외접원과 현

$\displaystyle \mathrm{\overline{DE}} = {\overline{BC}} - {\overline{BD}} - {\overline{CE}}$이므로 $\displaystyle \mathrm{\overline{DE}} = 8 - 3 - 3 = 2$ 직각삼각형 $\mathrm{ABD}$에서 $\displaystyle \mathrm{\overline{AB}} = \sqrt{3 ^{2} + \left( \sqrt{5} \right) ^{2}} = \sqrt{14}$ 직각삼각형 $\mathrm{ACD}$에서 $\displaystyle \mathrm{\overline{AC}} = \sqrt{5 ^{2} + \left( \sqrt{5} \right) ^{2}} = \sqrt{30}$ $\angle \mathrm{ACB} = \alpha$라 하면 $\displaystyle \sin \alpha = \frac{\mathrm{\overline{AD}}}{\mathrm{\overline{AC}}} = \frac{\sqrt{5}}{\sqrt{30}} = \frac{\sqrt{6}}{6}$ 삼각형 $\mathrm{ABC}$에서 사인법칙에 의하여 $\displaystyle \frac{\mathrm{\overline{AB}}}{\sin \alpha} = 2 \mathit{R}$, $\displaystyle \frac{\sqrt{14}}{\frac{\sqrt{6}}{6}} = 2 R$ $\therefore$ $\displaystyle R = \sqrt{21}$ 직각삼각형 $\mathrm{ADE}$에서 $\displaystyle \mathrm{\overline{AE}} = \sqrt{2 ^{2} + \left( \sqrt{5} \right) ^{2}} = 3$ $\angle \mathrm{DAE} = \beta$라 하면 $\displaystyle \sin \beta = \frac{2}{3}$ 삼각형 $\mathrm{AFG}$에서 사인법칙에 의해 $\displaystyle \frac{l}{\sin \beta} = 2 R$, $\displaystyle l = 2 \times \sqrt{21} \times \frac{2}{3} = \frac{4 \sqrt{21}}{3}$ $\therefore$ $\displaystyle R \times l = \sqrt{21} \times \frac{4 \sqrt{21}}{3} = 28$ [다른 풀이]

$\displaystyle \mathrm{\overline{DE}} = {\overline{BC}} - {\overline{BD}} - {\overline{CE}}$이므로 $\displaystyle \mathrm{\overline{DE}} = 8 - 3 - 3 = 2$ 이고 직각삼각형 $\mathrm{ADE}$에서 $\displaystyle \mathrm{\overline{AE}} ^{2} = {\overline{AD}} ^{2} + {\overline{DE}} ^{2} = 5 + 4 = 9$, $\displaystyle \mathrm{\overline{AE}} = 3$ 이므로 $\angle \mathrm{EAD} = \theta$라 하면 $\displaystyle \sin \theta = \frac{2}{3}$, $\displaystyle \cos \theta = \frac{\sqrt{5}}{3}$ 이때 두 현의 교차에 대한 성질에 의하여 $\displaystyle \mathrm{\overline{CE}} \times {\overline{BE}} = {\overline{AE}} \times {\overline{EG}}$, $\displaystyle \mathrm{3} \times 5 = 3 \times {\overline{EG}}$ $\therefore$$\displaystyle \mathrm{\overline{EG}} = 5$ $\displaystyle \mathrm{\overline{CD}} \times {\overline{BD}} = {\overline{AD}} \times {\overline{DF}}$, $\displaystyle \mathrm{5} \times 3 = \sqrt{5} \times {\overline{DF}}$ $\therefore$$\displaystyle \mathrm{\overline{DF}} = 3 \sqrt{5}$ 삼각형 $\mathrm{AFG}$에서 코사인법칙에 의하여 $\displaystyle l ^{2} = 8 ^{2} + \left( 4 \sqrt{5} \right) ^{2} - 2 \times 8 \times 4 \sqrt{5} \times \frac{\sqrt{5}}{3} = \frac{112}{3}$ 사인법칙에 의하여 $\displaystyle \frac{l}{\sin \theta} = 2 R$, $\displaystyle R = \frac{l}{2 \sin \theta}$ 이므로 $\displaystyle R \times l = \frac{l}{2 \sin \theta} \times l$$\displaystyle = \frac{l ^{2}}{2 \sin \theta}$$\displaystyle = \frac{\frac{112}{3}}{2 \times \frac{2}{3}} = 28$