Given the circle $O$ and the point $A$ in its plane. Construct the chord $M M^{\prime}$ of the circle, which is seen at a right angle from $O$ and $A$. Is the construction possible regardless of the position of point $A$?
All chords $M M^{\prime}$ of the circle $O$ with radius $r$, which are seen from $O$ at a right angle, are equal to the side of the inscribed square $(r \sqrt{2})$; the distance of these chords from $O$ is $\frac{1}{2} r \sqrt{2}$, and their midpoints lie on the circle $\gamma$ with radius $\frac{1}{2} r \sqrt{2}$ centered at $O$.

![](https://cdn.mathpix.com/cropped/2024_05_02_04316aef4f00f310dc1ag-1.jpg?height=441&width=423&top_left_y=289&top_left_x=840)

Now let $M M^{\prime}$ be such a chord, but also seen from $A$ at a right angle. Thus, $M 0 M^{\prime} A$ is a cyclic quadrilateral in a circle whose center is $\omega$; this center is equidistant from $O$ and $A$, and thus lies on the perpendicular bisector $B y$ of $O A$.

Accordingly, we can find $\omega$ and thus $M M^{\prime}$ by finding the common points of the circle $\gamma$ and the perpendicular bisector $B y$ of $O A$.

Such a common point exists if $O B \leqq O \omega$, that is, if $O A=a$,

$$
\frac{a}{2} \leqq \frac{1}{2} r \sqrt{2}, \quad \text { i.e., } \quad a \leqq r \sqrt{2}
$$

If $a < r \sqrt{2}$, there are two common points and thus two solutions (in a symmetric position relative to $O A$). $M M^{\prime}$ is tangent to the circle $\gamma$ at $\omega$ $\left(M M^{\prime} \perp O \omega\right)$.

If $a = r \sqrt{2}$, there is one common point and thus one solution. $\left(M M^{\prime} \perp O A\right)$.

If $a = 0$, that is, if $A$ and $O$ coincide, there are infinitely many solutions.

Béla Schreiber (izr. g. VIII. o. Bp.).

Note. The point $\omega$ is the common point of the circles with radius $\frac{1}{2} r \sqrt{2}$ drawn from $O$ and $A$.

$O A$ bisects the angle $M A M^{\prime}$, so $O A M \angle = O A M^{\prime} \angle = 45^{\circ}$. The construction can also be carried out based on this.