Meowth 寫到:--- phpBB : The Protected Message is not copied in this quote ---
see :)
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由 hsiaod 於 星期二 五月 13, 2003 11:56 am
see :)
由 yptsoi 於 星期一 五月 12, 2003 10:34 pm
see
由 ---- 於 星期一 五月 12, 2003 10:20 pm
Meowth 寫到:Kang means he jumps very high, runs very fast, and is still very cute.
can say nth
由 --- 於 星期一 五月 12, 2003 10:14 pm
Kang means he jumps very high, runs very fast, and is still very cute.
由 ---- 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...
由 --- 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.
Plane geo also works. U re right. I just knew you could, because you're Kang.
由 ---- 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.
由 --- 於 星期一 五月 12, 2003 9:59 pm
What's O?
由 ---- 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)
由 --- 於 星期一 五月 12, 2003 9:49 pm
How come "AOD similar to BOC "?
由 ---- 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?
(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?
由 yll 於 星期一 五月 12, 2003 9:32 pm
see
由 ---- 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...
由 --- 於 星期一 五月 12, 2003 8:58 pm
Don't just "see", try to solve it with plane geo, Kang.
由 ---- 於 星期一 五月 12, 2003 8:55 pm
see
由 E.T 於 星期一 五月 12, 2003 8:46 pm
see
由 heron0520 於 星期一 五月 12, 2003 8:18 pm
see see
由 --- 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
由 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」？？
由 Raceleader 於 星期一 五月 12, 2003 7:52 pm
不要火上加油