Given a circle, a line passing through its center, and a point $P$ in the plane of the circle.

1. Construct the point $Q$ on the given circle such that $Q Q^{\prime}=P Q^{\prime}$, where $Q^{\prime}$ is the orthogonal projection of the point $Q$ onto the given line.
2. What is the geometric locus of those points $P$ for which the problem is solvable?
I. Solution: We solve the problem using coordinate geometry. Let the center $O$ of the given circle of radius $r$ be the origin of the coordinate system, and the given line be the $X$-axis. Let the coordinates of the given point $P$ be $(u, v)$, and the coordinates of the point $Q$ on the circle be $\left(x_{1}, y_{1}\right)$, then the coordinates of the projection of $Q$, $Q^{\prime}$, are $\left(x_{1}, 0\right)$. Since $Q$ lies on the circle, it satisfies the equation:

$$
x_{1}^{2}+y_{1}^{2}=r^{2}
$$

and since $P Q^{\prime}=Q Q^{\prime}$, we have

$$
\left(u-x_{1}\right)^{2}+v^{2}=y_{1}^{2}
$$

From (1), solve for $y_{1}^{2}$ and substitute into (2), then rearrange and solve the resulting quadratic equation for $x_{1}$:

$$
\begin{gathered}
u^{2}-2 u x_{1}+x_{1}^{2}+v^{2}=r^{2}-x_{1}^{2} \\
2 x_{1}^{2}-2 u x_{1}+u^{2}+v^{2}-r^{2}=0 \\
x_{1}=\frac{2 u \pm \sqrt{4 u^{2}-8\left(u^{2}+v^{2}-r^{2}\right)}}{4}=\frac{u}{2} \pm \frac{1}{2} \sqrt{2 r^{2}-2 v^{2}+u^{2}}
\end{gathered}
$$

The distance $x_{1}$ (and thus the point $Q$) can now be constructed using the Pythagorean theorem.

The problem is solvable if the expression under the square root in the solution for $x_{1}$ is non-negative:

$$
2 r^{2}-2 v^{2}-u^{2} \geq 0
$$

or, dividing by $-2 r^{2}$,

$$
\frac{u^{2}}{2 r^{2}}+\frac{v^{2}}{r^{2}} \leq 1
$$

From this, it is clear that the points $(u, v)$ for which the construction is possible lie on or inside a central ellipse with axes aligned with the coordinate axes, whose semi-major axis is $r \sqrt{2}$ and semi-minor axis is $r$.

The number of solutions for $x_{1}$ is 2, 1, or 0, depending on whether the discriminant of the quadratic equation for $x_{1}$ is positive, zero, or negative, i.e., depending on whether $P$ lies inside, on, or outside the ellipse. For each $x_{1}$, there are two $y_{1}$ values, so the number of points $Q$ that can be constructed for a given $P$ is 4, 2, or 0 (each pair is symmetric with respect to the $X$-axis).

Miklós Horváth (Kaposvár, Táncsics g. III. o. t.)

II. Solution: Consider the problem as solved (see the diagram).

![](https://cdn.mathpix.com/cropped/2024_05_02_737a6c2c0a4fbce48dfag-1.jpg?height=468&width=703&top_left_y=1582&top_left_x=711)

In the right triangle $S Q R$, the height $Q Q^{\prime}$ is the geometric mean of the segments on the hypotenuse:

$$
\left(Q Q^{\prime}\right)^{2}=R Q^{\prime} \cdot Q^{\prime} S
$$

Since the problem states that $P Q^{\prime}=Q Q^{\prime}$, we have

$$
\left(P Q^{\prime}\right)^{2}=R Q^{\prime} \cdot Q^{\prime} S
$$

Draw a circle around the triangle $P R S$. (This is always possible if $P, R, S$ form a triangle; if $P$ lies on the diameter $R S$, the construction of $Q$ is simple, as $P, Q^{\prime}, Q$ form a 45° isosceles right triangle.) Let the line $P Q^{\prime}$ intersect this circle at point $Q^{\prime \prime}$. By the theorem of intersecting chords:

$$
R Q^{\prime} \cdot Q^{\prime} S=P Q^{\prime} \cdot Q^{\prime} Q^{\prime \prime}
$$

Comparing this with (1), we see that

$$
Q^{\prime} Q^{\prime \prime}=P Q^{\prime}
$$

Based on this, the construction is as follows. First, construct the point $Q^{\prime \prime}$ on the circumcircle of triangle $P R S$, for example, by measuring the distance $P S$ from $S$ on the extension of $P S$, and drawing a line parallel to $R S$ from the resulting point to intersect the circle at $Q^{\prime \prime}$. Connecting this to $P$ and drawing a perpendicular at the intersection with the diameter $R S$ gives the points $Q$ that satisfy the problem.

The correctness of the construction can be verified by reversing the above reasoning.

For a given point $P$, we get four solutions if the line parallel to $R S$ intersects the circle at two points, two solutions if it is tangent (since one solution point $Q$ and its reflection across the diameter $R S$ are both solutions), and no solutions if it does not intersect.

Let's examine for which points $P$ the problem has a solution. A solution is possible if the line parallel to $R S$ at least touches the circle. Since the tangent parallel to $R S$ touches the circle at the point $E$ where the angle bisector of $\angle R P S$ intersects the circle, the line is tangent for points $P$ such that $P T = T E$. Therefore, the problem is solvable if $P T \leq T E$. However, by a known theorem about the product of segments of a chord, $P T \cdot T E = R T \cdot T S$, so the solvability condition is:

$$
P T^{2} \leq R T \cdot T S
$$

In problem 698 (KML XII., 3. sz. 78-81. p.), we proved the following equality for the angle bisector of a triangle: the square of the angle bisector from a vertex is equal to the difference between the product of the sides meeting at the vertex and the product of the segments on the opposite side. Thus, the solvability condition becomes:

$$
P T^{2}=P R \cdot P S - R T \cdot T S \leq R T \cdot T S
$$

Rearranging gives:

$$
\frac{P R}{R T} \cdot \frac{P S}{T S} \leq 2
$$

Since $P T$ is the angle bisector, we have:

$$
\frac{P R}{R T}=\frac{P S}{T S}=\frac{P R+P S}{R S}
$$

The condition can be written as:

$$
\left(\frac{P R+P S}{R S}\right)^{2} \leq 2
$$

or

$$
P R+P S \leq R S \cdot \sqrt{2}
$$

Thus, the problem has a solution for points $P$ whose sum of distances from $R$ and $S$ is not greater than the constant value $R S \cdot \sqrt{2}$. The condition is satisfied by the points on or inside the ellipse with foci at $R$ and $S$, semi-major axis $\frac{R S}{2} \sqrt{2}$, and semi-minor axis $\sqrt{\frac{R S^{2}}{2}-\frac{R S^{2}}{4}}=\frac{R S}{2}$. We have thus arrived at the same result as in the first solution.

József Németh (Esztergom, Ferences g. III. o. t.)