## Task 21/81

Given an equilateral triangle $A B C$ in a plane. Determine the set of all points $P$ in the plane of the triangle for which $A P^{2}+B P^{2}=C P^{2}$.
We choose a right-angled Cartesian coordinate system such that the points $A, B$, and $C$ have the following coordinates: $A(-0.5 ; 0) ; B(0.5 ; 0), C(0 ; 0.5 \sqrt{3})$.

Obviously, then $A B=B C=C A=1$. The (variable) point $P$ has the coordinates $P(x ; y)$. Then we have

$$
A P^{2}=(x+0.5)^{2}+y^{2} \quad ; \quad B P^{2}=(x-0.5)^{2}+y^{2} \quad ; \quad C P^{2}=x^{2}+(y-0.5 \sqrt{3})^{2}
$$

and thus

$$
A B^{2}+B P^{2}=2 x^{2}+2 y^{2}+0.5=C P^{2}=x^{2}+y^{2}-y \sqrt{3}+0.75
$$

From this, it follows that

$$
x^{2}+y^{2}+y \sqrt{3}-0.25=0 \rightarrow x^{2}+(y+0.5 \sqrt{3})^{2}=1
$$

The set of points we are looking for is therefore a circle with the center $M(0 ;-0.5 \sqrt{3})$ and the radius $r=1$. The points $A$ and $B$ lie on the circle, and the center $M$ is symmetric to the point $C$ with respect to $A B$.