## Task B-1.7.

The lengths of the sides of an acute-angled triangle are three numbers, the largest of which is four greater than the smallest, and the middle one in size is the arithmetic mean of the remaining two. The height of the triangle drawn to the middle side divides the triangle into parts whose areas are in the ratio $3: 2$. Determine the perimeter of the given triangle.
## Solution.

Let the side lengths be $a-2, a, a+2$.

![](https://cdn.mathpix.com/cropped/2024_05_30_0c173970003c85556d5eg-07.jpg?height=545&width=711&top_left_y=681&top_left_x=593)

From the ratio of the areas, we get $\frac{x v / 2}{y v / 2} = \frac{x}{y} = \frac{3}{2}$, hence $x = \frac{3}{2} y$.

Apply the Pythagorean theorem to the right triangles $A N C$ and $N B C$ in the figure.

$$
v^{2} = (a+2)^{2} - x^{2}, \quad v^{2} = (a-2)^{2} - y^{2}
$$

After equating the right sides, we can write the resulting equation as $(a+2)^{2} - (a-2)^{2} = x^{2} - y^{2}$.

This simplifies to $8a = (x-y)(x+y)$, 1 point

and since $x + y = a$, we get $8a = (x-y)a$,

so $x - y = 8$.

From $x = \frac{3}{2} y$ and $x - y = 8$, we get $y = 16, x = 24$,

and $a = x + y = 40$.

The perimeter of the given triangle is $o = (a-2) + a + (a+2) = 3a = 120$.

Note: If the sides are $a, a+2, a+4$, we get:

$$
\begin{gathered}
v^{2} = (a+4)^{2} - x^{2} = a^{2} - y^{2} \\
(a+4)^{2} - a^{2} = x^{2} - y^{2} \\
8a + 16 = (x-y)(x+y) \\
8(a+2) = (x-y)(a+2)
\end{gathered}
$$

thus $x - y = 8$, and we proceed as in the above solution.

## COUNTY COMPETITION IN MATHEMATICS

## 2nd grade - high school - B variant

March 4, 2020.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.