Solution. Let the rectangle be $ABCD (BC < AB), AB = 12 \, \text{cm}$, and $AC$ be the diagonal along which the rectangle is folded, meaning that when the $\triangle ABC$ is reflected over $AC$, we get the $\triangle AB'C$. In the trapezoid $ACB'D$, $AD = DB' = B'C$ (note that $AD \neq AC$, because $AD < AC$, since in the right triangle $\triangle DAC$, the leg is smaller than the hypotenuse).

![](https://cdn.mathpix.com/cropped/2024_05_02_51ce44eec2c64e8a38e9g-1.jpg?height=512&width=507&top_left_y=351&top_left_x=798)

Consider the circle $k$ with $AC$ as its diameter, and its center at $O$.

$\angle ADC = 90^\circ$ (one of the angles of the given rectangle), by the converse of Thales' theorem, point $D$ lies on circle $k$. Since $\angle ABC = 90^\circ$, by reflection, $\angle AB'C = 90^\circ$. Therefore, points $B$ and $B'$ also lie on circle $k$. Based on the above and the folding of the rectangle, $A, D, B', C$ lie on the semicircle $k$ with diameter $AC$.

We know that $AD = DB' = B'C$, therefore

$$
\angle B'OC = \frac{\angle AOC}{3} = \frac{180^\circ}{3} = 60^\circ
$$

(in a circle, equal chords subtend equal central angles). By the theorem of inscribed and central angles:

$$
\angle B'AC = \frac{\angle B'OC}{2}, \quad \text{so} \quad \angle B'AC = \frac{60^\circ}{2} = 30^\circ,
$$

and thus $\angle CAB = 30^\circ$.

If one of the acute angles in a right triangle is $30^\circ$, then the opposite leg is half the hypotenuse. Let $BC = b$, then $AC = 2b$, and in the right triangle $\triangle ABC$, by the Pythagorean theorem: $(2b)^2 = 12^2 + b^2$, from which $b = 4\sqrt{3} \approx 6.93$.

Therefore, the shorter side of the given rectangle is approximately $6.93 \, \text{cm}$.
We folded a rectangular piece of paper along one of its diagonals. After the fold, the four vertices ended up at the four vertices of a trapezoid, three sides of which are of equal length. What is the length of the shorter side of the original rectangle if the longer side is 12 cm?