로그 원과 xy 최대최소

$\log _{3} x = X , \log _{3} y = Y$라 하면 $\left( \log _{3} x \right) ^{2} + \left( \log _{3} y \right) ^{2} = 2 \log _{9} x + 2 \log _{9} y = \log _{3} x + \log _{3} y$는 $X ^{2} + Y ^{2} = X + Y$, $\displaystyle \left( X - \frac{1}{2} \right) ^{2} + \left( Y - \frac{1}{2} \right) ^{2} = \frac{1}{2} \cdots \text{㉠}$ $\log _{3} xy = \log _{3} x + \log _{3} y = X + Y$에서 직선 $X + Y - \log _{3} xy = 0$과 ㉠이 만나야 하므로 $\displaystyle \frac{\left| \frac{1}{2} + \frac{1}{2} - \log _{3} xy \right|}{\sqrt{1 ^{2} + 1 ^{2}}} \leq \frac{1}{\sqrt{2}}$, $\left| 1 - \log _{3} xy \right| \leq 1$ $\therefore 0 \leq \log _{3} xy \leq 2$ $\therefore 1 \leq xy \leq 9$ ∴ $M + m = 9 + 1 = 10$