고3 킬러문항 4월 29번

$$\angle \mathbf{ OPR} =2\alpha , \; \angle \mathbf{ OPQ} =2\beta ,\; \mathbf{P} (0,a) $$라고 하자.

$$\tan \left( \dfrac {\pi } {2}-\alpha \right) =\cot \alpha =\dfrac {a} {2}\; \Rightarrow \;\tan \alpha =\dfrac {2} {a}\\ $$

$$ \tan \left( \dfrac {\pi } {2}-\beta \right) =\cot \beta =\dfrac {a} {1}\; \Rightarrow \;\tan \beta =\dfrac {1} {a}\\ $$

$$ \alpha +\beta =A$$ 라고 놓으면

$$\tan \theta =\tan 2\left( \alpha +\beta \right) =\tan 2A=\dfrac {2\tan A} {1-\tan ^{2}A}=\dfrac {4} {3} $$

 

$$ \tan A =t $$ 라고 놓으면

 

$$ \begin{aligned} 6t&=4-4t^{2} \\ 2t^2 +3t -2&=0\\ \left( t+2\right) \left( 2t-1\right)&=0 \; \Rightarrow \; t=-2 \; or \; t= \dfrac {1} {2} \end{aligned} $$

 

$$\tan 2A >0 $$ 이므로 $$0<2A< \dfrac {\pi} {2} \; \Rightarrow \; 0<A<\dfrac {\pi} {4} \; \Rightarrow \; \tan A>0$$

 

$$\therefore \tan A = \dfrac {1} {2} $$

 

$$\tan A = \theta =\tan \left( \alpha +\beta \right) = \dfrac {1} {2} $$ 이므로

 

$$ \dfrac {\tan \alpha +\tan \beta } {1-\tan \alpha \tan \beta }= \dfrac {\frac {2} {a}+\frac {1} {a}} {1-\frac {2} {a}\times \frac {1} {a}}=\dfrac {3a} {a^{2}-2}=\dfrac {1} {2} $$

 

위 식을 정리하면

 

$$ a^2 -6a -2=0 \; \Rightarrow \; a=3 \pm \sqrt{9+2} \; \Rightarrow \; \left(a-3 \right)^2 = \left( \pm \sqrt{11} \right)^2 =11 $$