Given the square $A B C D$. What is the geometric locus of points $P$ in the plane of the square for which $(P A+P C) / \sqrt{2}$ is as large as the larger of the distances $P B$ and $P D$?
We seek the geometric locus using coordinate geometry methods. Let the vertices of the square be $A (0; 1)$, $B(-1; 0)$, $C(0; -1)$, and $D(1; 0)$. Thus, we examine the question for a square with side length $\sqrt{2}$, but this does not affect the generality, since any two squares are similar.

The sought geometric locus is symmetric with respect to both axes. If $P(x, y)$ is a suitable point, then $P(-x, y)$ is also suitable, because in this case, the values of $P B$ and $P D$ merely swap, and this does not change the value of the right-hand side of the requirement

$$
\frac{P A + P C}{\sqrt{2}} = \max (P B, P D)
$$

Moreover, the value of the left-hand side remains unchanged if $P$ is replaced by $(x, -y)$. Therefore, it suffices to consider the case $x \geq 0$, in which the value of the right-hand side is

$$
P B = \sqrt{(x+1)^{2} + y^{2}}
$$

Thus, (1) becomes

$$
\sqrt{x^{2} + (y-1)^{2}} + \sqrt{x^{2} + (y+1)^{2}} = \sqrt{2} \cdot \sqrt{(x+1)^{2} + y^{2}} \quad x \geq 0
$$

After squaring both sides and discarding the common terms,

$$
\sqrt{x^{2} + (y-1)^{2}} \cdot \sqrt{x^{2} + (y+1)^{2}} = 2 x
$$

This can be rearranged after another squaring:

$$
x^{4} + x^{2}(2 y^{2} + 2) + (y^{2} - 1)^{2} = 4 x^{2}
$$

which simplifies to

$$
x^{4} + 2 x^{2}(y^{2} - 1) + (y^{2} - 1)^{2} = (x^{2} + y^{2} - 1)^{2} = 0
$$

and finally

$$
x^{2} + y^{2} - 1 = 0 \quad x \geq 0
$$

The obtained equation represents the right half of the unit circle centered at the origin, including the points on the $y$-axis. According to the above, every suitable point $P$ lies on the circumcircle of the initial square.

To prove the converse, we use elementary geometric reasoning to show that every point on the circumcircle satisfies $P$ as required. It suffices to examine the points on the $D A$ quarter of the circle. If $P$ is at the endpoints $D$ or $A$, then both sides of (1) are 2 and $\sqrt{2}$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_19f115c02b1f0f70805cg-2.jpg?height=1252&width=1138&top_left_y=136&top_left_x=488)

The length $P A / \sqrt{2}$ is represented by the segments $P K_{1} = A K_{1}$, where $K_{1}$ is the center of the square constructed outwardly on the leg $P A$ of the right triangle $P A C$. In other words, $P K_{1}$ and $A K_{1}$ are the legs of the isosceles right triangle $K_{1} P A$. Clearly, $P K_{1}$ is perpendicular to $P B$, and $K_{1} A$ is parallel to $P B$.

Similarly, $P C / \sqrt{2} = P K_{2} = K_{2} C = K^{*} A$, where $K_{2}$ is the center of the square constructed on the side $P C$ and $K^{*}$ is the reflection of $K_{2}$ over the origin, and the line $K_{1} K_{2}$ passes through $P$ and $D$ and is perpendicular to $P B$. We can also say that $K^{*}$ is the reflection of $K_{2}$ over the line $O K_{1}$, since the angles $K_{1} O K_{2} \angle = A P C \angle$ are right angles. Then,

$$
\frac{P A}{\sqrt{2}} + \frac{P C}{\sqrt{2}} = P K_{1} + P K_{2} = K_{1} K_{2} = K_{1} K^{*} = P B
$$

because the reflection implies that $K^{*} B$ is parallel to $K_{2} D$, and thus perpendicular to $P B$, since $B P D \angle$ is a right angle.