Folding square puzzle

Question

This problem is framed by me. Here I have posted to let others see the beauty of mathematics.

A square is folded as shown. $3$ squares are inscribed in $3$ triangles are formed. Area of blue and green squares are $2\,\mathrm{cm}^2$ and $3\,\mathrm{cm}^2$, respectively. Find the area of the red square.

Note: this question is not incorrect or incorrectly stated.

Answer 1

All segment lengths are marked in the diagram. Match below the total side lengths of the square,

$$\begin{align} L = & \sqrt2 (\sec a+\csc a+\tan a+1) \\ = & \sqrt2 (1+\cot a) + \sqrt3(\tan a+1) \\ = & \sqrt3(\sec a+\csc a) + x(\cot a+1)\\ \end{align}$$

Solve for $a$ from the first two expressions to get $\tan\frac{a}2 = \frac{\sqrt3}{\sqrt2}-1$ and then plug it into the last two expressions to otain $x =a-b\tan(\frac\pi4 - \frac a2) = \sqrt3 - \sqrt2$. Thus, the area of the red square is $$x^2 = 5-2\sqrt6$$

Answer 2

Let

• $\ell$ be the side of the big square $ABCD$.
• $x = \sqrt{2}$, $y = \sqrt{3}$ be the sides of the two green squares.
• $z$ be the side of the red square.

Choose a coordinate system where $A$ is the origin and $B$ lies on +ve $x$-axis.
In this coordinate system, square $ABCD$ becomes $[0,\ell]^2$.

Reflect everything across the fold line (the cyan dashed line). Square $ABCD$ get mapped to square $A'B'C'D'$ while the red square get mapped to the square $[\ell-z,\ell]^2$.

Let $E = (x,x)$, $F = (\ell-y,y)$, $G = (\ell-z,\ell-z)$ be the contact points of the two green and new red square with square $A'B'C'D'$.

Since $DF$ and $EG$ are line segments whose end points lie on opposite sides of square $A'B'C'D'$ and $DF \perp EG$, they are equal in length, i.e. $|DF| = |EG|$. This leads to $$(\ell - y)\sqrt{2} = (\ell - x - z)\sqrt{2} \quad\implies\quad z = y - x$$ As a result, the area of the original red square equal to $$z^2 = (\sqrt{3}-\sqrt{2})^2 = 5-2\sqrt{6}$$