3. Find all right-angled triangles with integer side lengths, whose inscribed circle has a radius of 2.

(Jaroslav Zhouf)
Solution. Let the right-angled triangle we are looking for be denoted as $ABC$ such that the right angle is at vertex $C$, and the lengths of its sides are denoted as $|BC|=a$, $|AC|=b$, and $|AB|=c$. Additionally, let $I$ be the center of the inscribed circle, and $T, U$, and $V$ be the points of tangency with the sides $BC, AC$, and $AB$ respectively (Fig. 1).

![](https://cdn.mathpix.com/cropped/2024_04_17_b2b0e89c199ea3581fafg-6.jpg?height=722&width=580&top_left_y=627&top_left_x=744)

Fig. 1

Quadrilateral $CUIT$ has right angles at vertices $C, T$, and $U$, and from the problem statement, we know that $|IT|=|IU|=2$, so it is a square. From the equality $|CT|=2$, we have $|BT|=a-2$ and from the equality of the segments of tangents from vertex $B$ to the inscribed circle, we have $|BV|=|BT|=a-2$. Similarly, we obtain the equalities $|AV|=|AU|=b-2$. The length of the hypotenuse $AB$ can thus be expressed both as $|BV|+|AV|=(a-2)+(b-2)=a+b-4$ and from the Pythagorean theorem as $|AB|=\sqrt{a^{2}+b^{2}}$. From this, we get the equation, which we square and then simplify step by step:

$$
\begin{aligned}
|AB|=a+b-4 & =\sqrt{a^{2}+b^{2}},\left.\quad\right|^{2} \\
(a+b-4)^{2} & =a^{2}+b^{2}, \\
a^{2}+b^{2}+2ab-8a-8b+16 & =a^{2}+b^{2}, \\
ab-4a-4b+8 & =0, \\
(a-4)(b-4) & =8 .
\end{aligned}
$$

The number 8 can be factored into the product of two integers as

$$
8=8 \cdot 1=4 \cdot 2=(-1) \cdot(-8)=(-2) \cdot(-4) \text {. }
$$

Without loss of generality, we can assume that $a \geq b$ or $a-4 \geq b-4$. Then we have the following four possibilities (we are looking for only positive solutions, as they represent the lengths of the sides of the triangle): ${ }^{2}$[^0]$\triangleright a-4=8$ and $b-4=1$, from which $a=12, b=5, c=a+b-4=13 ;$

$\triangleright a-4=4$ and $b-4=2$, from which $a=8, b=6, c=a+b-4=10$;

$\triangleright a-1=-1$ and $b-4=-8$, from which $a=0, b=4$, which does not satisfy the problem;

$\triangleright a-1=-2$ and $b-4=-4$, from which $a=-1, b=0$, which does not satisfy the problem.

Finally, we need to verify that the inscribed circle of both right-angled triangles with side lengths 5, 12, 13 and 6, 8, 10 indeed has a radius of length 2. If we write the radius $r$ in Fig. 1 everywhere instead of the number 2, we get the equality $c=a+b-2r$, from which the formula $r=\frac{1}{2}(a+b-c)$ follows, according to which for both triples $(5,12,13)$ and $(6,8,10)$, we indeed get $r=2$.

Answer. The sought triangle has side lengths 5, 12, 13 or 6, 8, 10.