For a certain triangle, we know the following:

the radius of the inscribed circle is one unit,

the radii of the excircles are integers,

the area of the triangle is also an integer.

Determine the sides of the triangle!
Before we solve the problem, let's review some relationships that exist between the radii of the inscribed and circumscribed circles of a triangle and the sides of the triangle.

# Lengths of the tangent segments

For the triangle \(ABC\) with the center \(O\) of the inscribed circle, draw perpendiculars from \(O\) to the sides. The foot of the perpendicular from \(O\) to \(AB\) is \(P\), to \(BC\) is \(Q\), and to \(AC\) is \(R\). Connecting any vertex of the triangle to \(O\), this is the axis of symmetry for the circle, and thus we have

\[
A P = A R = x, \quad B P = B Q = y, \quad C Q = C R = z
\]

The sum \(x + y + z\) is half the perimeter of the triangle and is denoted by \(s\).

## Area of the triangle

The area of triangle \(ABC\) can be obtained by summing the areas of triangles \(AOB\), \(BOC\), and \(COA\). Writing down the areas, if the radius of the inscribed circle is denoted by \(\varrho\) (in the diagram, \(r\) and \(r_b\) are used for \(\varrho\) and \(\varrho_b\)),

\[
t = \frac{(x+y) \varrho}{2} + \frac{(y+z) \varrho}{2} + \frac{(z+x) \varrho}{2} = (x+y+z) \varrho = s \varrho
\]

we obtain this relationship.

![](https://cdn.mathpix.com/cropped/2024_05_02_85d1ecd39c22d3df06f3g-1.jpg?height=629&width=1203&top_left_y=970&top_left_x=450)

## Radii of the excircles of the triangle

Draw the excircle of triangle \(ABC\) opposite side \(b\), i.e., the circle that touches side \(b\) and the extensions of sides \(a\) and \(c\). Denote the points of tangency on the line of side \(c\) by \(S\) and on the line of side \(b\) by \(T\).

Clearly, \(B S + B T = y + x + A S + y + z + C T = 2(x + y + z) = 2s\) because \(A S + C T = A C = z + x\).

But \(B T = B S = s\) and thus

\[
A S = z \quad \text{and} \quad C T = x
\]

Denote the center of the excircle opposite side \(b\) by \(O_b\) and its radius by \(\varrho_b\). The area of triangle \(ABC\) can be written as twice the area of triangle \(O_b S B\) minus twice the areas of triangles \(O_b S A\) and \(O_b T C\). Using the relationships from (3),

\[
t = \varrho_b (x + y + z) - z \varrho_b - x \varrho_b = \varrho_b y
\]

from which

\[
\varrho_b = \frac{t}{y}
\]

Similarly, we get

\[
\varrho_a = \frac{t}{x}, \quad \varrho_c = \frac{t}{z}
\]

Transitioning to the solution of our original problem,

\[
\frac{1}{\varrho_a} + \frac{1}{\varrho_b} + \frac{1}{\varrho_c} = \frac{x + y + z}{t'} = \frac{s}{t} = \frac{s}{s \varrho} = \frac{1}{\varrho} = 1
\]

We used the condition that the radius of the inscribed circle is 1, and substituted \(s \varrho\) for \(t\) from (2).

In our problem, we need to solve the equation from (5), using the fact that \(\varrho_a\), \(\varrho_b\), and \(\varrho_c\) are integers.

In exercise 1849, we showed that for \(n \geq 3\), the equation has solutions with distinct numbers. Now we need to find all solutions. It is not difficult to notice that at most one of \(\varrho_a\), \(\varrho_b\), and \(\varrho_c\) can be equal to 2, and the other two must be greater than 2. Moreover, it is clear that they cannot all be 4, because their sum would be less than 1. Therefore, there must be one that is less than 4, i.e., 2 or 3.

Without loss of generality, we can choose the notation such that \(\varrho_a \leq \varrho_b \leq \varrho_c\).

If \(\varrho_a = 2\), then \(\varrho_b = \varrho_c = 4\) or \(\varrho_b = 3\), \(\varrho_c = 6\). The other possibility is that \(\varrho_a = \varrho_b = \varrho_c = 3\).

Let's try to express the area of the triangle in terms of the radii of the inscribed and excircles.

Multiplying the equations from (4),

\[
\varrho_a \varrho_b \varrho_c = \frac{t^3}{x y z}
\]

From the similarity of triangles \(O_b S A\) and \(A O P\),

\[
\frac{\varrho_b}{x} = \frac{z}{\varrho}, \text{ and since } \varrho = 1, \text{ we get } \varrho_b = x z
\]

From (4),

\[
\varrho_b = \frac{t}{y} = x z, \quad \text{so} \quad t = x z y
\]

Substituting this into (6),

\[
\varrho_a \varrho_b \varrho_c = \frac{t^2}{x y z} \cdot x y z = t^2
\]

from which \(t = \sqrt{\varrho_a \varrho_b \varrho_c}\).

Substituting the possible values of \(\varrho_a\), \(\varrho_b\), and \(\varrho_c\),

\[
t = \sqrt{2 \cdot 4 \cdot 4} = \sqrt{32}, \quad t = \sqrt{2 \cdot 3 \cdot 6} = \sqrt{36} = 6, \quad t = \sqrt{27}
\]

We see that only one case gives an integer value, which is when \(t = 6\).

\[
\begin{aligned}
\varrho_a = 2, & \varrho_b = 3, \quad \varrho_c = 6. \\
\text{From (4),} \quad x = \frac{t}{\varrho_a} = 3, \quad y = \frac{t}{\varrho_b} = 2, & z = \frac{t}{\varrho_c} = 1.
\end{aligned}
\]

Thus, the sides of the triangle are

\[
a = y + z = 3, \quad b = z + x = 4, \quad c = x + y = 5
\]

The conditions are satisfied by the 3, 4, 5 unit right triangle.