Three small circles pave a hexagon

Consider a regular hexagon, i.e., a polygon with all 6 sides equal and all 6 angles equal, as shown above. Your task is very simple - Cover the entire hexagon with 3 identical circles of minimum possible radius. What is the value of that minimum radius, in terms of the side length of the hexagon? Let the side length be \(a\). Try to guess the answer before proceeding!

At first, let's find the radius of the smallest circle that completely contains a triangle. Well, maybe try yourself first :)
If it is an acute-angled triangle,
then it is its circumcircle as shown, because by considering any other circle containing the triangle, we can translate and rotate the triangle internally in the circle to contain two of its vertices on the circle and the third angle not on the circle will be larger than the angle subtended by the side on the corresponding arc of the circle and they are acute-angled. This means that radius of this circle is larger than the circumradius, due to the \(\frac{a}{2\sin{A}}\) formula, contradicting our requirement of minimality. If it is an obtuse-angled triangle,
the above argument does not hold true as the angles being obtuse, the value of radius would indeed compare in the opposite direction. So, consider the circle with diameter as the side opposite to obtuse angle, which is the largest side, and we claim that this circle is the smallest. Clearly, any smaller circle cannot contain the largest side because diameter is the largest line segment inside a circle. Now, the circle with diameter as the side opposite to obtuse angle, contains all the points on which the side subtends an obtuse angle and hence it contains the obtuse-angled vertex of the triangle as well and thus the entire triangle as shown.
Observe that these circles are uniquely determined, given the triangle. So, there is no dilemma in its determination.

Notice that there are \(6\) midpoints of sides of the hexagon. Now, there are two possibilities for the \(3\) circles to pave the hexagon, either each of these three circles contain exactly \(2\) unique midpoints (or) there exist a circle containing least \(3\) midpoints.

Case 1 - There exist a circle containing at least \(3\) midpoints
Let us assume that these \(3\) midpoints belong to consecutive sides to optimize the radius of the circle that is encompassing them, say the midpoints of \(BC, CD, DE\). The triangle formed by these \(3\) points is obtuse, with the measure of obtuse angle being \(120^{\circ}\) and hence the smallest circle encompassing these three points is the circle with diameter as the side opposite to the obtuse angle and that measures \(\frac{3a}{2}\) by basic trigonometry on the measure of midline of the trapezium \(BCDE\). Now, we can easily pave the hexagon with two more equal circles, considering the trapeziums \(DEFA, FABC\) as shown. Case 2 - Each of the three circles contain exactly \(2\) unique midpoints
In this case, we consider that none of these circles contain more than \(2\) vertices of the hexagon as otherwise it would entail a circle larger than the previous case-1 circle. Hence, each of these circles contain exactly \(2\) unique midpoints and vertices and ofcourse, they must be consecutive, as otherwise it will be larger than the case-1 circle. So such configuration of paving circles is formed by choosing the diameter (just to minimize the radius) with endpoints on alternate sides of the hexagon (\(P,Q,R\)) so that they contain exactly \(2\) unique midpoints and vertices as shown. It is clear upon inspection that the length of \(PQ\) (or) \(QR\) (or) \(RP\) would be larger than \(\frac{3a}{2}\) as some two of them would lie on alternate halves of lateral sides of the trapezium formed by the sides of hexagon. Hence, it is larger than the circle obtained in case-1 and thus rendering not optimal.

Therefore, the \(3\) identical circles with smallest possible radius covering a regular hexagon of side length \(a\) is as shown in case-1 and the value of the least radius is \(\frac{3a}{4}\).