## Task B-1.3.

The lengths of the sides of a triangle are three consecutive natural numbers, not less than 3. Calculate the difference in the lengths of the segments that the height to the middle side divides on that side.
## Solution.

Let the lengths of the sides of the triangle be $|A C|=n-1,|A B|=n$ and $|B C|=n+1$, as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_30_d4daff78382705a7c34eg-02.jpg?height=571&width=716&top_left_y=1750&top_left_x=590)

$\overline{C N}$ is the altitude to the side $\overline{A B}$, the middle side in length, and $x$ and $y$ are the lengths of the segments into which point N divides the side $\overline{A B}$. Then, from the right triangles $A N C$ and $N B C$, by the Pythagorean theorem, we have

$$
(n-1)^{2}-x^{2}=v^{2}
$$

$$
(n+1)^{2}-y^{2}=v^{2} \text {. }
$$

If we equate the obtained expressions for the square of the altitude, we get:

$$
\begin{array}{cc}
(n-1)^{2}-x^{2}=(n+1)^{2}-y^{2} & 1 \text { point } \\
n^{2}-2 n+1-n^{2}-2 n-1=x^{2}-y^{2} & \\
y^{2}-x^{2}=4 n . & 1 \text { point }
\end{array}
$$

By applying the formula for the difference of squares, we get

$$
(y+x)(y-x)=4 n
$$

Since $y+x=n$, it follows that

$$
(y+x)(y-x)=4 n
$$

or

$$
y-x=4
$$

The desired difference of the segments is 4.