## Task B-2.6.

A square $A B C D$ is given. In the same plane, a point $O$ is chosen such that $|O A|=|O B|=5 \mathrm{~cm}$ and $|O D|=\sqrt{13} \mathrm{~cm}$. What is the length of the side of this square?
## Solution.

First, let's sketch the given square $ABCD$ with side length $a$.

From the equality of the lengths of segments $|OA|=|OB|=5 \text{ cm}$, we conclude that triangle $ABO$ is isosceles with base $\overline{AB}$, which means that point $O$ lies on the perpendicular bisector of side $\overline{AB}$.

Let point $E$ be the intersection of the perpendicular bisector and side $\overline{CD}$, and point $F$ be the intersection of the perpendicular bisector and side $\overline{AB}$. Note that points $E$ and $F$ are the midpoints of the sides of the square to which they belong, i.e., $|AF|=|DE|=\frac{a}{2}$.

Since $|OD|=\sqrt{13} \text{ cm} < |OA|$, we conclude that point $O$ lies on the perpendicular bisector outside the square, closer to side $\overline{CD}$ than to side $\overline{AB}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b7629e40479ae2c2357g-14.jpg?height=553&width=243&top_left_y=1911&top_left_x=224)

Let $|OE|=x \text{ cm}$.

For the lengths of the sides of the right triangle $DEO$, we have $\left(\frac{a}{2}\right)^2 + x^2 = \sqrt{13}^2$, or $\frac{a^2}{4} + x^2 = 13$.

For the lengths of the sides of the right triangle $AFO$, we have $\left(\frac{a}{2}\right)^2 + (x + a)^2 = 5^2$, or $\frac{a^2}{4} + x^2 + 2ax + a^2 = 25$.

Let's solve the system of equations.

\[
\left\{
\begin{array}{l}
\frac{a^2}{4} + x^2 = 13 \\
\frac{a^2}{4} + x^2 + 2ax + a^2 = 25
\end{array}
\right.
\]

\[
13 + 2ax + a^2 = 25
\]

\[
2ax = 12 - a^2 \quad /: 2a
\]

\[
x = \frac{12 - a^2}{2a} = \frac{6}{a} - \frac{a}{2}
\]

Substituting $x$ into the first equation in the system, we get:

\[
\frac{a^2}{4} + \left(\frac{6}{a} - \frac{a}{2}\right)^2 = 13
\]

\[
\frac{a^2}{4} + \left(\frac{36}{a^2} - 6 + \frac{a^2}{4}\right) = 13
\]

\[
\frac{a^2}{2} - 6 + \frac{36}{a^2} = 13
\]

\[
\frac{a^2}{2} + \frac{36}{a^2} = 19 \quad / \cdot 2a^2
\]

\[
a^4 - 38a^2 + 72 = 0
\]

\[
(a^2 - 2)(a^2 - 36) = 0
\]

1. Case: $a^2 = 2$, the side length of the square is $a = \sqrt{2} \text{ cm}$.

Then, $x = \frac{12 - 2}{2\sqrt{2}} = \frac{10}{2\sqrt{2}} = \frac{5\sqrt{2}}{2} \text{ cm}$.

2. Case: $a^2 = 36$, the side length of the square is $a = 6 \text{ cm}$.

Then, $x = \frac{12 - 36}{12} = \frac{-24}{12} = -2 \text{ cm}$, which is impossible because the length of a side of a triangle cannot be a negative number.

We conclude that the desired side length is $\sqrt{2} \text{ cm}$.