由 --- 於 星期日 五月 18, 2003 9:21 am
平面上給定100個點,
==> 考慮距離最遠的兩點A, B 當直徑畫圓. D<1, R<1/2
其他點 M1, M2,...M98
"三角形ABMi是鈍角三角形"
(1) if angle A is obtuse
MiB^2= AB^2+AMi^2-2AB*AMi*cosA > AB^2,
MiB>AB
but we know, AB is the 最遠的兩點
so,不合
(2) with the same reason, angle B can't be obtuse.
(3) so, angle AMiB nust be obtuse
==> Mi 都是圓內角
==> 98 點 都在圓內,2點在圓周
so, we make another circle O' centered at the midpoint of AB, with radius 1/2, will cover all of the 100 points.