## Task 21/75

Given is the square $M_{1} M_{2} M_{3} M_{4}$ with side length 2a. In the plane of the square lies an arbitrary point $P$. Let $l_{i}$ denote the distance $P M_{i}$ $(i=1 ; 2 ; 3 ; 4)$.

1. The set of points $P$ is sought for which

$$
\sum_{i=1}^{4} l_{i}^{4}=4 n a^{4}
$$

where $n$ is a non-negative real number.

2. For which smallest $n$ is the sought set not empty?
3. The set is to be specified concretely for $n=13$ and for $n=24$.
The square is assumed, without loss of generality, to lie in a right-angled Cartesian coordinate system such that the center is at the origin and the vertices lie on the axes. The coordinates of the vertices are then

$$
M_{1}(a \sqrt{2} ; 0) \quad ; \quad M_{2}(0 ; a \sqrt{2}) \quad ; \quad M_{3}(-a \sqrt{2} ; 0) \quad ; \quad M_{4}(0 ;-a \sqrt{2})
$$

Let the coordinates of the point $\mathrm{P}$ be $P(x ; y)$. Then it follows (where for brevity $a \sqrt{2}=u$ is set)

$$
\begin{gathered}
\sum_{i=1}^{4} l_{i}^{4}=\left[(x-u)^{2}+y^{2}\right]^{2}+\left[x^{2}+(y-u)^{2}\right]^{2}+\left[(x+u)^{2}+y^{2}\right]^{2}+\left[x^{2}+(y+u)^{2}\right]^{2}= \\
=\left[x^{2}-2 x u+u^{2}+y^{2}\right]^{2}+\left[x^{2}+y^{2}-2 y u+u^{2}\right]^{2}+\left[x^{2}+2 x u+u^{2}+y^{2}\right]^{2}+\left[x^{2}+y^{2}+2 y u+u^{2}\right]^{2}
\end{gathered}
$$

Each of the square brackets contains the term $x^{2}+y^{2}+u^{2}$. Setting this to $\mathrm{w}$, we get

$$
\sum_{i=1}^{4} l_{i}^{4}=(w-2 x u)^{2}+(w-2 y u)^{2}+(w+2 x u)^{2}+(w+2 y u)^{2}
$$

from which, after a short calculation,

$$
\sum_{i=1}^{4} l_{i}^{4}=4 w^{2}+8 u^{2}\left(x^{2}+y^{2}\right)=4 w^{2}+8 u^{2}\left(w-u^{2}\right)=4 w^{2}+8 u^{2} w-8 u^{4}=4 n a^{4}
$$

follows. The solution of the resulting quadratic equation

$$
w^{2}+2 u^{2} w-2 u^{4}-n a^{4}=0
$$

is (since $w \geq 0$)

$$
w=\sqrt{3 u^{4}+n a^{4}}-u^{2}=x^{2}+y^{2}+u^{2}
$$

with $u=a \sqrt{2}$

$$
x+y=\sqrt{12 a^{4}+n a^{4}}+4 a^{2}=a^{2}(\sqrt{12+n}-4)
$$

The sought set of points is thus the circle around the coordinate origin with the radius $r=a \sqrt{\sqrt{12+n}-4}$. For $r$ to be real, it is necessary and sufficient that $n \geq 4$. The smallest $n$ is therefore $n=4$; for this value, $r=0$, so $P(x ; y)=P(0 ; 0)$ is the only element of the set.

For $n=13$, $r=a$, which is the incircle, and for $n=24$, $r=a \sqrt{2}$, which is the circumcircle of the square.