrm{O}_{n+1}\mathrm{O}_{n+2}\mathrm{O}_{n+3}.\qquad(8)$

〈그림 3〉에서와 같이 두 엇각 $\displaystyle \mathrm{O}_n\mathrm{O}_{n+1}\mathrm{O}_{n+2}$, $\displaystyle \mathrm{O}_{n+1}\mathrm{O}_{n+2}\mathrm{O}_{n+3}$의 크기가 같으므로, 직선 $\displaystyle \mathrm{O}_n\mathrm{O}_{n+1}$과 직선 $\displaystyle \mathrm{O}_{n+2}\mathrm{O}_{n+3}$은 서로 평행하다.