回覆 19# enlighten0626 的帖子
令 \( \displaystyle L_{1} : y = mx+3 , m < 0\)
\( \displaystyle B\left( \frac{36}{5-12m} , \frac{15}{5-12m} \right) , C\left( -\frac{3}{m} , 0 \right) \)
\( \displaystyle 30 = \overline{OB} + \overline{OC} + \overline{BC} = \frac{39}{5-12m} - \frac{3}{m} + \frac{15}{5 - 12m} \sqrt{ \frac{1}{m^2} + 1 } \)
\( \displaystyle 288m^3 - 120m^2 - 12m + 5 = (24m^2 - 1)(12m-5) = 0 \; \Rightarrow \; m = -\frac{1}{2\sqrt{6}} \)
\( \displaystyle \Rightarrow \; C(6,\sqrt{6} , 0) \; , \; \bigtriangleup OAC = \frac{1}{2} \cdot 3 \cdot 6\sqrt{6} = 9\sqrt{6} \)
