Let $ABCD$ be a square with unit sides. Which interior point $P$ will the expression $\sqrt2 \cdot AP + BP + CP$ have a minimum value, and what is this minimum?
1. **Reflecting Point $P$**:
   - Let $P'$ be the reflection of $P$ along $BC$.
   - Let $P''$ be the reflection of $P$ along $AB$.

2. **Transforming the Problem**:
   - Notice that the problem of minimizing $\sqrt{2} \cdot AP + BP + CP$ can be transformed into minimizing $AP'' + BP'' + CP'$.
   - This transformation is valid because the reflections preserve distances and the sum of distances from $P$ to the sides of the square.

3. **Collinearity Condition**:
   - The minimum value will occur when $B$, $P'$, and $P''$ are collinear.
   - This implies that $\triangle AP''B \cong \triangle CP'B$.

4. **Distance Calculation**:
   - From the congruence, we have $AP'' = CP' = BP' = BP''$.
   - Since $P$ is the midpoint of the diagonal $AC$, we have $AP = CP = \frac{1}{\sqrt{2}}$.

5. **Summing the Distances**:
   - Therefore, $AP'' + BP'' + CP' = 3 \cdot \frac{1}{\sqrt{2}}$.
   - Multiplying by $\sqrt{2}$, we get $\sqrt{2} \cdot (AP'' + BP'' + CP') = \sqrt{2} \cdot 3 \cdot \frac{1}{\sqrt{2}} = 3$.

Thus, the minimum value of $\sqrt{2} \cdot AP + BP + CP$ is $3$, and it occurs when $P$ is the midpoint of the diagonal $AC$.

The final answer is $\boxed{3}$