## Task 23/82

In a plane, let two points $A$ and $B$ be given. Using only a compass, construct two points $C$ and $D$ such that $A B C D$ is a square with side $A B=a$.
The task is solved when it has been succeeded to construct the segment $AC = a\sqrt{2}$. Then, namely, the point $C$ is found as the intersection of the circles around $A$ with $a\sqrt{2}$ and around $B$ with $a = AB$ as radii, and the point $D$ as the intersection of the circles around $A$ and $C$ with $a = AB$ as radius, which does not coincide with $B$. Except for the direction of rotation, the solution is thus unique.

To find $AC = a\sqrt{2}$, we perform an auxiliary construction.

Starting from $B$, we construct auxiliary points $B_2, B_3, B_4$, and $B_5$ such that $BB_2 = B_2B_3 = B_3B_4 = B_4B_5 = AB = a$ and $B_2, B_3, B_4, B_5$ lie in this order on the line through $A$ and $B$. This is possible by constructing equilateral triangles $ABA_1, A_1BA_2, A_2BB_2, A_2B_2A_3$, $A_3B_2B_3, A_3B_3A_4, A_4B_3B_4, A_4B_4A_5, A_4B_4B_5$. Then we draw

1. the circle $K_1$ around $A$ with $r_1 = AB_5 = 5a$
2. the circle $K_2$ around $B_3$ with $r_2 = AB_4 = 4a$
3. the circle $K_3$ around $B_4$ with $r_3 = AB_3 = 3a$.

One of the intersection points of $K_1$ and $K_2$ is $P$, and the intersection point of $K_1$ and $K_3$ that lies in the same half-plane as $P$ is $Q$.

Claim: $PQ = a\sqrt{2}$

Proof: We imagine $A$ as the origin of a right-angled Cartesian coordinate system, whose x-axis coincides with the line through $A$ and $B$ and which is oriented such that $P$ and $Q$ have positive coordinates.

Then $P$ has the coordinates $x_P = 3a, y_P = 4a$ (since $AP = 5a$, $\triangle AB_3P$ is right-angled at $B_3$), and $Q$ has the coordinates $x_Q = 4a, y_Q = 3a$ (for the same reason, $\triangle AB_4Q$ is right-angled at $B_4$). According to the formula for the distance between two points, we then have

$$
PQ = \sqrt{(4a - 3a)^2 + (3a - 4a)^2} = \sqrt{a^2 + (-a)^2} = a\sqrt{2}
$$

Thus, the task is solved.