6. The lengths of the sides of a triangle are expressed in meters by whole numbers. Determine them, if the triangle has a perimeter of $72 m$ and the longest side of the triangle is divided by the point of tangency of the inscribed circle in the ratio $3: 4$.
SOLUTION. We will use the general fact that the points of tangency of the inscribed circle divide the perimeter of the triangle into six segments, and that any two of them that emanate from the same vertex of the triangle are congruent. (Tangents from a given point to a given circle are symmetric with respect to the line connecting the given point to the center of the given circle.)

In our problem, the longest side of the triangle is divided into segments whose lengths we denote by $3x$ and $4x$, while the lengths of the segments from the vertex opposite the longest side are denoted by $y$ (see Fig. 1). The sides of the triangle thus have lengths $7x$, $4x+y$, and $3x+y$, where $x$ and $y$ are unknown positive numbers (we take lengths without units). For $7x$ to be the length of the longest side, it must be true that $7x > 4x + y$ or $3x > y$. Let us emphasize that the sought numbers $x$ and $y$ do not necessarily have to be integers, but according to the problem statement, this is true for the numbers $7x$, $4x+y$, and $3x+y$.

![](https://cdn.mathpix.com/cropped/2024_04_17_93ab0272c174b6cbb42cg-2.jpg?height=457&width=688&top_left_y=982&top_left_x=684)

Fig. 1

The information about the perimeter of the triangle can be written as the equation

$$
72 = 7x + (3x + y) + (4x + y) \text{ or } 36 = 7x + y
$$

Since $7x$ is an integer, $y = 36 - 7x$ is also an integer; and since according to the problem statement, the numbers $4x + y$ and $3x + y$ are also integers, $x = (4x + y) - (3x + y)$ is an integer as well. Therefore, from this point on, we are looking for pairs of positive integers $x$ and $y$ for which

$$
3x > y \quad \text{and} \quad 7x + y = 36.
$$

From this, it follows that $7x < 36 < 7x + 3x = 10x$, so $x \leq 5$ and simultaneously $x \geq 4$.

For $x = 4$, we have $y = 8$ and $(7x, 4x + y, 3x + y) = (28, 24, 20)$; for $x = 5$, we have $y = 1$ and $(7x, 4x + y, 3x + y) = (35, 21, 16)$. The sides of the triangle are therefore $(28, 24, 20)$ or $(35, 21, 16)$. (The triangle inequalities are clearly satisfied.)

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