String partitioning to enclose minimum total area

Question

Three strings totaling a length $U= 3 a + 4b + 2 \pi r$ cut into three parts together enclose minimum total area

$$ A= \frac{\sqrt3 a^2}{4} + b^2+\pi r^2,$$

when they are made into shapes of an equilateral triangle, square and circle respectively.

Please help determine $ (a,b,r)$ divisions in the case of these three polygons.

When there are only a square and a circle it is noted that the figures can be be drawn enclosed between parallels. Geometrical arrangement.

In the above it has been verbally indicated that first pairwise Lagrange Multiplier

$$\dfrac{ \dfrac{\partial{U}}{\partial{b}}}{ \dfrac{\partial{U}}{\partial{r}}}= \dfrac{ \dfrac{\partial{A}}{\partial{b}}}{ \dfrac{\partial{A}}{\partial{r}}}=\lambda$$

could be taken to determine $(b,r)$ relation:

$$ b=2r \tag1$$

Next $(a,r)$ pairwise relation is determined in a similar manner:

$$ \frac{a}{\sqrt3 }=2r \tag2$$

Idea was that after determining $ (a,b,r)$ we could check if there would be such pattern/regularity here as well for three figures of minimum total area (equilateral triangle,square,circle) in parallel line packing.

However it turns out that for an equilateral triangle base $a$ the ratio of altitudes

$$ \dfrac{\dfrac{a}{\sqrt3 }}{\dfrac{a \sqrt3 }{2} }= \dfrac32 $$

So no pattern is observed, it being seen that the arrangement with square/circle does not repeat with equilateral triangle and circle, further generalization for regular polygons with higher number of vertices $n=5,6,7..\infty \,$ &c..do not hold good on basis of total height of triangle. This conclusion has changed in view of the later observations.

On further examination the common top line for square and circle is found to match with the diameter of incircle $ 2 r_I=\dfrac{a}{\sqrt3}$ as shown and this common feature is further investigated for invariance finding for general regular polygons set... in the next answer area.

Answer 1

If i correctly understand the question, we consider the function $A$ defined on $I^3=[0,\infty)^3$ with values in $I=[0,\infty)$, defined by $$ A(a,b,r) = \frac{\sqrt3}4a^2+b^2+\pi r^2\ , $$ then fix a constant $U$, and want to minimize $A$ on the intersection of $I^3$ with the hyperplane given by the equation $$ C(a,b,r) = 3a+4b +2\pi r-U\ .$$ So we consider the function $F=F(a,b,r,\lambda)$ given by $$ \begin{aligned} F(a,b,r,\lambda) &= A(a,b,r) - \lambda C(a,b,r) \\ &= \frac{\sqrt3}4a^2+b^2+\pi r^2 -\lambda(3a+4b +2\pi r-U) \ . \end{aligned} $$ A local extremal value on the $2$-dimensional variety $C=0$ is a point $(a,b,r)$ in the interior of the triangle $C=0$ intersected with $I^3$, so that for a suitable $\lambda$ we have $F'(a,b,c,\lambda)=0$. This is a system of algebraic equations in the four variables $a,b,r,\lambda$: $$F'_a=F'_b=F'_r=F'_\lambda=0\ . $$ Explicitly: $$ \left\{ \begin{aligned} \frac{\sqrt 3}2a &= 3\lambda \ ,\\ 2b &= 4\lambda \ ,\\ 2\pi r &= 2\pi \lambda\ ,\\[3mm] 3a+4b +2\pi r &= U\ . \end{aligned} \right. $$ We obtain $$ U = 6\sqrt 3 \lambda + 8\lambda + 2\pi\lambda\ , $$ which determines $\lambda$, and from this $$ \frac U{6\sqrt 3 + 8 + 2\pi}=\lambda = \frac{\sqrt 3}6a=\frac 12 b=r\ . $$ It can be shown that this point is indeed the point where we have an absolute minimum for $A$.

The quick way to figure out this absolute minimum is by applying Cauchy-Schwarz: $$ \begin{aligned} \left( \frac {\sqrt 3}4a^2+b^2+\pi r^2 \right) \left( 3^2\cdot\frac 4{\sqrt 3} +4^2+2^2\pi \right) &\ge (3a+4b+2\pi r)^2 \\ &=U^2\ . \end{aligned} \tag{$CS$} $$ And the case of equality is also easily extracted from the above relation $(CS)$.

Now the relation $b=2r$, holding for the minimal point, allows to draw for this minimal point the square of side $b$, and the circle of diameter $r$ "between" the same two parallels, but the triangle does not fit between them.

The case with further regular polygons can be considered in the same manner, either using Lagrange multipliers, or directly the adapted $(CS)$ inequality.

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I typed the most of the above yesterday, and even if the question has lost the sun shine in the mean time, i completed the answer, and will submit it now...

Answer 2

Generalization

$ r $ in-radius, $ R $ circum-radius.

We try finding an invariant for minimum total area of regular polygons made out of a given string length minimizing total area for a general polygon pair.

$$ U= 2nR \sin \frac{\pi}{n}+ 2 \pi r;\, A = n \frac{R^2}{2}\sin\frac{2\pi}{n}+ \pi r^2$$

$$\dfrac{ \dfrac{\partial{U}}{\partial{R}}}{ \dfrac{\partial{U}}{\partial{r}}}= \dfrac{ \dfrac{\partial{A}}{\partial{R}}}{ \dfrac{\partial{A}}{\partial{r}}}$$

$$ \dfrac{2n\sin \dfrac{\pi}{n}}{2 \pi}=\dfrac{nR\sin \dfrac{2 \pi}{n}}{2 \pi R}$$ that simplifies to $$ R=\dfrac{r} {\cos \frac{\pi}{n}} $$

and we have it getting simplified to in-radius

$$ R \cos \frac{\pi}{n}= r = r_{I} $$

which is the constant common (half) distance sought between the parallels by Lagrange evaluation and this is found to be independent of n for any arbitrary pair. In other words the same parallels fit between them an equilateral triangle and a regular octagon pair, for example. Or for that matter between two concentric circle of radial difference $2r_{I}=d. $

Since all regular polygons are related to the circle this way, it means that we can combine arbitrarily any number of polygons to have a common invariant in-circle diameter or parallel distance $d$ among all or any of them.

Lemma

If a string length L is partitioned into an arbitrary number of segments each forming a polygon $ (n,m,p,...,\infty) $ boundary then the total enclosed area is minimised with a common distance $d$ as common circum-diameter $ 2r_{I}$ that should satisfy:

$$ d=2r_I =\dfrac{L}{ n \tan \dfrac{\pi}{n} +m \tan \dfrac{\pi}{m} + p \tan \dfrac{\pi}{p}+..\pi} $$

where the last $\pi$ term is for the circle $ \lim_{z\rightarrow \infty\,} z\tan \dfrac{\,\pi}{z}\rightarrow \,\pi $

To consider a numerical example from above relation...

If a string 100 units length is desired to be divided into 5 parts to enclose minimum total area among : an equilateral triangle, regular pentagon, hexagon,octagon and a circle, then these must be enclosed/fitted between parallel lines 5.33383 units apart and should be cut to approximate lengths ( 27.7153, 19.3763,18.4769,17.6747, 16.7567 ) units. The parallel lines spaced at $d$ are tangent to each of the regular polygon inscribed circles as shown for a different example: