Find the minimum possible area of a hexagon circumscribed about the unit circle in \(\mathbb R^2\).
Solution
Let the six sides be tangent to the unit circle at points \[ T_1,\ldots,T_6 \] in cyclic order, and let \[ \alpha_i \] be the positive central angle from \(T_i\) to \(T_{i+1}\), with indices taken modulo \(6\).
1. One has\[ 0<\alpha_i<\pi \]for every \(i\), and\[ \sum_{i=1}^6\alpha_i=2\pi. \]
The tangency points occur in cyclic order around the circle, so the six positive central gaps sum to a full angle \(2\pi\).
If some gap were at least \(\pi\), the two corresponding consecutive tangent lines would be parallel or would meet on the side opposite the bounded intersection containing the circle. A bounded circumscribed polygon therefore has every consecutive gap strictly less than \(\pi\).
2. Let \(V_i\) be the vertex where the tangent lines at \(T_i\) and\(T_{i+1}\) meet. Then the tangent segment from \(V_i\) to either tangencypoint has length\[ \tan\frac{\alpha_i}{2}. \]
The two tangent segments from \(V_i\) to the circle have equal length. Hence the line from the center \(O\) to \(V_i\) bisects the angle between the radii \(OT_i\) and \(OT_{i+1}\). In the right triangle \[ OV_iT_i, \] the angle at \(O\) is \(\alpha_i/2\), the adjacent leg \[ OT_i \] has length \(1\), and the opposite leg is \(V_iT_i\). Therefore \[ V_iT_i = \tan\frac{\alpha_i}{2}. \]
3. The perimeter \(P\) of the hexagon is\[ P = 2\sum_{i=1}^6 \tan\frac{\alpha_i}{2}. \]
2. Every side of the hexagon is the union of one tangent segmentfrom each of its two endpoint vertices. Thus summing all side lengthscounts exactly two tangent segments at every vertex, giving the formula.
4. The area \(A\) of the hexagon is\[ A = \sum_{i=1}^6 \tan\frac{\alpha_i}{2}. \]
Join the center of the unit circle to all six vertices. This decomposes the hexagon into six triangles, each having one side of the hexagon as base and altitude equal to the inradius \(1\). Hence \[ A = \frac12P. \] Apply step 3.
5. The function\[ \tan x \]is strictly convex on\[ (0,\pi/2). \]
Its second derivative is \[ 2\sec^2x\tan x, \] which is strictly positive on that interval.
6. One has\[ A \geq 6\tan\frac{\pi}{6} = 2\sqrt3. \]
By step 1, \[ 0<\frac{\alpha_i}{2}<\frac{\pi}{2} \] and \[ \sum_{i=1}^6\frac{\alpha_i}{2} = \pi. \] Jensen’s inequality for the strictly convex function from step 5 gives \[ \frac16 \sum_{i=1}^6 \tan\frac{\alpha_i}{2} \geq \tan\left( \frac16\sum_{i=1}^6\frac{\alpha_i}{2} \right) = \tan\frac{\pi}{6}. \] Use step 4 and \[ \tan(\pi/6)=\frac1{\sqrt3}. \]
7. Equality is attained by the regular hexagon circumscribed about theunit circle.
For the regular circumscribed hexagon, the six tangency points are equally spaced, so \[ \alpha_i=\frac{\pi}{3} \] for every \(i\). Every Jensen inequality in step 6 is then an equality, giving area \(2\sqrt3\).
8. The minimum possible area is\[ \boxed{2\sqrt3}. \]
Step 6 gives the lower bound and step 7 realizes it.
9. Q.E.D.
Step 8 answers the problem.