Progressing through the regular polyhedra from triangles to squares to pentagons to hexagons, pentagons are the only ones that don’t tile a plane. How do you figure they’re the best for buildings?
this post was submitted on 27 Apr 2025
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pentagon
plane
oh no
e: well I expected this low-effort comment to get buried within seconds so just to be clear, it's purely a mathematical theory where if you had two pentagons and an extra vertex, then only 9/11 of the edges would tesselate correctly. I'm not sure what that is in Radians, but it's definitely an American 77 units for sure.
I expected this low-effort comment to get buried within seconds
But the Earth isn't a plane.
Sure, human scaled patches of the Earth's surface can be approximated by a similarly sized patch of a plane, but if we're talking about tiling the entire surface of the Earth with buildings, it can actually be done using twelve pentagons or twenty isosceles triangles. We just need buildings whose footprints are roughly 1/12th and 1/20th the Earth's surface respectively.
For the pentagon, that'd be around 510.07 × 10^12 m² divided by 12 = 42.505 × 10^12 m². With the Pentagon building having seven floors, one such building would have roughly 297.541 × 10^12 m² of floor space.
For the triangle, that'd be around 510.07 × 10^12 m² divided by 20 = 25.503 × 10^12 m². Assuming this building has seven floors like the Pentagon building does, it'd have roughly 178.524 × 10^12 m² of floor space.
The good thing about dividing into triangles, however is that it can be subdivided into four similar isosceles triangles, which can be applied recursively down to a far more realistic scale.
Doing that, we can subdivide the original triangles sixteen times yielding the following:
25.503 × 10^12 m² / (4^16) = 5.937 × 10^3 m²
And since the area of an isosceles triangle is equal to s²(√3)/4 we can rearrange things to find the side length of a compound with area of 5 937m²
s = √(4A/(√3)) = 117.103 m
I think that's a human-enough scale for buildings.
In total, there's 85 899 345 920 such buildings, covering the Earth.
If one such building has 7 floors, it'd have at most 41 559 m² of floor space.
EDIT:
Hit enter too soon. Additional proofreading.
Damn, I discovered a small mistake in the calculations partway through. Corrected.
That makes me think of the last episode of Severance, where two characters who have lost their memory of the external world are debating whether the equator is a continent or a continent-sized building.
Designated park space
Technically it's only regular pentagons that don't tile the plane.
I know everything about planes because I am shorter than my brothers. When we were young, they called me Tattoo and if I pointed at something with my finger, they screamed in unison, "Da plane! Da plane!"
My high ass was like ‘wait they put tile on aircraft?’ and after accepting that I just assumed it messed with airflow or the Bernoulli effect.
Then I was like, ‘Oh’.
No!
Hexagons are the Bestagons!
Close… no offense to CGP, but Texagons are the bestagons.
Funny for once, but highly inefficient. Hexagons remain bestagons.
Hexagons are the bestagons
Came into the thread for this comment.
Canada should build an Hexagon as a flex.
And just call it “The Bestagon”. And it’s used for their MoD, as well as to play hockey (but I repeat myself)
What do you intend to do with the gaps then?
Room for parks!
Parks are in the middle, Roman domus style.
Yeah, that should work. If you stick with the definition provided by OP, “small buildings” would be technically allowed as well.
If the gasp are a problem, you could go with alternate pentagons too.
Another thing about the initial conditions. OP defined that there would be only three way intersections, but that’s clearly not the case here.
You might never see the sun again!
Far point although it might be better for driving during dawn and dusk