## Task A-3.4.

In the plane of square $A B C D$, but outside it, there is a point $P$. If

$$
|P A|=\sqrt{5}, \quad|P B|=\sqrt{26} \quad \text { and } \quad|P D|=\sqrt{20}
$$

determine the length of the side of the square.
## Solution.

Place the square in the coordinate system so that its vertices are $A(0,0), B(a, 0), C(a, a)$, and $D(0, a)$, and let $P(-x,-y)$.

![](https://cdn.mathpix.com/cropped/2024_05_30_a439f3fdcb63d942eea2g-18.jpg?height=708&width=826&top_left_y=1371&top_left_x=612)

The formula for the distance between any two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is

$$
\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}
$$

Using this formula, we can express the conditions of the problem as follows:

$$
\begin{aligned}
x^{2}+y^{2} & =5 \\
x^{2}+(y+a)^{2} & =20 \\
(x+a)^{2}+y^{2} & =26
\end{aligned}
$$

By subtracting the first equation from the second and third, we get

$$
\begin{aligned}
& a^{2}+2 a y=15 \\
& a^{2}+2 a x=21
\end{aligned}
$$

Subtracting these two equations, we get $6 + 2 a y = 2 a x$. Simplifying and squaring this equation, we obtain

$$
9 + 6 a y + a^{2} y^{2} = a^{2} x^{2} = a^{2} (5 - y^{2}) = 5 a^{2} - a^{2} y^{2}
$$

Substituting $a^{2} = 15 - 2 a y$ into the last equation, and letting $t = a y$, we get

$$
9 + 6 t + t^{2} = 5(15 - 2 t) - t^{2} \Longrightarrow t^{2} + 8 t - 33 = 0
$$

The solutions to the quadratic equation are $t = 3$ and $t = -11$.

For $t = a y = -11$, from $a^{2} + 2 a y = 15$ we get $a = \sqrt{37}$. From the equations $a^{2} + 2 a y = 15$ and $a^{2} + 2 a x = 21$, we then have

$$
x = -\frac{8}{\sqrt{37}} \quad \text{and} \quad y = -\frac{11}{\sqrt{37}}
$$

In this case, point $P$ is inside the square $ABCD$, which contradicts the conditions of the problem.

For $t = a y = 3$, we similarly get $a = 3$, and $x = 2$ and $y = 1$. Direct verification shows that this configuration satisfies the conditions of the problem.

Therefore, the length of the side of the square is 3.

Note: The system in the above solution can also be solved in another way. Squaring and adding the equations $a^{2} - 15 = 2 a y$ and $a^{2} - 21 = 2 a x$, we get

$$
2 a^{4} - 72 a^{2} + 666 = 4 a^{2} (x^{2} + y^{2}) = 4 a^{2} \cdot 5 = 20 a^{2}
$$

From this, we obtain the biquadratic equation

$$
a^{4} - 46 a^{2} + 333 = 0
$$

which gives $a^{2} \in \{9, 37\}$, and thus, due to the positivity of $a$, we have $a \in \{3, \sqrt{37}\}$. Complete the analysis as in the previous solution.