Given a circle $k$, a point $P$, and a segment $d$ with its direction. Construct a triangle $ABC$ inside $k$ such that the side $AB$ is parallel and equal to $d$, and the angle bisector of $\angle ACB$ passes through $P$.
The side $AB$ of the sought triangle will be a chord of length $d$ of the circle $k$. It is known that the set of lines (geometric locus) that cut out chords of a given length $d$ from a given circle $k$ are the tangents of a circle $k'$ concentric with $k$ - provided that $d$ is less than the diameter $2r$ of $k$. For now, we only examine this case. In this case, the suitable positions of $AB$ are cut out from $k$ by the tangents of $k'$ parallel to $d$.

On the other hand, in a given circle, equal inscribed angles have equal arcs between their sides, so the sides of the two equal parts of the angle $ACB=\gamma$ bisected by the line $PC=f$ are also equal. Since one of the arcs $AB$ of $k$ lies between the sides of $\gamma$, $f$ must pass through the midpoint $F$ of the corresponding arc $AB$. Therefore, $f$ can only be the line connecting $F$ and $P$.

Based on this, the construction is as follows: We get a point of $k'$ if we construct a chord $LM=d$ of arbitrary length in $k$ and its midpoint $N$; from this, the radius of $k'$ is $KN$, where $K$ is the center of $k$. Then we construct the line perpendicular to the direction of $d$ through $K$, which intersects $k'$ at $E$ and $k$ at $F$. The tangent to $k'$ at $E$ cuts out $A, B$ from $k$, and the line $PF$ also cuts out $C$ from $k$. We get two points for both $E$ and $F$, so the same applies to the pairs of points $A, B$ and $C$, hence the number of solutions is at most $2 \cdot 2=4$. (In the situation of Figure 1, there are 4 solutions: $A_{1} B_{1} C^{\prime}$, $A_{1} B_{1} C^{\prime \prime}$, $A_{2} B_{2} C^{\prime}$, $A_{2} B_{2} C^{\prime \prime}$.)

![](https://cdn.mathpix.com/cropped/2024_05_02_d6f942acc6b9a7f6a5a1g-1.jpg?height=694&width=547&top_left_y=732&top_left_x=759)

Figure 1

The triangle $ABC$ only satisfies the conditions if the line $PF$ intersects the segment $AB$ at an internal point of $AB$, in other words: if $P$ is either inside the interior of the angle domain $AFB$ with $K$, or inside the interior of the angle domain of the opposite angle to $AFB$. The number of solutions is already smaller if one or more of the lines $FA$, $FB$ determined by the points $F, A, B$ pass through $P$. (Figure 2 - which, for clarity, does not show the omitted $d$ - shows that the number of solutions when $P$ is inside one of the regions of the plane divided by the 4 lines $FA$, $FB$ is as follows. The 8 lines are pairwise parallel, and 4 lines pass through each of the $F$s, so the number of plane regions is 27.)

![](https://cdn.mathpix.com/cropped/2024_05_02_d6f942acc6b9a7f6a5a1g-1.jpg?height=740&width=819&top_left_y=1847&top_left_x=642)

Figure 2

If $d=2r$, then the only possible position of $AB$ is the diameter of $k$ parallel to $d$, so the number of solutions is at most 2. Finally, if $d>2r$, there are no solutions.

András Pór (Budapest, Corvin Mátyás g. II. o. t.)

[^0]:    ${ }^{1}$ See, for example, Lajos Vigassy: Geometric locus of lines, Example 2. K. M. L. 20 (1960) p. 82.