원에 내접하는 사각형 빈칸

[출제의도] 삼각함수를 이용하여 추론하기 $\displaystyle {\overline{\mathrm{AB}}}$$\displaystyle = {\overline{\mathrm{AD}}}$$= k$라 할 때 두 삼각형 $\mathrm{ABC}$, $\mathrm{ACD}$에서 각각 코사인법칙에 의하여 $\cos \left( \angle \mathrm{ACB} \right)$$\displaystyle = \frac{10 ^{2} + {\overline{\mathrm{BC}}} ^{2} - k ^{2}}{2 \times 10 \times {\overline{\mathrm{BC}}}}$$\displaystyle = \frac{1}{20} \left( {\overline{\mathrm{BC}}} + \frac{{\square {100 - k ^{2}}}}{{\overline{\mathrm{BC}}}} \right)$, $\cos \left( \angle \mathrm{DCA} \right)$$\displaystyle = \frac{10 ^{2} + {\overline{\mathrm{CD}}} ^{2} - k ^{2}}{2 \times 10 \times {\overline{\mathrm{CD}}}}$$\displaystyle = \frac{1}{20} \left( {\overline{\mathrm{CD}}} + \frac{{\square {100 - k ^{2}}}}{{\overline{\mathrm{CD}}}} \right)$ 이다. 이때 두 호 $\mathrm{AB}$, $\mathrm{AD}$에 대한 원주각의 크기가 같으므로 $\cos \left( \angle \mathrm{ACB} \right)$$= \cos \left( \angle \mathrm{DCA} \right)$이다. $\displaystyle \frac{1}{20} \left( {\overline{\mathrm{BC}}} + \frac{100 - \mathit{k} ^{2}}{{\overline{\mathrm{BC}}}} \right)$$\displaystyle = \frac{1}{20} \left( {\overline{\mathrm{CD}}} + \frac{100 - k ^{2}}{{\overline{\mathrm{CD}}}} \right)$ $\displaystyle {\overline{\mathrm{BC}}} - {\overline{\mathrm{CD}}}$$\displaystyle = \left( 100 - k ^{2} \right) \times \left( \frac{1}{{\overline{\mathrm{CD}}}} - \frac{1}{{\overline{\mathrm{BC}}}} \right)$ $\displaystyle {\overline{\mathrm{BC}}} - {\overline{\mathrm{CD}}}$$\displaystyle = \left( 100 - k ^{2} \right) \times \left( \frac{{\overline{\mathrm{BC}}} - {\overline{\mathrm{CD}}}}{{\overline{\mathrm{BC}}} \times {\overline{\mathrm{CD}}}} \right)$ $\displaystyle {\overline{\mathrm{AC}}}$$= 10 < 2 R$이므로 $\displaystyle {\overline{\mathrm{BC}}}$$\displaystyle \ne {\overline{\mathrm{CD}}}$ 그러므로 $\displaystyle {\overline{\mathrm{BC}}} \times {\overline{\mathrm{CD}}}$$= 100 - k ^{2}$ 사각형 $\mathrm{ABCD}$의 넓이는 두 삼각형 $\mathrm{ABD}$, $\mathrm{BCD}$의 넓이의 합과 같으므로 $\displaystyle \frac{1}{2} k ^{2} \sin \left( \angle \mathrm{BAD} \right) + \frac{1}{2} \times {\overline{\mathrm{BC}}} \times {\overline{\mathrm{CD}}} \times \sin \left( \pi - \angle \mathrm{BAD} \right)$ $\displaystyle = \frac{1}{2} \left\{ k ^{2} + \left( 100 - k ^{2} \right) \right\} \sin \left( \angle \mathrm{BAD} \right)$ $= 50 \sin \left( \angle \mathrm{BAD} \right)$$= 40$ 에서 $\sin \left( \angle \mathrm{BAD} \right)$$\displaystyle = {\square {\frac{4}{5}}}$이다. 따라서 삼각형 $\mathrm{ABD}$에서 사인법칙에 의하여 $\displaystyle \frac{{\overline{\mathrm{BD}}}}{\sin \left( \angle \mathrm{BAD} \right)}$$= 2 R$에서 $\displaystyle {\overline{\mathrm{BD}}}$$\displaystyle = \frac{8}{5} R$이므로 $\displaystyle {\overline{\mathrm{BD}}} : R$$\displaystyle = {\square {\frac{8}{5}}} : 1$ 따라서 $f \left( k \right)$$= 100 - k ^{2}$, $p$$\displaystyle = \frac{4}{5}$, $q$$\displaystyle = \frac{8}{5}$이므로 $\displaystyle \frac{f \left( 10 p \right)}{q}$$\displaystyle = \left( 100 - 8 ^{2} \right) \times \frac{5}{8}$$\displaystyle = \frac{45}{2}$