The vertices of a deltoid are $A(0 ; 0), B(1 ; 2), C(4 ; 0), D(1 ;-2)$. Consider those circles whose circumference is equidistant from all four vertices. What is the radius of the largest circle?
We can easily see that a circle cannot be circumscribed around a deltoid, that is, it is not a cyclic quadrilateral. Indeed, if it were a cyclic quadrilateral, then by symmetry, this would mean that there is a right angle at points $B$ and $D$. However, this is not true. The direction tangent of line $AB$ (easily readable from the coordinates) is 2, while that of line $BC$ is $-2/3$, meaning the two lines are not perpendicular to each other. Therefore, there is no circle that contains all four points inside it (or outside it) and is equidistant from them. If such a circle existed, then by scaling down (or up) from the center, the circle would pass through all four vertices of the deltoid.

Thus, only the following is possible:

(a) 3 points are in one (the inner) and 1 point is in the other (the outer) part of the plane, or

(b) 2 points are inside the circle, and 2 points are outside the circle.

In case (a), we have three possibilities:

1. if point $A$ is outside,
2. if point $B$ is outside (by symmetry, this is the same as if point $D$ were outside),
3. if point $C$ is outside.

In the first case, we first find the equation of the circle passing through points $B, C, D$. The coordinates of the center of the circle are $O_{1}\left(\frac{11}{6} ; 0\right)$ and the radius is $r=\frac{13}{6}$. It is clear that point $A$ is inside the circle, since $O_{1} A=\frac{11}{6}<\frac{13}{6}$. We need to shrink this circle from point $O_{1}$ so that it is equidistant from all four points. This is satisfied when

$$
r_{1}=\frac{\frac{11}{6}+\frac{13}{6}}{2}=2
$$

units are chosen as the radius of the circle.

In the second and third cases, the calculations can be performed similarly; here we only provide the results.

In the second case, the coordinates of the center of the circle are $O_{2}\left(2 ;-\frac{1}{4}\right)$, and the radius is $r_{2}=\frac{\sqrt{65}+\sqrt{97}}{8} \approx 2.23889$.

In the third case, $r_{3}=2$. (The coordinates of $O_{3}$ are $\left(\frac{5}{2} ; 0\right)$)

(b) Finally, there are 2 cases where the circle separates the points.

Let the separated points be $A B$ and $C D$. Write the equation of the perpendicular bisector of each segment. The intersection of these will be equidistant from $A$ and $B$, and from $C$ and $D$. Calculate these distances.

The coordinates of the midpoint of segment $A B$ are $\left(\frac{1}{2} ; 1\right)$, the direction tangent of line $AB$ is 2, and the equation of the perpendicular bisector $f_{1}$ is $y-1=-\frac{1}{2}\left(x-\frac{1}{2}\right)$; the coordinates of the midpoint of segment $C D$ are $\left(\frac{5}{2} ;-1\right)$, the direction tangent of line $CD$ is $\frac{2}{3}$, and the equation of the perpendicular bisector $f_{2}$ is $y+1=-\frac{3}{2}\left(x-\frac{5}{2}\right)$. The intersection of $f_{1}$ and $f_{2}$ has coordinates $O_{4}\left(\frac{3}{2} ; \frac{1}{2}\right)$.

$$
\overline{O_{4} A}=\overline{O_{4} B}=\sqrt{\left(\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{\sqrt{10}}{2}, \quad O_{4} C=O_{4} D=\sqrt{\left(\frac{5}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{\sqrt{26}}{2}
$$

If we draw a circle around $O_{4}$ with radius $r_{4}=\frac{\frac{\sqrt{10}}{2}+\frac{\sqrt{26}}{2}}{2} \approx 2.065$, it will be equidistant from all four points, with $A, B$ inside the circle and $C, D$ outside. (By axial symmetry, the separation of $A, D$ and $B, C$ gives the same result).

Finally, let $B, C$ and $A, D$ be the separated point pairs, and similarly write the intersection of the perpendicular bisectors of the segments. The result is

$$
O_{5}(2 ; 0) \quad \text { and } \quad r_{5}=\frac{2+\sqrt{5}}{2} \approx 2.118
$$

We can see that the largest radius is $r_{2}=2.23889$; this was obtained when point $B$ was outside the circle, and points $A, C, D$ were inside.

Méder Áron (Budapest, Táncsics M. Gimn., II. o.t.)

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