A triangle has sides of integer lengths, the radius of its inscribed circle is one unit. What is the radius of the circumscribed circle?
The area of the triangle $t=r \cdot s$, where $r$ is the radius of the inscribed circle, and $s$ is the semiperimeter. Since $r=1$, we have $t=s$. Using Heron's formula:

$$
s=\sqrt{s \cdot(s-a) \cdot(s-b) \cdot(s-c)}
$$

which simplifies to

$$
s=(s-a)(s-b)(s-c) \cdot(1) \text{ Hence } 8 s=(2 s-2 a)(2 s-2 b)(2 s-2 c) \cdot(2)
$$

Since $a, b, c$ are integers, $2 s$ is also an integer, and for example, $2 s-2 a$ has the same parity as $2 s$. The same can be said about every factor on the right side of (2). Considering that the left side of (2) is even, the factors on the right side must also be even, so the integers $x=s-a, y=s-b, z=s-c$ are all integers. Given that $s-a+s-b+s-c=s$, equation (1) becomes:

$$
x+y+z=x \cdot y \cdot z
$$

We can assume that among $x, y, z$, $z$ is the largest, and then

$$
x+y+z \leq 3 z \cdot(4) H a
$$

If $x, y, z$ were all at least 2, then $x \cdot y \cdot z \geq 4 z .(5) From (3), (4), and (5), we get 3z \geq 4z$, which is a contradiction for $z>0$. Therefore, at least one of $x, y, z$ must be 1. Assuming $x=1$, from (3) we get $1+y+z=y \cdot z$, or $(y-1)(z-1)=2$. Since $y$ and $z$ are positive integers and $z \geq y, y=2$ and $z=3$. Thus, $s=6$ and $a=5, b=4, c=3$.

By the converse of the Pythagorean theorem, the triangle is a right triangle, and by Thales' theorem, the radius of the circumscribed circle is half the length of the hypotenuse, which is $\frac{5}{2}$.

It can be easily verified that the radius of the inscribed circle of a triangle with sides $3, 4, 5$ is indeed 1 unit.