Three small circles pave a hexagon

Read time - 15 minutes
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 given triangle. maybe try yourself first :)

If it is an acute-angled triangle,
then it is its circumcircle as shown below, 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 corresponding arc of the circle and they both are acute-angled. This means that radius of this circle is larger than the circumradius, due to the \(R=\frac{a}{2\sin{A}}\) 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 longest side, and we claim that this circle is the smallest. Clearly, any smaller circle cannot contain the longest side because diameter is the longest 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 above.
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 at 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 which measures \(\frac{3a}{2}\), because \(CD=a,\ BE=2a\) and the measure of midline of the trapezium \(BCDE\) is the average of them. Now, we can easily pave the hexagon with two more equal circles, considering the trapeziums \(DEFA, FABC\) as shown. Thus, the minimum radius required to cover a hexagon in this case is \(\frac{3a}{4}\). 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 (inorder to minimize the radius for encompassing \(\triangle PCQ\) and \(\triangle PDQ\), which are obtuse) with endpoints on alternate sides of the hexagon (\(P,Q,R\)) so that they contain exactly \(2\) unique midpoints and vertices as shown. Denote the midpoints of sides \(AB, BC, CD, DE, EF, FA\) as \(M_{AB},M_{BC},M_{CD},M_{DE},M_{EF},M_{FA}\) respectively. Consider trapezium \(BCDE\) which is clearly isoceles as \(BC=DE=a\) and WLOG, let points \(P, Q, R\) lie on \(BC, DE, FA\) respectively. Let the points \(P, Q\) be at distances \(x,y\) (can be negative) from the midpoints \(M_{BC}, M_{DE}\) respectively in opposite orientations as shown below. We consider \(P\) closer to the longer parallel side and \(Q\) closer to the shorter parallel side. Drop perpendicular from \(Q\) to \(BC\) to intersect \(BC, CD\) at \(X,Y\) respectively. \(\triangle YCX\) is a \(30-60-90\) triangle and \(\triangle DYQ\) is isosceles with \(\angle DYQ=\angle DQY=30^{\circ}\). Thus, \[DQ=DY=\frac{a}{2}-y\Rightarrow YC=\frac{a}{2}+y\Rightarrow XY=\frac{\sqrt{3}}{2}\left( \frac{a}{2}+y\right)\ ;\ XC=\frac{a}{4}+\frac{y}{2}\] and \(CP=\frac{a}{2}+x\). Now, \(QY=2DQ\cos{30^{\circ}}=\sqrt{3}\left(\frac{a}{2}-y\right)\). Thus, \begin{align*} QX&=\frac{3\sqrt{3}a}{4}-\frac{\sqrt{3}y}{2}\ ;\ XP=\frac{3a}{4}+x+\frac{y}{2}\\ \text{By Pythagoras}&\text{ in }\triangle PXQ,\\ PQ&=\sqrt{\left(\frac{3\sqrt{3}a}{4}-\frac{\sqrt{3}y}{2}\right)^2+\left(\frac{3a}{4}+x+\frac{y}{2}\right)^2}\\ \text{Upon simplification,}&\\ PQ&=\sqrt{x^2+y^2+\frac{3a}{2}(x-y)+xy+\frac{9a^2}{4}}\\ \text{If }x\geq y&\Rightarrow PQ\geq\sqrt{\frac{9a^2}{4}}=\frac{3a}{2} \end{align*}

Observe here that, if points \(P,Q,R\) occur mutually in opposite orientation of the midpoints with respect to the trapeziums, i.e., values of \(x,y\) are both positive for every two pairs of points, then one of the \(3\) pairs of points, must have \(x\geq y\) (just consider \(x\) as the largest among them and take the trapezium accordingly) and thus we have the corresponding length greater than (or) equal to \(\frac{3a}{2}\), completing our proof.

Suppose if the points \(P,Q,R\) do not mutually occur in opposite orientation, then some two of them must occur in such a way that \(x\) is positive and \(y\) is negative (\(P\) and \(R\) as shown below), i.e., both points are closer to the longer parallel side of the trapezium. In this case, it is clear that \[\frac{3a}{2}(x-y)>0\ ;\ x^2+y^2+xy=(x+y)^2-xy>0\Rightarrow \text{one side}>\frac{3a}{2}\]

Therefore, one of the sides of the \(\triangle PQR\) must be greater than (or) equal to \(\frac{3a}{2}\). The equality holds only if \(x=y=0\Rightarrow P,Q,R\) are the midpoints of sides \(AB, CD, EF\) respectively, which is same as the case-1 configuration of the points \(P,Q,R\).
Therefore, the minimum radius required to cover the hexagon entirely with three identical circles is \(\boxed{\frac{3a}{4}}\) as in the case-1.