원 위 점과 중선 길이 곱 최댓값

$\displaystyle \mathrm{\overline{AB}}$가 지름이므로 $\mathrm{\angle} P = 90 ^\circ$이다. $\displaystyle \therefore \overline{\mathrm{PA}} ^{2} + \overline{\mathrm{PB}} ^{2} = a ^{2}$ $\mathrm{\triangle} APN$에서 $\displaystyle \mathrm{\overline{PA}} ^{2} + \mathrm{\overline{PN}} ^{2}$$\displaystyle = \mathrm{\overline{PA}} ^{2} + \left( \frac{1}{2} \mathrm{\overline{PB}} \right) ^{2} = \mathrm{\overline{AN}} ^{2}$ $\mathrm{\triangle} BPM$에서 $\displaystyle \mathrm{\overline{PB}} ^{2} + \mathrm{\overline{PM}} ^{2}$$\displaystyle = \mathrm{\overline{PB}} ^{2} + \left( \frac{1}{2} \mathrm{\overline{PA}} \right) ^{2} = \mathrm{\overline{BM}} ^{2}$

따라서, $\displaystyle \mathrm{\overline{AN}} ^{2} + \mathrm{\overline{BM}} ^{2}$$\displaystyle = \overline{\mathrm{PA}} ^{2} + \overline{\mathrm{PB}} ^{2} + \frac{1}{4} \left( \overline{\mathrm{PA}} ^{2} + \overline{\mathrm{PB}} ^{2} \right) = \frac{5}{4} a ^{2}$ 에서 $\displaystyle \mathrm{\overline{AN}} ^{2} + \mathrm{\overline{BM}} ^{2} \geq 2 \overline{AN} \cdot \overline{BM}$이므로 $\displaystyle \mathrm{\overline{AN}} \bullet \overline{BM}$의 최댓값은 $\displaystyle \frac{5}{8} a ^{2}$이다.