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Re: [數學]摺紙問題

發表 G@ry 於 星期日 四月 29, 2007 8:59 am

☆ ~ 幻 星 ~ ☆ 寫到:左鍵: 點擊縮放; 右鍵: 觀看原圖 #ed_op#br#ed_cl##ed_op#br#ed_cl#如圖,正方形ABDC中#ed_op#br#ed_cl#CG=GH=HA#ed_op#br#ed_cl#DE=EF=FB#ed_op#br#ed_cl#若將H摺到GE上,同時讓A在BD上#ed_op#br#ed_cl#證明#ed_op#br#ed_cl#DA:AB=2^(1/3):1 #ed_op#div#ed_cl# #ed_op#/div#ed_cl#
#ed_op#br#ed_cl#解題得出下圖:#ed_op#br#ed_cl##ed_op#img style="font-family: Wingdings;" src="richedit/upload/2kf5d256c92c.png" alt="image file name: 2kf5d256c92c.png" border="0"#ed_cl##ed_op#br#ed_cl#設AH=A'H=x, 正方形邊長=3x, JB=y, JA=JA'=3x-y;#ed_op#br#ed_cl#A'B = √[(3x-y)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-y#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#] = √[3x(3x-2y)]; ∠EH'A'=∠BA'J=θ => EA' = xsinθ = xy/(3x-y)#ed_op#br#ed_cl##ed_op#br#ed_cl#由於以下開始的計算比較複雜,故暫設x=1來簡化計算,下面另有保留x的版本:(下面設a#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#=2, a=2#ed_op#sup#ed_cl#1/3#ed_op#/sup#ed_cl#)#ed_op#br#ed_cl#2=EB=EA'+A'B=y/(3-y)+√[3(3-2y)]=> [(2-y)√3]#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#=[(3-y)√(3-2y)]#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#=> y#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#-6y#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+12y-15/2=0#ed_op#br#ed_cl#設 z=y-2, (z+2)#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#-6(z+2)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+12(z+2)-15/2=0  =>  z#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#=-1/2, z=-1/a  =>  y=2-1/a;#ed_op#br#ed_cl#A'B = √[3(3-2y)] = √[3(3-4+2/a)] = √[3(a#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-1)(a#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#+a#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+1)] / √(a#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#+a#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+1)#ed_op#br#ed_cl#      = √[3(a#ed_op#sup#ed_cl#6#ed_op#/sup#ed_cl#-1)] / √(2a+a#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+1) = √[3(4-1)] / √(a+1)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# = 3/(a+1)#ed_op#br#ed_cl#DA':A'B = (DB-A'B)/A'B = DB/A'B-1 = 3/[3/(a+1)]-1 = a = 2#ed_op#sup#ed_cl#1/3#ed_op#/sup#ed_cl#:1#ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl#有x的版本:(下面設a#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#=2, a=2#ed_op#sup#ed_cl#1/3#ed_op#/sup#ed_cl#)#ed_op#br#ed_cl#2x = EB = EA'+A'B = xy/(3x-y) + √[3x(3x-2y)]  =>  [(2x-xy)√3x]#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# = [(3x-y)√(3x-2y)]#ed_op#sup#ed_cl#2#ed_op#br#ed_cl##ed_op#/sup#ed_cl#=>  y#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#-6xy#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+12x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#y-15x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#/2 = 0#ed_op#br#ed_cl#設 z=y-2x, (z+2x)#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#-6x(z+2x)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+12x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#(z+2x)-15x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#/2=0  =>  z#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#=-x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#/2, z=-x/a  =>  y=2x-x/a;#ed_op#br#ed_cl#A'B = √[3x(3x-2y)] = √[3x(3x-4x+2x/a)] = √[3x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#(a#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-1)(a#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#+a#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+1)] / √(a#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#+a#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+1)#ed_op#br#ed_cl#      = √[3x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#(a#ed_op#sup#ed_cl#6#ed_op#/sup#ed_cl#-1)] / √(2a+a#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+1) = √[3x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#(4-1)] / √(a+1)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# = 3x/(a+1)#ed_op#br#ed_cl#DA':A'B = (DB-A'B)/A'B = DB/A'B-1 = 3x/[3x/(a+1)]-1 = a = 2#ed_op#sup#ed_cl#1/3#ed_op#/sup#ed_cl#:1 #ed_op#br#ed_cl##ed_op#br#ed_cl#

[數學]摺紙問題

發表 ☆ ~ 幻 星 ~ ☆ 於 星期六 四月 28, 2007 7:34 pm

左鍵: 點擊縮放; 右鍵: 觀看原圖 #ed_op#BR#ed_cl##ed_op#BR#ed_cl#如圖,正方形ABDC中#ed_op#BR#ed_cl#CG=GH=HA#ed_op#BR#ed_cl#DE=EF=FB#ed_op#BR#ed_cl#若將H摺到GE上,同時讓A在BD上#ed_op#BR#ed_cl#證明#ed_op#BR#ed_cl#DA:AB=2^(1/3):1 #ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl#