Origami Triangle

Label the diagram as shown. The area of the square is equal to $x^{2}$ . From the picture it is clear that length DG is $x$ (it's AD folded over). By Pythagoras' theorem, Length DB is $\sqrt{x^{2}+x^{2}}$ , that is $\sqrt{2}x$ . Hence, length GB is $\sqrt{2}x-x$ . Triangle BGF is isoceles, so GF is also $\sqrt{2}x-x$ . EF is double this, that is $2(\sqrt{2}x-x)$ . Hence, the area of triangle DEF is $0.5\times 2( \sqrt{2}x-x)\times x$ (from $0.5\times base\times height)$ . When we divide this by $x^{2}$ , we get $\frac{x(\sqrt{2}x-x)}{x^{2}}$ . Finally, by cancelling x, the area is $\sqrt{2}-1$ .

Let $\left[ \triangle XYZ \right]$ denote the area of $\triangle XYZ$ .

From the picture we note that $\angle ADB = 45°$ and due to similar triangles, $\angle ADE = \angle EDG = 22.5°$ .

Thus the length of $\overline { AE }$ is $\tan { 22.5° }$ and $\left[ \triangle AED \right] = \dfrac{\tan { 22.5° }}{2}$ , $\left[ \triangle DEF \right] =\tan { 22.5° }$ . Therefore the area of the $\triangle DEF$ divided by the area of the square ( $1$ ) is $\tan { 22.5° }$ .

To work out the exact value of $\tan { 22.5° }$ we can start by recalling that $\tan { 45° } =1$ . Therefore we are trying to find $x$ where $\tan { 2x } =1$ . Let $y=\tan {x }$ , and via the $\tan { (A+B) }$ rule we have:

$\dfrac{2y}{1-{y}^{2}}=1~ ~\Rightarrow ~~{y}^{2}+2y-1=0$

Putting this into the quadratic formula gives $y=-1 \pm \sqrt{2}$ . Obviously our answer must be positive so we have $\left[ \triangle DEF \right] =\large\color{#20A900}{\boxed{\sqrt{2}-1}}$ .

Let the square's side length be 1. The square's area will also be 1, so the area of the triangle divided by the area of the square is equal to the area of the triangle in this case. We know the height of the triangle in question is 1, since it is determined by folding over a side of the square. The height of the little triangle (in the top right) is the length of the diagonal subtracted by the height of the big triangle, which is $\sqrt{2}-1$ . Imagine this little triangle split into two smaller triangles by the diagonal line. These are isosceles triangles, with angles 90/45/45. Thus, the length of the base of our big triangle is $2(\sqrt{2}-1)$ . The area of the large triangle is $\frac{1}{2}*2(\sqrt{2}-1)*1$ ( $\frac{1}{2}*base*height$ ), so the area of the triangle divided by the area of the square is $\sqrt{2}-1$ .