一條花了九牛二虎之力+看完答案也不懂的題目

幻藍

ABCDEF is a convex hexagon. Each of the three diagonals AD , BE , CF bisects the area of the hexagon. Show that the three diagonals intersect at a common point.
[HINT : The areaof a triangle of sides of lengh a, b and included angle θ is equal to 1/2 ab sinθ . ]

oscar

我從題目中看到這句：「ABCDEF is a convex hexagon. Each of the three diagonals AD , BE , CF bisects the area of the hexagon.」
一個十分有用的資料！
我要利用這資料來證明ABCDEF是正六邊形。

幻藍

Originally posted by oscar at 2005-6-3 06:06 PM:

但是答案說3個對邊都能相等分割...
這就只有正六邊形才做到！


兔兒只是中三...(對不對？)
vector是CE程度的...
用別的方法吧...

[ Last edited by oscar on 2005-6-3 at 06:08 PM ]

我捉字蝨而已~XD

我是中三的.....
ps:這題數據說是中三數來的......=.="

oscar

Originally posted by 可愛的兔兒 at 2005-6-2 22:54:

但是答案說3個對邊都能相等分割...
這就只有正六邊形才做到！

Originally posted by 落雷 at 2005-6-2 23:45:

所以要先證明那是Regular Convex Hexagon
然後很簡單地就可以解決

我會用vector做一次

Let O be the centre
A,B,C,D,E,F are end points in clockwise direction
Assume AD,CF are straight lines and prove BE is straight line as well

By parrellogram law of vector
b=a+c
e=d+f

b+e=(a+d)+(c+f)
b+e=0+0
b=(-1)e
BO//OE
BE is straight line passes through O
Three lines are concurrent

兔兒只是中三...(對不對？)
vector是CE程度的...
用別的方法吧...

[ Last edited by oscar on 2005-6-3 at 06:08 PM ]

抹茶可樂

Originally posted by oscar at 2005-6-2 05:46 PM:

紅色的step表明了ABCDEF一定是一個正六邊形，
因為一個不規則六邊形的對邊不可以把兩面的面積分割成一樣的。
所以問題應該是問Regular Convex Hexagon。

所以要先證明那是Regular Convex Hexagon
然後很簡單地就可以解決

我會用vector做一次

Let O be the centre
A,B,C,D,E,F are end points in clockwise direction
Assume AD,CF are straight lines and prove BE is straight line as well

By parrellogram law of vector
b=a+c
e=d+f

b+e=(a+d)+(c+f)
b+e=0+0
b=(-1)e
BO//OE
BE is straight line passes through O
Three lines are concurrent

幻藍

Originally posted by oscar at 2005-6-2 05:46 PM:

紅色的step表明了ABCDEF一定是一個正六邊形，
因為一個不規則六邊形的對邊不可以把兩面的面積分割成一樣的。
所以問題應該是問Regular Convex Hexagon。

oscar

Originally posted by 可愛的兔兒 at 2005-6-1 21:12:

我也不太明的.....

Note that △ABX and △DEX have the same area , because together with BCDX each occupies a half of the hexagon .
Similarly , △BCY and △EFY have the same area , △CDZ and AFZ have the same area .Hence
ab=(a'+a"")(b'+b'')
b''c''=(b+b')(c+c')
a''c=(a+a')(c'+c'') .
Multiply c to the first throughout to get abc=a'b'c+a'b''c+a''b'c+a''b''c .
Multiply a to the second throughout to get ab''c''=abc+abc'+ab'c+ ab'c' .
Multiply b'' to the third throughout to get a''b''c=ab''c'+ab''c''+a'b''c'+a'b''c'' .
Add up the three obtain 0=a'b'c+a'b''c+a''b'c+a''b''c+abc+abc'+ab'c+ ab'c'+ab''c'+ab''c''+a'b''c'+a'b''c'' .
In particular , a'b'c=0 , but c cannot be zero , hence a'=0 or b'=0 .
I either case means AD , BE , CF are concurrent .
Alternatively , we can mu;tiply the three to obtain
abca''b''c''=(a+a')(b+b')(c+c')(a''+a')(b''+b')(c''+c')>abca''b''c'' if one of a' , b' , c' is positive
Therefore , a'=b'=c'=0 and AD , BE , CF are concurrent

紅色的step表明了ABCDEF一定是一個正六邊形，
因為一個不規則六邊形的對邊不可以把兩面的面積分割成一樣的。
所以問題應該是問Regular Convex Hexagon。

幻藍

Originally posted by 大胃 at 2005-5-31 09:58 AM:

都一星期0羅和say個答案出來啦喂:em27:

我也不太明的.....

Note that △ABX and △DEX have the same area , because together with BCDX each occupies a half of the hexagon .
Similarly , △BCY and △EFY have the same area , △CDZ and AFZ have the same area . Hence
ab=(a'+a"")(b'+b'')
b''c''=(b+b')(c+c')
a''c=(a+a')(c'+c'') .
Multiply c to the first throughout to get abc=a'b'c+a'b''c+a''b'c+a''b''c .
Multiply a to the second throughout to get ab''c''=abc+abc'+ab'c+ ab'c' .
Multiply b'' to the third throughout to get a''b''c=ab''c'+ab''c''+a'b''c'+a'b''c'' .
Add up the three obtain 0=a'b'c+a'b''c+a''b'c+a''b''c+abc+abc'+ab'c+ ab'c'+ab''c'+ab''c''+a'b''c'+a'b''c'' .
In particular , a'b'c=0 , but c cannot be zero , hence a'=0 or b'=0 .
I either case means AD , BE , CF are concurrent .
Alternatively , we can mu;tiply the three to obtain
abca''b''c''=(a+a')(b+b')(c+c')(a''+a')(b''+b')(c''+c')>abca''b''c'' if one of a' , b' , c' is positive
Therefore , a'=b'=c'=0 and AD , BE , CF are concurrent

細薯

我在school intranet問得出的answer....

Let ABCDEF counts anti-clockwisely and...
AD passes through points X,Y
BE passes through points Y,Z
CF passes through points Z,X

As...
Area of ABCD = Area ADEF -> Area of ABZX + BCZ + CDYZ + XYZ = Area of DEY + EFXY + FAX
Area of BCDE = Area EFAB -> Area of BCZ + CDYZ + DEY = Area of EFXY + FAX + ABZX + XYZ
Area of CDEF = Area FABC -> Area of CDYZ + DEY + XYZ + EFXY = Area of FAX + ABZX + BCZ

then try to prove Area of XYZ = 0 then means XYZ is located on same point.
Furthermore it means the 3 diagonals are concurrent.

我總覺得有點錯.....

大胃

Originally posted by JOHNLAM17 at 2005-5-4 09:46 PM:
過左一week la
講answer la =.=

都一星期0羅和say個答案出來啦喂:em27: