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If I have a triangle with sides of length a, b, c and I have a rope of length L, what is the maximum surface of a boundary I can form with that rope that is entirely inside the triangle.

Normally, if the L > a + b + c, then the surface is the surface of the triangle.

Any ideas ?

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by "maximum surface of a boundary", do you mean something like the maximum area of a convex region with the rope as its boundary? – AakashM Apr 23 '13 at 8:07
@AakashM The region does not have to be convex. The boundry can have any form. – vlad Apr 23 '13 at 8:08
so you are trying to maximize are for given perimeter for a triangle in Euclidean plane? – Santosh Linkha Apr 23 '13 at 8:14
There is no need to specify convexity, but the maximal region will of course be convex. – John Bentin Apr 23 '13 at 8:16
Interesting question! So, as long as $L$ is smaller than the length of the inscribed circle of the triangle, the maximal surface will be given by a circle. Since it is possible to explicitly give the radius of the inscribed circle in terms of $a, b, c$, we also have a lower bound for $L$ to become interesting: If $L<\alpha = \pi\sqrt{\frac{(-a+b+c)(a-b+c)(a+b-c)}{a+b+c}}$, nothing interesting happens. If $\alpha<L<a+b+c$, my gut feeling is that one obtains the desired result by 'pushing' the inscribed circle into the corners. Any thoughts on this? – HSN Apr 23 '13 at 11:09

I think HSN is on the right track. When L > a+b+c or when L can form a circle inside the triangle the problem is easy. Now for the interesting case in between:

(I think) L should cut the corners of the triangle in straight lines such that the length of these lines are equal for all three corners. The reason is that if these shortcuts have different length, one could push the longest one towards the original corner of triangle by epsilon and pull on the shorter edge and increase the area. Note that these edges should leave out Isosceles triangles at each corner (with the similar argument). Lets $e$, $f$ and $g$ denote the length of the side of these triangles and let $h$ be their base (equal for all 3).

We can write the area for the first small triangle in one corner as: $$Area_1 = \frac{1}{2} h^2 \sqrt{h^2/e^2 - 1/4}$$ and the total area surrounded by L as: $$A_{rope} = Area_{abc} - Area_{1} - Area_{2} - Area_{3}$$ here $Area_{abc}$ is the total area of triangle abc with $p = (a+b+c)/2$: $$Area_{abc} =\sqrt{p(p-a)(p-b)(p-c)} $$ On the other hand you can write L as: $$L = a+b+c+3 h - 2(e + f+g)$$ Now we can remove $e$, $f$ and $g$ from the two equations for area and $L$, and write them based on $h$ and the angles of triangle ($A,B,C$) using the cosine rule. For example: $$a^2 = b^2 + c^2 - 2bc \cos(A)$$ Here $A$ is the angle opposite to the edge $a$.

Now we solve the formula for L to obtain $h$ as a function of $L$ and substitute it in the formula for $A_{rope}$, which gives us the area for this $L$ (corresponding $h$).

The only part that is missing (except for details) is a proof that maximum area formed using the rope will form straight shortcuts. A first thought may suggest that it should be curvy (like an expanding circle) but it seems to me this is more optimal. I may be wrong. On the other hand if it is straight, our key observation that the length of these shortcuts are equal (i.e. $h$) gives us the rest.

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Let's go with @Ross' assertion that the optimal figure has equal-radius "caps" in the corners of the triangle, and derive a formula for the maximal area in terms of the triangle's perimeter and area.

Given a triangle $\triangle ABC$, with inradius $R$ and perimeter $P$, and recall that its area, $T$, must be given by $2T = RP$. Let the centers of the caps be $X$, $Y$, $Z$ (with $X$ near $A$, etc.); let the common cap radius be $r$.

Slice off the corners of $\triangle ABC$ by making cuts from the caps' centers to their points of tangency with the triangle's edges. We get three quadrilaterals (with diagonals $AX$, $BY$, $CZ$) that we can assemble into a smaller triangle similar to $\triangle ABC$ ---we stole the angles from $\triangle ABC$, after all--- such that the three "caps" form the new incircle: the smaller circle's inradius is $r$. Writing $p$ for its perimeter and $t$ for its area, we must have $$p = P \cdot \frac{r}{R} = \frac{P^2 r}{2T} \qquad t = T \cdot \left(\frac{r}{R}\right)^2 = \frac{P^2r^2}{4T}$$

Now ...

