중점을 잇는 선분 길이 합

[출제의도] 무한등비급수를 활용하여 문제 해결하기 (ⅰ) 삼각형의 중점연결정리에 의하여 $\displaystyle \mathrm{\overline{M _{1} M _{2}}} = \frac{1}{2} \overline{AB} = 2$ $\displaystyle \mathrm{\overline{M _{3} M _{4}}} = \frac{1}{2} \overline{M _{1} M _{2}} = 1$ $\displaystyle \mathrm{\overline{M _{5} M _{6}}} = \frac{1}{2} \overline{M _{3} M _{4}} = \frac{1}{2}$ $\vdots$ $\displaystyle \mathrm{\overline{M _{1} M _{2}}} + \overline{M _{3} M _{4}} + \overline{M _{5} M _{6}} + \cdots = \frac{2 _{}}{1 - \frac{1}{2}} = 4$ (ⅱ) 삼각형의 중점연결정리에 의하여 $\displaystyle \mathrm{\overline{M _{2} M _{3}}} = \frac{1}{2} \overline{CM _{1}} = 2$ $\displaystyle \mathrm{\overline{M _{4} M _{5}}} = \frac{1}{2} \overline{M _{2} M _{3}} = 1$ $\displaystyle \mathrm{\overline{M _{6} M _{7}}} = \frac{1}{2} \overline{M _{4} M _{5}} = \frac{1}{2}$ $\vdots$ $\displaystyle \mathrm{\overline{M _{2} M _{3}}} + \overline{M _{4} M _{5}} + \overline{M _{6} M _{7}} + \cdots = \frac{2 _{}}{1 - \frac{1}{2}} = 4$ (ⅲ) $\mathrm{\triangle} ABC$에서 $\displaystyle \mathrm{\cos} B = \frac{{4 ^{2} + 8 ^{2} - 6 ^{2}} _{}}{{2 \cdot 4 \cdot 8} ^{}} = \frac{11 _{}}{16 ^{}}$이므로 $\displaystyle \mathrm{\overline{AM _{1}}} {} ^{2} = 16 + 16 - 32 \cos B = 10$ $\displaystyle \therefore \mathrm{\overline{AM _{1}}} = \sqrt{10}$ 따라서, (ⅰ). (ⅱ), (ⅲ)으로부터 $\displaystyle \mathrm{\overline{AM}} _{1} + \overline{M _{1} M _{2}} + \overline{M _{2} M _{3}} + \cdots = 8 + \sqrt{10}$