\section*{Problem 6B - 121246B}

In the plane, let two non-intersecting circles \(k_{1}\) and \(k_{2}\) and a point \(A\) outside both circles be given.

We are looking for an equilateral triangle \(\triangle A B C\) such that \(B\) lies on \(k_{1}\) and \(C\) lies on \(k_{2}\).

a) Justify and describe a construction of such triangles.

b) Determine the largest number that can occur as the number of such triangles \(\triangle A B C\) in cases where there are not infinitely many such triangles.
Assume there exists a \(B \in k_{1}\) and a \(C \in k_{2}\) with \(A B=B C=C A\), then I obtain the point \(C\), if \(B\) is already fixed, by rotating the segment \(A B\) around \(A\) by \(60^{\circ}\).

If I choose an arbitrary point \(B\) on \(k_{1}\) and rotate \(A B\) around \(A\) by \(60^{\circ}\), it is likely that the endpoint of the segment \(A B\) after the rotation lies on the circle \(k_{2}\). If I rotate the entire circle \(k_{1}\) around \(A\) by \(60^{\circ}\), and if there is such a triangle, then the rotated circle \(k_{1}\) intersects the circle \(k_{2}\) at a point \(C^{\prime}\) that satisfies the conditions of the problem; for the original point \(B\) of the rotation around \(A\) by \(60^{\circ}\) from \(C^{\prime}\) lies on \(k_{1}\), and it is certainly true that

\[
\angle C^{\prime} A B^{\prime}=60^{\circ} \quad \text{and} \quad C^{\prime} A=B^{\prime} A
\]

Thus, \(\triangle A B^{\prime} C^{\prime}\) is the desired triangle.

Construction: The circle \(k_{1}\) is rotated around \(A\) once in the mathematically positive direction and once in the mathematically negative direction by \(60^{\circ}\). This results in the circles \(k_{1}^{\prime}\) and \(k_{1}^{\prime \prime}\).

If points \(C\) exist with \(C \in k_{2} \cap k_{1}^{\prime}\) or \(C \in k_{2} \cap k_{1}^{\prime \prime}\), then these points are solutions to the problem. The corresponding points \(B\) are obtained by constructing the equilateral triangle over \(A C\) and the condition \(B \in k_{1}\).

Determination: The maximum number of solutions can be four equilateral triangles, and this occurs precisely when \(k_{1}^{\prime} \cap k_{2}\) and \(k_{1}^{\prime \prime} \cap k_{2}\) each contain exactly two points.

\(k_{1}^{\prime}\) and \(k_{1}^{\prime \prime}\) can each intersect \(k_{2}\) in at most two points, as it was assumed that \(k_{2}\) is not identical to \(k_{1}^{\prime}\) or \(k_{1}^{\prime \prime}\).

The case where four equilateral triangles occur is possible, as shown below (figure).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1948.jpg?height=848&width=580&top_left_y=1164&top_left_x=744)

I choose the point \(A\) on the line segment connecting \(M_{1}\) and \(M_{2}\) and will prove that \(a_{1}\) and \(a_{2}\) can be chosen such that \(M_{2} M_{1}^{\prime}0\) is found, for which \(\left(M_{1} M_{2}\right)^{2}60^{\circ}\) and \(\angle B M_{1}^{\prime} M_{1}>60^{\circ}\) also hold (by the theorem that the larger side in a triangle has the larger angle opposite it).

Thus, the sum of the angles in this triangle would be greater than \(180^{\circ}\), leading to a contradiction.

Therefore, four is the maximum number of solutions unless there are infinitely many solutions.

Adapted from \([3]\)

\subsection*{9.15 XIII. Olympiad 1973}

\subsection*{9.15.1 First Round 1973, Class 12}