Given four points on a plane, which are not on a circle and no three are on a straight line. Construct a circle that is equidistant from all four points. What is the maximum number of such circles?
Let's denote the points by $P_{1}, P_{2}, P_{3}$, and $P_{4}$, the circle that satisfies the condition by $k$, its radius by $r$, and the distance from the points by $d$. None of the four points can be outside the circle $k$, because then their distance from the center of $k$ would be $r+d$, which would mean that the four points lie on a circle, contradicting the condition. Similarly, all four points cannot be inside the circle $k$ either. Therefore, the points are positioned relative to the circle $k$ in one of the following ways: either three are inside the circle and one is outside (or vice versa), or two are inside and two are outside. Let's examine the number of possible solutions in both cases.

First, choose 3 points, since they are not collinear, they determine a circle $k_{1}$ with radius $r_{1}$. Then we need to construct a circle that is equidistant from $k_{1}$ and the fourth point. This will clearly be concentric with $k_{1}$, and its radius will be $r_{1}+d$ if $P_{4}$ is outside $k_{1}$ (Figure 1), or $r_{1}-d$ if $P_{4}$ is inside $k_{1}$, where $d$ is half the distance from $P_{4}$ to the circle $k_{1}$. The circle $k$ thus obtained satisfies the condition, as it is at a distance $d$ from each point. In this way, we get as many solutions as there are ways to choose 3 points out of 4. This gives us 4 possibilities.
![](https://cdn.mathpix.com/cropped/2024_05_02_a53213155fc713fe998eg-1.jpg?height=478&width=1064&top_left_y=682&top_left_x=516)

Figure 1

Now let's consider the case where two points lie outside the circle $k$, let these be $P_{1}, P_{2}$, and two inside, $P_{3}, P_{4}$. The distance from $P_{1}$ and $P_{2}$ to $k$ is now $r+d$, and from $P_{3}$ and $P_{4}$ it is $r-d$. The center of a circle passing through two points is always on the perpendicular bisector of the segment defined by the two points. If we draw the perpendicular bisectors of the two chosen segments, their intersection (if it exists) gives the center $O$ of the circle $k$, since $k$ is now concentric with the circles passing through $P_{1}, P_{2}$ and $P_{3}, P_{4}$, and its radius is the arithmetic mean of the distances from $O$ to the endpoints of the segments (Figure 2). The 4 points define 3 different pairs of segments, so we can get at most 3 more solutions to the problem. If the 4 points form a parallelogram, there is only one solution—unless it is a square or rectangle, but the problem statement excludes this—because the perpendicular bisectors of the parallel sides do not intersect. In the case of a trapezoid, we get 2 solutions, except if the trapezoid is symmetric, in which case the perpendicular bisectors coincide and there are infinitely many solutions. However, since a circle can be circumscribed around a symmetric trapezoid, this case is also excluded by the problem statement.

In total, therefore, there can be at most 7 solutions.

![](https://cdn.mathpix.com/cropped/2024_05_02_a53213155fc713fe998eg-1.jpg?height=526&width=572&top_left_y=1756&top_left_x=755)

József Mármarosi (Kisbér, Táncsics M. Gymnasium, 2nd grade)