[數學]不等式(2)

[數學]不等式(2)

--- 於 星期一 五月 12, 2003 4:17 pm


Let ABCD be a cyclic quadrilateral(圓內接四邊形).
Prove that | AB - CD | + | AD -BC | >= 2 | AC -BD |.

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--- 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]

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scsnake 於 星期一 五月 12, 2003 7:49 pm


ccccc
sisisi
吸吸吸

scsnake
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Raceleader 於 星期一 五月 12, 2003 7:52 pm


灰灰的  灰灰的  灰灰的

Raceleader
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Raceleader 於 星期一 五月 12, 2003 7:52 pm


不要火上加油

Raceleader
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scsnake 於 星期一 五月 12, 2003 7:54 pm


奇怪,我怎麼不能編輯/刪除文章??我的狀態也變成「離線中」??

scsnake
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--- 於 星期一 五月 12, 2003 8:13 pm


let t<=q, p>=s,不失一般情況  ............... 不等式技巧

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heron0520 於 星期一 五月 12, 2003 8:18 pm


see see
當把握........

heron0520

 
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E.T 於 星期一 五月 12, 2003 8:46 pm


see
Click this : [>> 有趣的科學(Science's World) <<]

Let's go to discuss ~*

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---- 於 星期一 五月 12, 2003 8:55 pm


see

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--- 於 星期一 五月 12, 2003 8:58 pm


Don't just "see", try to solve it with plane geo, Kang.

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---- 於 星期一 五月 12, 2003 9:28 pm


Meowth 寫到:Don't just "see", try to solve it with plane geo, Kang.


my lvl is much lower than your lvl, i don't think i can...

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yll 於 星期一 五月 12, 2003 9:32 pm


see

yll
帥哥良~
帥哥良~
 
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---- 於 星期一 五月 12, 2003 9:47 pm


can this be counted as a proof?
(修改:有些地方寫得不清楚,故補充一下)
[hide:8177502da2]
Let O be the intersection of AC and BD.
angle OAD=angle OBC (angle in same seg)
angle ODA=angle OCB (angle in same seg)
angle AOD=angle BOC (vert opp. angles)
AOD similar to BOC (AAA)
AOD similar to BOC
AO/BO=DO/CO=AD/BC=(AO-DO)/(BO-OC)
AD=(AO-DO)/(BO-OC) * BC
AD-BC = (AO-DO-BO+OC)/(BO-OC) *BC
=(AC-BD)/(BO-OC) *BC
Similarly,
AB-CD=(AC-BD)/(DO-CO) * CD

|AD-BC|+|AB-CD|
=|AC-BD||BC/(BO-OC) + CD/(DO-CO)| (BO-OC <>0, DO-CO<>0)
By triangle inequality,
BC>BO-OC, CD>DO-CO
So
>2|AC-BD|

Equality case holds when BO=OC=DO, easily checked.
[/hide:8177502da2]
is it right?

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--- 於 星期一 五月 12, 2003 9:49 pm


How come "AOD similar to BOC "?

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---- 於 星期一 五月 12, 2003 9:56 pm


Meowth 寫到:How come "AOD similar to BOC "?

angle OAD=angle OBC (angle in same seg)
angle ODA=angle OCB (angle in same seg)
angle AOD=angle BOC (vert opp. angles)
AOD similar to BOC (AAA)

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--- 於 星期一 五月 12, 2003 9:59 pm


What's O?

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---- 於 星期一 五月 12, 2003 10:04 pm


sorry, i didn't type it. O is the intersection of AC and BD.

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--- 於 星期一 五月 12, 2003 10:04 pm


I see. Our O are different.
Plane geo also works. U re right. I just knew you could, because you're Kang.

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---- 於 星期一 五月 12, 2003 10:11 pm


Meowth 寫到:I see. Our O are different.
Plane geo also works. U re right. I just knew you could, because you're Kang.


....................... kang doesn't mean anything...

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