OE⊥AB通过O点RT△EBO BO = BE = 1/2 * AB =√3/2
距离D = OE = 1 /√(2 + B 2)
得到的毕达哥拉斯定理:(3/4)+(1 /(?2 + b的2))= 1. ∴2 + b 2分配= 4
设x = 0时,为y = 1 / b的,设y = 0,x = 1时/
面积s = 1/2 * | 1 / | * | 1 / B = 1 /(2 | A | * | B |)≥1 / [| A | + | B | 2] = 1/4
当且仅当| A | = | B | =√2的至少四分之一.