Consider a rectangular piece of paper ABCD, with AB = √2 and AD = 1. The rectangle is folded so that:
 B coincides with the point X on CD and:
 The resulting crease is AY which passes through the corner A.
Determine the length of the three sides of the triangle DXY.
Consider the rectangle ABDC, A at the origin, B at (sqrt(2),0), C at (Sqrt(2),1), D at (0,1). By definition
The fold of B to point X on CD causes BX to be perpendicular to the crease AY. Also, ABY & BCX are similar triangles, and AY bisects BX.
Consider line AY and define angle q as the slope of AY. The equation of AY then becomes y=tan(q)x. Similarly
the equation for line BX can be written (it's slope is perpendicualr to that of AY) y=cot(q)(xsqrt(2)).
Again, by the definition of the fold, these two lines intersect at some x, where y=1/2. Solving yields
cot(q)+tan(q)=2*sqrt(2), or q=pi/8 (22.5 deg).
From there, using the similar triangles and trig functions, the coodinates of the points can be found and are:
X (1,1)
Y (sqrt(2), 2sqrt(2))
D (given) (0,1)
Using the distance formula, the line segment lengths then are
DX = 1
XY = SQRT(64*sqrt(2))
DY = SQRT(52*sqrt2))
Edited on May 21, 2014, 2:51 pm
Edited on May 21, 2014, 2:52 pm

Posted by Kenny M
on 20140520 17:35:09 