4. A rectangle $A B C D$ has been cut out of paper. A point $E$ is chosen on side $\overline{A B}$ such that $|A E|=8$ and $|B E|=17$. A point $F$ is chosen on side $\overline{C D}$ such that $|C F|=3$. After folding the paper along the line $\overline{E F}$, vertex $B$ coincides with a point $S$ on side $\overline{A D}$. Determine the length of side $\overline{B C}$ of rectangle $A B C D$ and write it as an irreducible fraction.
First solution.

By folding the paper according to the conditions of the problem, we get the following:

![](https://cdn.mathpix.com/cropped/2024_05_30_2e894aa57f8bdfbd97cfg-30.jpg?height=822&width=803&top_left_y=1822&top_left_x=635)

Quadrilateral $EBCF$ is mapped to its congruent quadrilateral $SEFT$ (quadrilateral $SEFT$ is the reflection of quadrilateral $EBCF$ across the line $EF$), i.e., triangle $EBF$ is mapped to its congruent triangle $SEF$, so

$$
|SE| = |BE| = 17
$$

By applying the Pythagorean theorem to triangle $SAE$, we get

$$
\begin{aligned}
|SA|^2 = 17^2 - 8^2 & = 289 - 64 = 225 \\
|SA| & = 15
\end{aligned}
$$

Let $|AD| = |BC| = x$.

Notice that $|DS| = x - 15$ and $|DF| = 22$.

1 POINT

![](https://cdn.mathpix.com/cropped/2024_05_30_2e894aa57f8bdfbd97cfg-31.jpg?height=834&width=934&top_left_y=971&top_left_x=538)

Since, due to the congruence of triangles $EBF$ and $SEF$, $|BF| = |SF|$,

1 POINT

by applying the Pythagorean theorem to triangle $FDS$ and triangle $BCF$, we get:

$$
\begin{gathered}
|SD|^2 + |DF|^2 = |FC|^2 + |BC|^2 \\
(x - 15)^2 + 22^2 = 3^2 + x^2 \\
x^2 - 30x + 225 + 484 = 9 + x^2 \\
30x = 700 \\
x = \frac{70}{3}
\end{gathered}
$$

##