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)(x-sqrt(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), 2-sqrt(2))
D (given) (0,1)

Using the distance formula, the line segment lengths then are

DX = 1
XY = SQRT(6-4*sqrt(2))
DY = SQRT(5-2*sqrt2))

Edited on May 21, 2014, 2:51 pm

Edited on May 21, 2014, 2:52 pm

Posted by Kenny M on 2014-05-20 17:35:09