Let $RICE$ be a quadrilateral with an inscribed circle $O$ such that every side of $RICE$ is tangent to $O$. Given taht $RI=3$, $CE=8$, and $ER=7$, compute $IC$.
1. Let $X$, $Y$, $Z$, and $W$ be the points of tangency on sides $RI$, $IC$, $CE$, and $ER$ respectively.
2. By the property of tangents from a point to a circle, we have:
   - $RX = RW$
   - $IY = IX$
   - $CZ = CY$
   - $EW = EZ$
3. Let $RX = RW = a$, $IY = IX = b$, $CZ = CY = c$, and $EW = EZ = d$.
4. The lengths of the sides of the quadrilateral can be expressed in terms of these segments:
   - $RI = RX + IX = a + b = 3$
   - $IC = IY + CY = b + c$
   - $CE = CZ + EZ = c + d = 8$
   - $ER = EW + RW = d + a = 7$
5. According to Pitot's Theorem, for a quadrilateral with an inscribed circle, the sum of the lengths of opposite sides are equal:
   \[
   RI + CE = IC + ER
   \]
6. Substituting the given values:
   \[
   3 + 8 = IC + 7
   \]
7. Solving for $IC$:
   \[
   11 = IC + 7
   \]
   \[
   IC = 11 - 7
   \]
   \[
   IC = 4
   \]

The final answer is $\boxed{4}$