두 원의 접선 각

[출제의도] 삼각함수의 덧셈정리를 활용하여 문제해결하기

원점을 $\mathrm{O}$, $\angle \mathrm{OPA} = \alpha$, $\angle \mathrm{BPO} = \beta$라 하자. 삼각형 $\mathrm{POA}$와 삼각형 $\mathrm{PQA}$가 서로 합동이고 삼각형 $\mathrm{PRB}$와 삼각형 $\mathrm{POB}$가 서로 합동이므로 $\angle \mathrm{APQ} = \angle \mathrm{OPA}$, $\angle \mathrm{RPB} = \angle \mathrm{BPO}$ $\angle \mathrm{RPQ} = \theta = 2 \left( \alpha + \beta \right)$이므로 $\tan \theta$$= \tan 2 \left( \alpha + \beta \right)$ $\displaystyle = \frac{\tan \left( \alpha + \beta \right) + \tan \left( \alpha + \beta \right)}{1 - \tan \left( \alpha + \beta \right) \times \tan \left( \alpha + \beta \right)} = \frac{4}{3}$ 이때, $\tan \left( \alpha + \beta \right) = t$라 두면 $\displaystyle \frac{2 t}{1 - t ^{2}} = \frac{4}{3}$, $2 t ^{2} + 3 t - 2 = 0$ $\displaystyle a > \sqrt{2}$에서 $0 < \theta < \pi$이므로 $\displaystyle 0 < \alpha + \beta < \frac{\pi}{2}$ $\therefore$ $\displaystyle t = \tan \left( \alpha + \beta \right) = \frac{1}{2}$ $\displaystyle \tan \alpha = \frac{1}{a}$, $\displaystyle \tan \beta = \frac{2}{a}$ $\displaystyle \tan \left( \alpha + \beta \right) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{3 a}{a ^{2} - 2} = \frac{1}{2}$ $a ^{2} - 6 a - 2 = 0$, $\displaystyle a = 3 + \sqrt{11}$ 따라서 $\left( a - 3 \right) ^{2} = 11$