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I need to find the minimal area of a convex hexagon with all of its sides 1 unit away from a point. Here's what I've done so far.

I think this is equivalent to asking for the minimal area of a hexagon circumscribed around a circle with radius 1, since then all of its sides will be 1 unit away from the center of the circle. Intuitively, it looks like a regular hexagon would minimize the area, and playing around in GeoGebra seems to support this. However, I'm having trouble proving this result formally.

I've tried two things so far:

One, starting with a regular hexagon and moving one of the points of tangency along the circle. If we overlap the regular hexagon with this new hexagon, there is a triangle of added area and a triangle of lost area. I've tried to show that the lost area is greater than the gained area, but I haven't been able to find a way to get the area of either triangle.

Second, I split the hexagon into 12 triangles. In a regular hexagon, each triangle would have an area of $\frac{\tan 30 ^{\circ}}{2} = \frac{\sqrt{3}}{6}$. I tried to show that if we change the central angle, the area will decrease, but this gave me the opposite of what I wanted - if the angle is $45^{\circ}$, for example, the area of the triangle would be $\frac{1}{2}$, which is greater than the previous result.

I'm a high school student and haven't taken calculus yet, so please try to give a proof using only trigonometry or other elementary methods, if that is possible. However, if there is a more elegant proof using more advanced concepts, please point me to any relevant resources.

Dan
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agous
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  • For this question, does the 1-unit distance mean the perpendicular distance to the extended line segment? Or alternatively if the foot of altitude from the point to the extended line is outside the line segment, then the 1-unit is the distance to the nearest endpoint? – peterwhy Dec 18 '23 at 05:07
  • @peterwhy I should have clarified - the distance is the perpendicular distance from the point to the extended line, even if the intersection falls outside the line segment. – agous Dec 18 '23 at 05:13
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    If you split the hexagon into six triangles, each of them formed by one (tangential) edge and the center, then they all have the same height, namely the radius of the circle. Therefore their areas will be proportional to the lengths of the edges, and the sum of areas proportional to the sum of edges. So you can reduce the problem to minimizing the circumference, which may or may not be a simpler problem to tackle. It is also something you might find good explanations for in a web search, if you can't make progress on your own, as optimality of honeycombs is discussed frequently. – MvG Dec 18 '23 at 07:35
  • I wish you hadn't edited convexity into this question. The original version was much more interesting! – TonyK Dec 18 '23 at 23:03
  • @TonyK Without the requirement of convexity, wouldn't the minimum area just be 0, as in Prem's answer? – agous Dec 18 '23 at 23:08
  • I like Prem's answer! Be honest $-$ would you have thought of it yourself? – TonyK Dec 19 '23 at 01:17

2 Answers2

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We will show that the hexagon of minimum area is the regular hexagon.

enter image description here

In the diagram, vertices $A$ and $B$ are variable. Assume the other vertices $F,G,H,I$ are fixed. That is, $2\theta$ is variable, and $2\alpha$ is fixed.

We will show that $\text{Area}_{\text{green+blue}}$ is minimized when $\theta=0$.

$\begin{align} \text{Area}_{\text{green}}&=\text{Area}_{\triangle ADO}+\text{Area}_{\triangle ACO}-\text{Area}_{\text{minor sector }CDO}\\ &=\frac12 \tan(\alpha-\theta)+\frac12 \tan(\alpha-\theta)-\frac12(2\alpha-2\theta)\\ &=\tan(\alpha-\theta)-(\alpha-\theta) \end{align}$

Similarly, $\text{Area}_{\text{blue}}=\tan(\alpha+\theta)-(\alpha+\theta)$.

$\therefore \text{Area}_{\text{green+blue}}=\tan(\alpha-\theta)+\tan(\alpha+\theta)-2\alpha$

We need to show that $\tan{(\alpha-\theta)}+\tan{(\alpha+\theta)}$ is minimized when $\theta=0$. Without calculus, as requested.

Let:
$a=\tan(\alpha-\theta)$
$b=\tan(\alpha+\theta)$

$\tan(2\alpha)=\tan((\alpha-\theta)+(\alpha+\theta))=\dfrac{a+b}{1-ab}$

$1-\dfrac{a+b}{\tan(2\alpha)}=ab\le\left(\dfrac{a+b}{2}\right)^2$ ( the inequality is because $0\le\left(\frac{a-b}{2}\right)^2=\left(\frac{a+b}{2}\right)^2-ab$ )

Multiply by $4\tan(2\alpha)$ and rearrange.

$4\tan(2\alpha)(a+b)^2 +(a+b)-4\tan(2\alpha)\ge0$

Solving this quadratic inequality gives $a+b\ge2\tan\alpha$.

That is, $\tan{(\alpha-\theta)}+\tan{(\alpha+\theta)}\ge2\tan\alpha$. Note that equality is reached when $\theta=0$.

Thus, $\tan{(\alpha-\theta)}+\tan{(\alpha+\theta)}$ is minimized when $\theta=0$.

So $\text{Area}_{\text{green+blue}}$ is minimized when $\theta=0$, that is, when the angles at vertices $A$ and $B$ are equal.

By repeating this argument with the other pairs of neighboring vertices of the hexagon, we see that the hexagon of minimum area is the regular hexagon.

This implies that the minimum area of the hexagon is $2\sqrt3$.

