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...such that

  1. There's no overlapping
  2. No more such triangles can be added without overlapping.

Let $r$ be, on average, the ratio of the area covered by triangles with respect to the area which is not.

Q1: What is the minimum/infimum value of $r$?

Q2: If we use squares instead of equilateral triangles, how does $r$ change?

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  • $\begingroup$ Does overlapping of edges or vertices count or only interiors? $\endgroup$
    – Ross Millikan
    Oct 31 at 2:28
  • $\begingroup$ @RossMillikan Only interiors. $\endgroup$
    – Eric
    Oct 31 at 2:36
  • $\begingroup$ Can't both equilateral triangles and squares tile the plane? $\endgroup$
    – Someone
    Oct 31 at 12:16
  • $\begingroup$ @Someone The question is asking for the minimum area that can be covered such that no additional triangles can be placed without overlap. $\endgroup$
    – Daniel Mathias
    Oct 31 at 12:53
  • $\begingroup$ Oh, that makes more sense. $\endgroup$
    – Someone
    Oct 31 at 13:09

3 Answers 3

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Upper bound for the infimum:

$$r = \frac {\sqrt{3}} {2(\sqrt{2} + \sqrt{3})} = \frac 3 2 - \sqrt \frac 3 2 \approx 0.28$$ Correction: By OP's definition the value of interest is $$R=\frac r {1-r} = \frac {2 \sqrt 6 -3} 5 \approx 0.38$$ Thanks to @Kris Van Bael for spotting the mistake.

Pattern:

enter image description here

The red triangle is not part of the pattern but explains how rows are spaced.

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    $\begingroup$ By the OP’s definition of R, your solution is approx .38. $\endgroup$
    – Kris Van Bael
    Nov 1 at 9:19
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Triangle upper bound (worse than loopy walt's)

$r = 8/19 \approx 42.10\%$

remark: As Kris noted in comments, I miscalculated, and some open areas can hold a triangle. I thought my solution was a local optimum, but it is not. Although that means the solution can 'easily' be fixed by rotating the green triangles, I am pretty sure the solution will get increasingly better until the triangles are rotated 30 degrees. And then (after resizing/repositioning), we end up in loopy walts solution. So my triangle solution does not add anything useful to the question after all.

enter image description here
note: the black grid size is 27/4 triangles, which can be improved a tiny bit since the blueish triangles can be tilted a bit.

Square upper bound

$r = \frac {\sqrt{2}} {(3-1/2\sqrt{2}) \times(1 + \sqrt{2})-\sqrt{2}} \approx 34.31\% $

pattern

enter image description here

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  • $\begingroup$ By the OP’s definition, your triangle value for R is actually approx 0.42. $\endgroup$
    – Kris Van Bael
    Nov 1 at 9:34
  • $\begingroup$ Based on my calculation, I believe that your triangle solution isn’t valid. The drawing is not to proportion, there is actually room for a triangle left and right of the green triangles. $\endgroup$
    – Kris Van Bael
    Nov 1 at 9:50
  • $\begingroup$ Kris 1 fixed; 2 Ill have a look when I have more time $\endgroup$
    – Retudin
    Nov 1 at 11:39
  • $\begingroup$ I obtained the same result for your square solution. $\endgroup$
    – Weather Vane
    Nov 4 at 21:41
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The minimum uncovered area is ZERO , when the Plane is tightly tiled with triangles ...

When we shrink the triangles a little , while keeping the centers unchanged , then no new triangles can be added yet uncovered area is not Zero and can increase ...

Eventually , when the shrinkage is enough , then we can add a new triangle which will partially share a base side 2 triangles while the other two sides will touch 2 triangles at the base vertexes near the top ...
This gives the maximum uncovered area ...

[[ I will add images & calculations a while later. ]]

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    $\begingroup$ Why is that particular configuration of triangles optimal for the task? $\endgroup$
    – Bass
    Oct 31 at 7:11
  • $\begingroup$ You're free to use these images. I had to mock it up to be sure what you meant. I'd be interested in the math behind this being optimal. Starting with this, you make them smaller like this and keep making them smaller until more triangles fit in between. I think that last picture is past that point, actually, even though the triangles are a little larger than half the size of the starting point. I would not expect this to beat the current record of only 38% coverage. $\endgroup$
    – Engineer Toast
    Nov 1 at 20:10
  • $\begingroup$ Thank you for the Image Set , I will use it over the week end , @EngineerToast , I am currently stuck up with some thing else which makes me unable to make Images to update my Post. Due to that , I am unable to meaningfully reply to user Bass. $\endgroup$
    – Prem
    Nov 3 at 7:26

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