1. A square piece of paper ABCD is folded such that A coincides with the midpoint M of the side BC.
2. The resulting crease intersects AB in X and CD in Y.

Determine the ratio |AX|:|DY|

This ratio will be true for any square so for simplicity let
A = (0,0) B=(0,1) C=(1,1) D=(1,0)
M=(.5,1)
Midpoint AM=(.25,.5)
Slope AM=2
The crease is the perpendicular bisector of AM
y-.5=-.5(x-.25)
y=-.5x+6.25
X=(0,.625)
AX=.625
If x=1, y=.125
Y=(1,.125)
DY=.125
AX/DY=.625/.125=5

Posted by Jer on 2014-04-07 00:27:59