Find the locus of points $P$ in the plane of a square $ABCD$ such that $$\max\{ PA,\ PC\}=\frac12(PB+PD).$$

[i](Anonymous314)[/i]
1. **Set up the coordinate system:**
   Let the vertices of the square \(ABCD\) be \(A(-1, 1)\), \(B(1, 1)\), \(C(1, -1)\), and \(D(-1, -1)\). Let \(P(x, y)\) be any point in the plane.

2. **Understand the condition:**
   The condition given is:
   \[
   \max\{PA, PC\} = \frac{1}{2}(PB + PD)
   \]
   This means that the maximum distance from \(P\) to either \(A\) or \(C\) is equal to half the sum of the distances from \(P\) to \(B\) and \(P\) to \(D\).

3. **Simplify the problem:**
   Note that the line \(x = y\) is the perpendicular bisector of \(AC\). Therefore, comparing \(PA\) and \(PC\) is equivalent to comparing the \(x\) and \(y\) coordinates of \(P\). If \(x < y\), then \(PA < PC\), and if \(x > y\), then \(PA > PC\).

4. **Assume \(P\) is closer to \(A\) than \(C\):**
   Without loss of generality, assume \(P\) is closer to \(A\) than \(C\), so \(x \leq y\). The condition now reads:
   \[
   PC = \frac{1}{2}(PB + PD)
   \]
   This translates to:
   \[
   2\sqrt{(x-1)^2 + (y+1)^2} = \sqrt{(x-1)^2 + (y-1)^2} + \sqrt{(x+1)^2 + (y+1)^2}
   \]

5. **Expand and simplify:**
   Expanding the distances, we get:
   \[
   2\sqrt{(x-1)^2 + (y+1)^2} = \sqrt{(x-1)^2 + (y-1)^2} + \sqrt{(x+1)^2 + (y+1)^2}
   \]
   Let \(a = x^2 + y^2 + 2\), \(b = y - x\), and \(c = x + y\). Then:
   \[
   4(a + 2b) = 2a + 2\sqrt{(a - 2c)(a + 2c)}
   \]
   Simplifying further:
   \[
   (a + 4b)^2 = a^2 - 4c^2
   \]
   Since \(a^2 \leq (a + 4b)^2\), equality must hold, implying \(c = 0\) and \(b = 0\).

6. **Solve for \(x\) and \(y\):**
   Since \(c = 0\) and \(b = 0\), we have:
   \[
   x + y = 0 \quad \text{and} \quad y - x = 0
   \]
   Solving these equations, we get \(x = 0\) and \(y = 0\).

7. **Conclusion:**
   The only point \(P\) that satisfies the given condition is the center of the square \(ABCD\), which is \((0, 0)\).

The final answer is \(\boxed{(0, 0)}\).