A rectangular sheet was folded along its diagonal so that the four vertices form an isosceles trapezoid, three of whose sides are equal, and the fourth side is $10 \sqrt{3}$. What are the dimensions of the original rectangle?
I. Solution. Bisect the rectangle $ACBD$ along the diagonal $AB$. Since the angles at the vertices of the rectangle are $90^\circ$, $\angle ACB$ and $\angle ADB$ are right angles, and thus $C$ and $D$ lie on the Thales circle with $AB$ as the diameter.

Draw the radii from points $C$ and $D$; this forms 3 isosceles triangles: $\triangle AOD$, $\triangle DOC$, and $\triangle COB$. The third sides of these triangles are also equal in length, so the angles between the radii are $60^\circ$, making the triangles equilateral:

$$
a = r = 5 \sqrt{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_c164404bf45c2d2ae13dg-1.jpg?height=442&width=374&top_left_y=446&top_left_x=865)

Since the angles of $\triangle DAB$ are $90^\circ$, $60^\circ$, and $30^\circ$,

$$
\sin 60^\circ = \frac{b}{10 \sqrt{3}}
$$

from which $b = 15$.

The sides of the rectangle are: $a = 5 \sqrt{3}$ and $b = 15$.

II. Solution. Let the fourth vertex of the trapezoid formed by the bisection be $A'$. The triangle $\triangle DCA'$ is isosceles, and $F$ is the midpoint of $DC$, so $A'F$ is the axis of symmetry of the triangle, and thus $\angle A'FC = 90^\circ$. $\angle A'CF = \angle ABD$ because they are corresponding angles, and similarly, $\angle ADB = \angle FA'C$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c164404bf45c2d2ae13dg-1.jpg?height=477&width=526&top_left_y=1388&top_left_x=797)

$\triangle DAB \sim \triangle A'FC$ because their angles are pairwise equal. This implies that the ratios of their corresponding sides are equal: $FC : A'C = AB : BD$, so

$$
\frac{a}{2} : b = a : 10 \sqrt{3}, \quad 2b = 10 \sqrt{3}, \quad b = 5 \sqrt{3}
$$

In $\triangle ABD$, by the Pythagorean theorem, $a^2 + 75 = 300$, so $a = 15$.