問數~關於空間幾何~

Let angle BAE=t, drop a perpendicular from B to AE and let the intersection of be F. Let BF prolonged (before folding) to meet AD (or AD prolonged) at G.
Then BF=AB(sin t)=4sin t, FG=(4cos t)/(tan t), AG=4/tan t.
Fold up along AE, then the 二面角 = angle BFG, and write the cosine of this angle as k.
consider triangle BFG, by cosine law
BG^2=BF^2+FG^2-2BFxFGcos(angle BFG)
= 16sin^2 t+(16cos^2 t)/(tan^2 t) -32kcos^2 t

consider triangle BAG,
cos(angle BAG)=(AB^2+AG^2-BG^2)/(2ABxAG)
=(16sin^2 t + 16/tan^2 t -(16sin^2 t+(16cos^2 t)/(tan^2 t) -32kcos^2 t )) / (32/tan t)
=(1/tan^2 t - cos^2 t/ tan^2 t + 2kcos^2 t) / (2/tan t)
=(sin^2 t/tan^2 t +2kcos^2 t) / (2/tan t)
=(cos^2 t + 2kcos^2 t)(tan t) / 2
=((1+2k)/2) sin t cos t

consider triangle BAD
BD^2=AB^2+AD^2-2ABxADcos (angle BAG) (note angle BAG=angle BAD)
= 16+49-56(((1+2k)/2) sin t cos t)
=65-28(1+2k)sin t cos t
=65-14(1+2k)sin(2t)

For a fixed 二面角, k is fixed, and BD is minimum (i.e., BD^2 is minimum) when sin(2t) is maximum. The maximum attains when sin(2t)=1, so t=45 degree.

Hence, when BD is minimum, BE=ABtan t=AB=4.

Remark: in fact, the answer does not depend on the value of the 二面角.

2008-01-17 23:21:19 補充：
如果不需要嚴格證明，有一個更簡單的方法去計。當紙片沿AE折至完全重疊時，(即兩面角=0)，B點和ACD在同一個平面上，考慮三角形ABD，跟據三角不等式BD >= AD-AB=3，最小值當ABD三點組成同一直線時發生。要折疊後B點在AD線上，則角BAE必須是45度，所以BE=AB=4.

E係BC中間.....