由 --- 於 星期一 五月 12, 2003 7:48 pm
[hide:be235fff44]my method:
let O be 外接圓心; r be 外接圓心半徑
let p=角AOB/2,q=角BOC/2,s=角COD/2, t=角DOA/2
p+q+t+s=pi
let t<=q, p>=s,不失一般情況
(p+s)/2< pi/2
t+(p+s)/2=(p+q+s+t)/2+(t-q)/2=pi/2+(t-q)/2<=pi/2
length AB=2rsinp, BC=2rsinq, CD=2rsins, DA=2rsint
AC=2rsin(p+q)=2rsin(s+t)
BD=2rsin(q+s)=2rsin(t+p)
| AB - CD | - | AC -BD |.
=|2rsinp-2rsins|-|2rsin(s+t)-2rsin(t+p)|
=4rsin((p-s)/2)*cos((p+s)/2)-4rsin((p-s)/2)*cos(t+(p+s)/2)
=8rsin((p-s)/2)*sin(t/2)*sin((t+p+s)/2)
>=0
同理,| AD -BC | - | AC -BD | >=0
so, | AB - CD | + | AD -BC | >= 2 | AC -BD |.## [/hide:be235fff44]