Dan
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  • How did you get the areas of the orange, green, and blue regions? Also, why does showing that $A_{\triangle ABC}$ is maximized when $\theta = 0$ show that the area of the hexagon is minimzed under that condition? – agous Dec 18 '23 at 20:01
  • @agous For example, consider the orange+green+blue region. Let $D$ be the point where $AB$ extended touches the circle, and let $E$ be the point where $AC$ extended touches the circle. Let the centre of the circle be $O$. $AD=AE=\tan(2\alpha)$. So right triangle $ADO$, and right triangle $AEO$, each have area $\frac12 \tan(2\alpha)$. So quadrilateral $ADEO$ has area $\tan(2\alpha)$. Minor sector $ADE$ has area $2\alpha$. So the area of the orange+green+blue region is $\tan(2\alpha)-2\alpha$. – Dan Dec 18 '23 at 22:22
  • @agous As for your second question in your comment, $BC$ is a variable side of the hexagon; the other vertices of the hexagon are fixed. When $A_{\triangle ABC}$ is maximized, we have poisitioned $BC$ so that the area of the hexagon is minimized. – Dan Dec 18 '23 at 22:22
  • I think I get it- since the area of the orange + green + blue region is always the same, when the orange region ($A_{\triangle ABC}$) is maximized the blue + green region is minimized. Since this is the part that contributes to the area of the hexagon, then the area of the hexagon is minimized. Is my understanding correct? – agous Dec 18 '23 at 22:54
  • @agous Yes, you got it! – Dan Dec 18 '23 at 22:55
  • Also, I might be mistaken, but why does $\tan(2 \alpha) = \tan ((\alpha - \theta)-(\alpha + \theta))$? Shouldn't $\tan ((\alpha - \theta)-(\alpha + \theta)) = - \tan (2 \theta)$? – agous Dec 18 '23 at 23:02
  • Let us continue this discussion in chat. – agous Dec 18 '23 at 23:04
  • @agous I have editted my answer. I hope the new version is more clear. (Some of your comments may now be obsolete, so you may consider deleting them to avoid confusing future readers.) – Dan Dec 19 '23 at 03:51
2

BEFORE OP INCLUDED CONVEXITY :

"[A] I need to find the minimal area of a hexagon with all of its sides 1 unit away from a point [...] [B] I think this is equivalent to asking for the minimal area of a hexagon circumscribed around a circle [...] [C] Intuitively, it looks like a regular hexagon would minimize the area [...] [D] having trouble proving this result formally [...]"

[A] It is a tough one. Elementary Calculus might be essential. I think , Calculus of Variations involving Differential Equations & trigonometry might be required here.

[B] It is not Equivalent to Circumscribed Circle Case.
Image given shows a "Smaller Area" when Hexagon is outside the Circle.

[C] Intuition seen going in wrong Direction. I think Circumscribed Circle Case might give Maximal Area & regular Hexagon might give non-optimally minimum area.

[D] It might involve Calculus of Variations & might not be trivial.

Image is highlighting the Area of "Non-Convex Hexagon" outside the Circle with the Area much less than the Area of Circle.
I used tangents to the Circle , hence all the Hexagon Sides are at Constant Distance from the Center of the Circle.

HEXAGON

When I rotate couple of tangents this way or that way , I think I can arbitrarily reduce the Hexagon Area. I think , minimal area is almost ZERO , like the Black Area in this Image , where the Arrow Directions indicate where the tangents move :

HEXAGON

AFTER OP INCLUDED CONVEXITY :

To be Convex , the Hexagon has to be around the Center of the Circle & it has to be around the Circle , which indicates that the minimum area of the Hexagon must exceed area of the Circle.

Every Point of tangency will always occur within each side of the Hexagon.

[[ OP Comment to user peterwhy "the distance is the perpendicular distance from the point to the extended line, even if the intersection falls outside the line segment" will never occur in Convex Case ]]

HEXAGON

Let the angles at the Center be $\theta_1,\cdots,\theta_6$ , where $\Sigma \theta_n = 2 \pi$

Area of Hexagon is $A = \Sigma [ 2 \times [1 \times \tan(\theta_n/2)/2] ]$
Hence Area is minimum when $\Sigma \tan(\theta_n/2)$ is minimum.
OBSERVATION : Area is maximum ($\infty$) , when one $\theta$ angle is $\pi$ , giving Parallel lines.

When $\theta_n$ angles are not Equal , we can consider reducing the largest angle while increasing the smallest angle by same amount , keeping the Angle Summation Constant.
Intuitively , that will reduce the Area Summation because the decrease [ in the larger $\tan()$ ] is larger than the increase [ in the smaller $\tan()$ ] to give reduced total.
Proof can be made rigorous via combinatoric analysis on the Summation.
By Jensen Inequality with Convex $\tan()$ , we have $\Sigma \tan(\theta_n/2) \ge 6 \tan ( [ \Sigma \theta_n/2 ] / 6 )$
With $\Sigma \theta_n / 2 = \pi$ , we get :
$6 \tan ( \pi / 6 ) \le \Sigma \tan(\theta_n/2)$ to get the minimum Area.

ADDENDUM :

(1) When using Jensen Inequality , we have to show that $\tan()$ is convex within $(0,\pi/2)$
Proof is elementary & uses AM-GM Inequality with the expansion of $\tan(A+B)$

Details :
https://planetmath.org/convexityoftangentfunction

(2) Jensen Inequality works on Convex functions to compare $f(A+B)$ with $f(2A)$ & $f(2B)$ , where $f()$ must be Convex & $\tan()$ in indeed Convex.

Details :
https://en.wikipedia.org/wiki/Jensen%27s_inequality
Minimize the sum of tangents when sum of angles are constants

Prem
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