An origami problem for January 3rd

Let $ABCD$ be a unit square. The folding procedure gives $60^{\circ}$ angles, so both triangles are equilateral. The height of $\triangle AKH$ is $\frac{1}{2}$ and the height of $\triangle DIC$ is $\frac{\sqrt{3}}{2}$ . The ratio of the squares of these heights is $\boxed{\frac{1}{3}}$ .

Happy Thirdsday

Let the paper be a unit square. Consider just after the second fold. Let $\angle B'CG = \theta$ . Since $\triangle BCG$ and $\triangle B'CG$ are congruent right triangles, $\angle B'GC = \angle BGC = 90^\circ - \theta$ , and $\angle AGB' = 180^\circ - 2(90^\circ - \theta) = 2\theta$ . Since $B'C = 1$ , $B'G = \tan \theta$ . We note that:

$\begin{aligned} B'G \sin \angle AGB' & = \frac 12 \\ \tan \theta \color{#3D99F6} \sin 2 \theta & = \frac 12 & \small \color{#3D99F6} \text{Note that }\sin 2 \theta = 2 \sin \theta \cos \theta \\ 2 \sin^2 \theta & = \frac 12 \\ \sin^2 \theta & =\frac 14 \\ \sin \theta & = \frac 12 \\ \implies \theta & = 30^\circ \end{aligned}$

Then we note that both $\triangle AKH$ and $\triangle DIC$ are equilateral triangles. $\triangle AKH$ has an altitude of $\frac 12$ and $\triangle DIC$ has an altitude of (\frac {\sqrt 3}2). Since the area of similar figure is directly proportional to the square of its linear dimension. The ration of the areas of $\triangle AKH$ and $\triangle DIC$ is given by:

$\begin{aligned} \frac {[\triangle AKH]}{[\triangle DIC]} & = \frac {\left(\frac 12\right)^2}{\left(\frac {\sqrt 3}2\right)^2} = \boxed {\dfrac 13}\end{aligned}$