  • The area inside the smaller triangle, but outside its incircle, is given by $t - \pi r^2$. This is precisely the area of not covered by the three-capped figure in $\triangle ABC$. Therefore, the area, $K$, that is covered by the three-capped figure is $$K = T- \left(t-\pi r^2\right) = T-\frac{P^2-4\pi T}{4T} r^2$$

  • The perimeter of the smaller triangle is precisely that portion of $\triangle ABC$'s perimeter not bounding the three-capped figure. Therefore, the portion that does bound the figure is $P-p$, so that the total perimeter, $L$, of the three-capped figure is $$L = P-p+2\pi r = P-\frac{P^2-4\pi T}{2T}r$$

Solving the perimeter equation for $r$ yields

$$r = \frac{ 2T(P-L) }{P^2-4\pi T}$$

so that the maximal area is

$$K = T - \frac{ T(P-L)^2 }{P^2-4\pi T}$$

which agrees with @JohnBentin's result, since, using his angle definitions, $$P = 2 R \left( \cot \alpha + \cot \beta + \cot\gamma \right)$$

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(+1) Yes, that's neater than using the half-angles. – John Bentin Apr 25 '13 at 10:51

In the range of interest, at each corner, you want a circular arc with the same radius. Any other way of handling the corners will be less area due to the isoperimetric inequality. Let the radius of the circle be $r$ and the perimeter of the triangle be $P$. The length in contact with the sides of the triangle is then $L-2\pi r$, so the tangency points are $\frac 16( P-L+2\pi r)$ from each corner. We can find $r$ by drawing a triangle in angle $A$ from the corner to the center of the arc to the point of tangency. The sides are $\sqrt{r^2+\frac 1{36}( P-L+2\pi r)^2},r,\frac 16( P-L+2\pi r)$ and we can use the law of sines to get $\frac r{\sin \frac A2}=\frac {\frac 16( P-L+2\pi r)}{\sin (\frac \pi 2-\frac A2)}$

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I don't agree that the tangency points are $\frac16 (P−L+2πr)$ from each corner. To see why, consider a triangle of sides $1$, $1$, and $√2$, so that $P=2+√2$, and let $L=2πr$, where $r=1−1/√2$ is the radius of the incircle (make $L$ slightly bigger if you like). Then the tangency points are distant either $1/√2$ or $1−1/√2$ (or slightly less) from each corner. – John Bentin Apr 24 '13 at 19:13

This is more of an extended comment on Siamak's answer than a proper answer. Siamak says: "The only part that is missing (except for details) is a proof that maximum area formed using the rope will form straight shortcuts." Unfortunately, I don't think that such a proof will be forthcoming. For a counterexample, let's take a rectangular corner $A$ and fix units so that the length of the shortcut $MN$, where $M$ is on $AB$ and $N$ is on $AC$, is $2$ (here $|AM|=|AN|=\sqrt 2$). Choose the parabola that fits symmetrically inside the corner of the triangle touching the sides $AB$ and $AC$ at $P$ and $Q$ respectively such that the arc length of the parabola from $P$ to $Q$ equals $|PM|+|MN|+|NQ|$. According to my calculation, following this parabola from $P$ to $Q$ will extend the area enclosed by the loop by ~0.19426 square units compared with following the piecewise straight path $PMNQ$.

Edit: (Just to finish off the correct part of Ross's answer) The optimal loop may be constructed by fitting circular arcs of radius $$r=\dfrac{P-L}{2(\cot\alpha+\cot\beta+\cot\gamma-\pi)},$$where $\alpha$, $\beta$, and $\gamma$ are the semi-angles of the triangle, tangentially into the corners of the triangle. The resulting enclosed area is$$\Delta-\dfrac{(P-L)^2}{4(\cot\alpha+\cot\beta+\cot\gamma-\pi)},$$where $\Delta$ is the area of the triangle.

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This is good to know. would be even more interesting to see the sketch of calculations that shows the area does indeed increase. – Siamak May 8 '13 at 16:53
@Siamak: The calculation for the parabola is elementary, but not short, and it would be laborious to write out in $\LaTeX$. I won't bother, because the parabola is suboptimal. The optimal solution with circular arcs is well treated in Blue's answer. – John Bentin May 9 '13 at 6:06

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