# Put infinitely many equilateral triangles of equal size on the plane

...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?

• Does overlapping of edges or vertices count or only interiors? Oct 31, 2023 at 2:28
• @RossMillikan Only interiors.
– Eric
Oct 31, 2023 at 2:36
• Can't both equilateral triangles and squares tile the plane? Oct 31, 2023 at 12:16
• @Someone The question is asking for the minimum area that can be covered such that no additional triangles can be placed without overlap. Oct 31, 2023 at 12:53
• Oh, that makes more sense. Oct 31, 2023 at 13:09

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:

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

• By the OP’s definition of R, your solution is approx .38. Nov 1, 2023 at 9:19

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.

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

• By the OP’s definition, your triangle value for R is actually approx 0.42. Nov 1, 2023 at 9:34
• 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. Nov 1, 2023 at 9:50
• Kris 1 fixed; 2 Ill have a look when I have more time Nov 1, 2023 at 11:39
• I obtained the same result for your square solution. Nov 4, 2023 at 21:41

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. ]]

• Why is that particular configuration of triangles optimal for the task?
– Bass
Oct 31, 2023 at 7:11
• 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. Nov 1, 2023 at 20:10
• 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.
– Prem
Nov 3, 2023 at 7:26