Paper Folding 1

Consider the picture shown, the folded triangle has sides $\frac{1}{2}$ and $1$ and hypotenuse $\frac{\sqrt{5}}{2}$ , now from the small right triangle we have $\cos(2a)=\frac{y}{1/2}$ hence $x=\frac{1}{2}-y=\frac{1}{2}-\frac{1}{2}\cos(2a)=\frac{1}{2}-\frac{1}{2}(\cos^2a-\sin^2a)$ replacing we get $x=\frac{1}{2}-\frac{1}{2}\left(\left(\frac{2}{\sqrt{5}}\right)^2-\left(\frac{1}{\sqrt{5}}\right)^2\right)=\frac{1}{2}-\frac{3}{10}=\frac{1}{5}$ so $\boxed{x=.2}$

Draw a line parallel to the lower side of the square through the red point. Because the right angle vertex is located at this point the two right triangles formed to the left and right of the point are similar with scale factor 1:2. Thus the proportion ((1/2) - d)/sqrt(1^ - (1 - d)^2) = 1/2 holds, which is equivalent to 5d^2 - 6d + 1 = (5d - 1)(d - 1) = 0. Since d < 1, it follows that d = 1/5.

The vector (-1,0) has been rotated by 53.4 degrees. multiply this by the rotation matrix that accounts for the rotation.

x = cos(53.4) * -1

y = sin(53.4) * -1

y is approximately -.8.

therefore the difference is approximately .2