## Task 2 - 281232

Given a point $A$ in a plane $\epsilon$ and a length $a$.

Determine the set of all points $C$ in $\epsilon$ for which there exist points $B$ and $D$ such that $A B C D$ is a parallelogram with $A B=a$ and $A C: A B=B D: A D$.
Let $A$ be at the origin and point $B$ on the $x$-axis at $(a, 0)$. Point $C$ has coordinates $(x, y)$ and point $D$ is at $(x-a, y)$. The following should hold:

$$
\frac{\sqrt{x^{2}+y^{2}}}{a}=\frac{\sqrt{(a-(x-a))^{2}+y^{2}}}{\sqrt{(x-a)^{2}+y^{2}}}
$$

We square and multiply by the denominators:

$$
\left(x^{2}+y^{2}\right)\left((x-a)^{2}+y^{2}\right)=a^{2}\left((x-2 a)^{2}+y^{2}\right)
$$

$$
\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}-2 a x+a^{2}\right)=a^{2}\left(x^{2}+y^{2}-4 a x+4 a^{2}\right)
$$

We substitute $r^{2}=x^{2}+y^{2}$:

$$
\begin{gathered}
r^{2}\left(r^{2}-2 a x+a^{2}\right)=a^{2}\left(r^{2}-4 a x+4 a^{2}\right) \\
r^{4}-\left(2 a x-a^{2}\right) r^{2}-a^{2} r^{2}+\left(4 a x-4 a^{2}\right) a^{2}=0 \\
r^{4}-2 a x r^{2}+\left(4 a x-4 a^{2}\right) a^{2}=0 \\
r^{2}=a x \pm \sqrt{a^{2} x^{2}-4 a^{3} x+4 a^{4}} \\
r^{2}=a x \pm a \sqrt{x^{2}-4 a x+4 a^{2}} \\
r^{2}=a x \pm a(x-2 a)
\end{gathered}
$$

Therefore, there are 2 possible solutions.

1. The "minus" solution:

$$
r^{2}=a x-a x+2 a^{2}=2 a^{2}
$$

This is a circle around $A$ with radius $\sqrt{2} a$.

2. The "plus" solution:

$$
\begin{gathered}
r^{2}=x^{2}+y^{2}=2 a x-2 a^{2} \\
(x-a)^{2}-a^{2}+y^{2}=-2 a^{2} \\
(x-a)^{2}+y^{2}=-a^{2}
\end{gathered}
$$

This equation is not solvable in $\mathbb{R}$. The only solution for the locus of point $C$ is the circle around $A$ with radius $r=\sqrt{2} a$.