4. Given a square $M N P Q$ with a side length of $1 m$. On the sides of this square, points $A, B, C$ are marked:

- point $A$ is at $\frac{1}{3}$ of the side $N P$ from point $N$,
- point $B$ is at $\frac{2}{3}$ of the side $M Q$ from point $M$,
- point $C$ is the midpoint of the side $M N$.

Triangle $A B C$ is not a right triangle. On which side and by how much should point $C$ be moved so that triangle $A B C$ becomes a right triangle?
Solution. Let's assume that point $C$ needs to be moved to position $C^{\prime}$ so that triangle $A B C^{\prime}$ becomes a right triangle (diagram on the right). Let $x = C C^{\prime}$ be the length that needs to be determined. Further, let $D$ be the midpoint of segment $M B$. Then $B D = \frac{1}{3} = A N$ and $M N =$

![](https://cdn.mathpix.com/cropped/2024_06_05_501b6f3ef6775dab6656g-4.jpg?height=327&width=356&top_left_y=1687&top_left_x=1132)
$N C = \frac{1}{2}$. From the right triangles $A B D, A N C^{\prime}, M B C^{\prime}, A B C^{\prime}$, we get

\[
\begin{aligned}
& A B^{2} = 1^{2} + \left(\frac{1}{3}\right)^{2} = \frac{10}{9}, \quad A C^{\prime 2} = \left(\frac{1}{2} - x\right)^{2} + \left(\frac{1}{3}\right)^{2} \\
& B C^{\prime 2} = \left(\frac{1}{2} + x\right)^{2} + \left(\frac{2}{3}\right)^{2} \text{ and } A C^{\prime 2} + B C^{\prime 2} = A B^{2}
\end{aligned}
\]

By substituting the first three into the fourth equation and simplifying, we get $2 x^{2} = \frac{1}{18}$, from which we find $x^{2} = \frac{1}{36} = \left( \pm \frac{1}{6}\right)^{2}$. Therefore, $x = \frac{1}{6}$ or $x = -\frac{1}{6}$, which means that point $C$ needs to be moved left or right.