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?
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 $6.93 \, \text{cm}$.