In an equilateral triangle, we draw circles around each vertex such that none of the circles intersect each other or the opposite side of the triangle. How should we choose the radius of each circle so that the three circles together cover the largest possible part of the triangle?
Solution: Let the side of the equilateral triangle $ABC$ be of unit length, and let the radii of the three circles be denoted by $R_{a}, R_{b}$, and $R_{c}$, where $R_{a} \geq R_{b} \geq R_{c}$. The area of the part of the triangle covered by the circles is one-sixth of the total area of the circles. The radius of the largest circle is not less than $\frac{1}{2}$, half the side of the triangle, and not greater than $\frac{\sqrt{3}}{2}$, the height of the triangle. $R_{b}$ and $R_{c}$ are at most $1-R_{a}$, otherwise the circles centered at $B$ and $A$, and the circles centered at $C$ and $A$ would intersect. The area of the covered regions can be increased by setting the radii $R_{b}$ and $R_{c}$ to $1-R_{a}$. In this case, the two circles with radius $1-R_{a}$ touch the circle centered at $A$, but do not touch each other or the opposite sides, since their radius is not greater than $\frac{1}{2}$. The total area thus covered is:

$$
T=\frac{\pi}{6}\left(R_{a}^{2}+R_{b}^{2}+R_{c}^{2}\right)=\frac{\pi}{6}\left(R_{a}^{2}+\left(1-R_{a}\right)^{2}+\left(1-R_{a}\right)^{2}\right)
$$

After squaring and rearranging, we obtain the following quadratic expression:

$$
T=\frac{\pi}{6}\left(3 R_{a}^{2}-4 R_{a}+2\right)
$$

This is maximized when the expression in parentheses is maximized. We need to find the value of $R_{a}$ in the interval $\left[\frac{1}{2} ; \frac{\sqrt{3}}{2}\right]$ where the value of $3 R_{a}^{2}-4 R_{a}+2$ is the largest. The extremum of the quadratic expression can be easily determined by completing the square:

$$
3\left(R_{a}-\frac{2}{3}\right)^{2}+\frac{2}{3}
$$

From this, it is clear that the maximum is obtained at the endpoint of the closed interval $\left[\frac{1}{2} ; \frac{\sqrt{3}}{2}\right]$ that is farthest from $\frac{2}{3}$. This value is $\frac{\sqrt{3}}{2}$, the height of the triangle. The covered area is thus maximized when one of the circles is as large as possible, touching the opposite side of the triangle and the other two circles. The radii of the circles are then $\frac{\sqrt{3}}{2}, 1-\frac{\sqrt{3}}{2}, 1-\frac{\sqrt{3}}{2}$.

Remark. László Fejes Tóth, the founder of the Hungarian school of discrete geometry, investigated the densest packing of congruent circles in the plane. Interesting problems arise when considering the packing of circles of different radii. The arrangement obtained in our problem is also an optimal solution for several packing problems. For example, the densest packing of two circles of different radii in the plane. A similar problem: consider a "water molecule" in the plane, where the two smaller atoms are in the same proportion to the larger one as the circles in our problem; what will be the density of the densest packing of such molecules, and what will be the structure of the packing?[^0]

[^0]: ${ }^{1}$ For a memorial to László Fejes Tóth, who recently passed away, see page 258 of our journal.