☆ ~ 幻 星 ~ ☆ 寫到: 如圖,正方形ABDC中
CG=GH=HA
DE=EF=FB
若將H摺到GE上,同時讓A在BD上
證明
DA:AB=2^(1/3):1
解題得出下圖:
設AH=A'H=x, 正方形邊長=3x, JB=y, JA=JA'=3x-y;
A'B = √[(3x-y)
2-y
2] = √[3x(3x-2y)]; ∠EH'A'=∠BA'J=θ => EA' = xsinθ = xy/(3x-y)
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
3=2, a=2
1/3)
2=EB=EA'+A'B=y/(3-y)+√[3(3-2y)]=> [(2-y)√3]
2=[(3-y)√(3-2y)]
2=> y
3-6y
2+12y-15/2=0
設 z=y-2, (z+2)
3-6(z+2)
2+12(z+2)-15/2=0 => z
3=-1/2, z=-1/a => y=2-1/a;
A'B = √[3(3-2y)] = √[3(3-4+2/a)] = √[3(a
2-1)(a
4+a
2+1)] / √(a
4+a
2+1)
= √[3(a
6-1)] / √(2a+a
2+1) = √[3(4-1)] / √(a+1)
2 = 3/(a+1)
DA':A'B = (DB-A'B)/A'B = DB/A'B-1 = 3/[3/(a+1)]-1 = a = 2
1/3:1
有x的版本:(下面設a
3=2, a=2
1/3)
2x = EB = EA'+A'B = xy/(3x-y) + √[3x(3x-2y)] => [(2x-xy)√3x]
2 = [(3x-y)√(3x-2y)]
2
=> y
3-6xy
2+12x
2y-15x
3/2 = 0
設 z=y-2x, (z+2x)
3-6x(z+2x)
2+12x
2(z+2x)-15x
3/2=0 => z
3=-x
3/2, z=-x/a => y=2x-x/a;
A'B = √[3x(3x-2y)] = √[3x(3x-4x+2x/a)] = √[3x
2(a
2-1)(a
4+a
2+1)] / √(a
4+a
2+1)
= √[3x
2(a
6-1)] / √(2a+a
2+1) = √[3x
2(4-1)] / √(a+1)
2 = 3x/(a+1)
DA':A'B = (DB-A'B)/A'B = DB/A'B-1 = 3x/[3x/(a+1)]-1 = a = 2
1/3:1