一條花了九牛二虎之力+看完答案也不懂的題目

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。

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

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 ]

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

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。