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We are given a triangle $T$ on a plane $P$, with sidelengths $a$, $b$ and $c$, where $c \ge b \ge a > 0$. A straight line $L$ on $P$ is selected uniformly at random from the set of all the horizontal and vertical straight lines cutting $T$. Note that a.s. there is $1$ and only $1$ uncut side of $T$.


Question: What maximum expected length (as a function of $a$, $b$ and $c$) of the uncut side of $T$ over all possible triangles $T$ on $P$, where the expectation is taken over the random selection of $L$?

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    $\begingroup$ "What minimum expected length (as a function of a, b and c)" and "over all orientations of T on the plane and all possible choices of a, b and c" are incompatible. I assume the latter is just a typo (otherwise the story is trivial: put $a=b=c=0$), right? $\endgroup$ – fedja Jul 30 '20 at 15:47
  • $\begingroup$ Thank you fedja. I am correcting the question. $\endgroup$ – Penelope Benenati Jul 30 '20 at 15:54
  • $\begingroup$ How does the randomization between horizontal and vertical cuts arise? And what is the significance of the uncut length? Both of those are different from typical geometric setups. $\endgroup$ – Matt F. Jul 30 '20 at 16:27
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    $\begingroup$ Also, can you give an example with a numerically calculated expectation for some scalene triangle? It looks complicated no matter how you do it. $\endgroup$ – Matt F. Jul 30 '20 at 16:42
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    $\begingroup$ Seconding what Matt F is saying — the question 'makes sense' in the sense that it's a quantity one can calculate for any given triangle easily enough, and probably for any orientation of the given triangle with some effort. But understanding why you're trying to maximize this quantity may help with giving more guidance and avoiding XY problems. $\endgroup$ – Steven Stadnicki Jul 30 '20 at 16:46
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Let the vertices of the triangle be $A$, $B$, and $C$, which we also use for the angle measures, opposite the sides of lengths $a$, $b$ and $c$ respectively.

Suppose we know that in the ideal configuration, a horizontal line cuts the triangle at $A$, and a vertical line cuts the triangle at $B$. Let $\theta$ be the angle between the horizontal line and side $AB$.

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Then the measure of the possible set of horiztonal lines is $c \sin(\theta)+b \sin(A - \theta)$, and the expected length of the side uncut by horizontal lines is $$h=\frac{bc \sin(\theta)+bc \sin(A - \theta)}{c \sin(\theta)+b \sin(A - \theta)}$$

Similarly the measure of the possible set of vertical lines is $c\sin(\pi/2-\theta)+a \sin(B-(\pi/2-\theta))$, or $c\cos(\theta)-a\cos(B+\theta)$, and the expected length of the side uncut by horizontal lines is $$v=\frac{ac\cos(\theta)-ac\cos(B+\theta)}{c\cos(\theta)-a\cos(B+\theta)}$$

So the problem asks to maximize $\ell=(h+v)/2$ over all possible $\theta$. [Update: that was for the version of the problem with equal likelihood of horizontal and vertical lines. On one interpretation of the current version, $\ell$ might instead be the sum of the numerators of the above $h$ and $v$, divided by the sum of their denominators. In any case:] The maximum has a closed-form expression, but it would be unilluminating.

So the full procedure here is:

  • solve for $A$, $B$, and $C$.
  • find $\theta$ and $\ell$ assuming that the horizontal line cuts $A$ and the vertical line cuts $B$ as above
  • find $\theta$ and $\ell$ in the other eight cases for which vertex is cut by the horizontal line and which vertex is cut by the vertical line
  • see which of these nine possibilities for $\theta$ and $\ell$ actually represent geometrically sensible arrangements
  • choose the geometrically sensible possibility with highest $\ell$.
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  • $\begingroup$ Thank you very much for your answer Matt. There are other similar ways to obtain a list of cases whose analysis would provide the final solution. I found right now a solution method with only three cases. Anyway, the problem is always that, for at least one of these cases, the maximization is not trivial (I also tried to use Mathematica, which seems to take too much time - perhaps it's my fault because I just started to use it, but I am not sure). Hence, the problem here is how to have a small number of manageable expressions which can all be maximized feasibly, to get the final answer. $\endgroup$ – Penelope Benenati Aug 1 '20 at 11:45
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This is also too long for a comment but it shows where the real problem lies.

With the new formulation, the complementary expectation of the sum of two cut sides is just of the form $$ \frac {\sum_i a_i^2(|\cos\theta_i|+|\sin\theta_i|)}{\max_i(a_i|\cos\theta_i|)+\max_i(a_i|\sin\theta_i|)} $$ where $a_i$ are the sides and the angles $\theta_i$ that the sides $a_i$ make with one of the axis can be thought of as the angle $\theta$ by which the triangle is rotated plus some fixed offsets. Now, as long as no axis is parallel to one of the sides, both the numerator and the denominator are just linear combinations of $\sin\theta$ and $\cos\theta$, i.e., are proportional to $\cos(\theta+\rm{something})$. Shifting $\theta$ by that something in the denominator, we see that our function is of the form $\alpha\frac{\cos(\theta+\xi)}{\cos\theta}=\alpha(\cos\xi-\sin\xi\tan\theta)$. Since $\tan\theta$ is monotone between singularities, we conclude that the extremal coordinate systems are just the ones with one axis parallel to one of the sides.

The real question is then "to which side?". I suspect that the maximum of the original expectation is attained when one of the axes is parallel to the largest side but I'm not sure yet.

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  • $\begingroup$ Thank you very much fedja, this sounds principled and very useful. $\endgroup$ – Penelope Benenati Aug 1 '20 at 14:10
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A bit too long for a comment, here's one suggestion: take $a\lt b\lt c$ and use $(0,0)$ and $(c,0)$ as two points of the triangle. The third point $(x,y)$ (taking $y\gt 0$ WLOG) can be found in whatever usual way you prefer. Now, instead of rotating the triangle, rotate the lines: we can parametrize the pencils of lines as being in the directions $(\cos\theta, \sin\theta)$ and $(-\sin\theta, \cos\theta)$ for $0\leq\theta\leq\frac\pi2$ (by symmetry). Finding the EV in question is a straightforward if annoying calculation, and then maximizing it over $\theta$ should be similarly straightforward. One problem with this approach — and a problem I suspect is innate to the question — is that the answer is likely to be sensitive to specific conditions/regions on $a,b,c$; it's hard to pin down a better one, but this parametrization feels very 'unnatural' for answering questions that are somewhat innately coordinate-centric.

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  • $\begingroup$ Thank you Steven. I was thinking about something similar. I think the main problem here is to find a very concise way to avoid complicate calculations through an analysis by cases, by finding a very small number of distinct conditions/cases to simplify the calculation. $\endgroup$ – Penelope Benenati Jul 30 '20 at 19:17
  • $\begingroup$ All the "symmetric" cases should be grouped together to avoid redundant calculations. $\endgroup$ – Penelope Benenati Jul 30 '20 at 19:35

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