$ABCD$ is a square side $1$. $P$ and $Q$ lie on the side $AB$ and $R$ lies on the side $CD$. What are the possible values for the circumradius of $PQR$?
1. **Identify the positions of points \(P\), \(Q\), and \(R\):**
   - \(P\) and \(Q\) lie on side \(AB\) of the square \(ABCD\), which has side length 1.
   - \(R\) lies on side \(CD\) of the square.

2. **Determine the shortest possible circumradius:**
   - The shortest possible circumradius occurs when \(P\) is arbitrarily close to \(Q\) and \(RQ\) is perpendicular to \(AB\).
   - Let \(P\) and \(Q\) be very close to each other on \(AB\), say \(P = (x, 1)\) and \(Q = (x + \epsilon, 1)\) where \(\epsilon\) is a very small positive number.
   - Let \(R = (x, 0)\) on \(CD\).
   - The triangle \(PQR\) is almost a right triangle with \(PQ \approx \epsilon\) and \(QR = 1\).
   - The circumradius \(R\) of a right triangle with legs \(a\) and \(b\) is given by \(R = \frac{\sqrt{a^2 + b^2}}{2}\).
   - Here, \(a = \epsilon\) and \(b = 1\), so \(R \approx \frac{\sqrt{\epsilon^2 + 1}}{2}\).
   - As \(\epsilon \to 0\), \(R \to \frac{1}{2}\).
   - Therefore, the lower bound of the circumradius is \(\frac{1}{2}\) exclusive.

3. **Determine the largest possible circumradius:**
   - The largest possible circumradius occurs when \(P\) is diametrically opposed to \(R\) and \(RQ\) is perpendicular to \(AB\).
   - Let \(P = (0, 1)\), \(Q = (1, 1)\), and \(R = (1, 0)\).
   - The triangle \(PQR\) is a right triangle with legs \(PQ = 1\) and \(QR = 1\).
   - The hypotenuse \(PR\) is \(\sqrt{1^2 + 1^2} = \sqrt{2}\).
   - The circumradius \(R\) of a right triangle with hypotenuse \(c\) is given by \(R = \frac{c}{2}\).
   - Here, \(c = \sqrt{2}\), so \(R = \frac{\sqrt{2}}{2}\).
   - Therefore, the upper bound of the circumradius is \(\frac{\sqrt{2}}{2}\) inclusive.

4. **Conclusion:**
   - The range of possible values for the circumradius \(R\) of triangle \(PQR\) is \(\left(\frac{1}{2}, \frac{\sqrt{2}}{2}\right]\).

The final answer is \(\boxed{\left(\frac{1}{2}, \frac{\sqrt{2}}{2}\right]}\).