Hyperbolic tiling, 7:1

Higher-dimensional geometry (previously "Polyshapes").

Hyperbolic tiling, 7:1

Postby Marek14 » Sun Apr 24, 2022 9:37 am

I wanted to share this beautiful tiling that was recently found, based on a equality that I discovered:
image4.png
(835.82 KiB) Not downloaded yet

image5.png
(829.5 KiB) Not downloaded yet

image6.png
(826.11 KiB) Not downloaded yet

image7.png
(901.55 KiB) Not downloaded yet


It is made from triangles, squares, and large triangles with 7 times the edge of the small polygons. All are regular. The edge of small polygons is the same as the edge of {5,4}.
Marek14
Pentonian
 
Posts: 1191
Joined: Sat Jul 16, 2005 6:40 pm

Re: Hyperbolic tiling, 7:1

Postby ICN5D » Thu Apr 28, 2022 12:07 am

I like #6 and #7. It's crazy to see those triangles get warped into hooks and spikes like that.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: Hyperbolic tiling, 7:1

Postby mr_e_man » Wed Feb 08, 2023 12:58 am

Just now I made sure that this exists.

Denoting the short edge as s, and the long edge as 7s, we have cosh(s/2) = φ/√2 and cosh(7s/2) = (13φ+8)/√2 = φ⁷/√2. The angles are

2arcsin(φ⁻¹) ≈ 76.3454°, in a square;
2arcsin(φ⁻¹/√2) ≈ 51.8273°, in a small triangle;
2arcsin(φ⁻⁷/√2) ≈ 2.7910°, in a large triangle.

So what needed to be verified was

4*2arcsin(φ⁻¹) + 2arcsin(φ⁻¹/√2) + 2arcsin(φ⁻⁷/√2) = 360°.

(This can be done using some trig identities and φ arithmetic.)
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
Tetronian
 
Posts: 538
Joined: Tue Sep 18, 2018 4:10 am

Re: Hyperbolic tiling, 7:1

Postby Marek14 » Wed Feb 08, 2023 6:30 am

It can really be shown through the identities? I was barely able to show that {5,6} has twice the edge of {5,4}...

Here's another interesting feature of this tiling I found recently:

The big triangle has seven times the edge of the small triangle, yes. BUT, it ALSO has seven times its area!

This can be actually proven quite easily:

Angle of small triangle = t
Angle of square = s
Angle of large triangle = x

Area of small triangle = π-3t

4t + 2s = 2π
t+4s+x = 2π

Multiplying the first equation by 2 and subtracting gives us:

8t + 4s = 4π
-t-4s-x = -2π
-----
7t - x = 2π

x = 7t - 2π

So area of the large triangle is:
π-3x = π - 3(7t-2π) = π - 21t + 6π = 7π - 21t = 7(π-3t)
Marek14
Pentonian
 
Posts: 1191
Joined: Sat Jul 16, 2005 6:40 pm

Re: Hyperbolic tiling, 7:1

Postby mr_e_man » Wed Feb 08, 2023 4:24 pm

First let's verify cosh(7s/2)=φ⁷/√2, given that cosh(s/2)=φ/√2. The basic addition laws are

cosh(x+y) = cosh(x)cosh(y) + sinh(x)sinh(y),
sinh(x+y) = sinh(x)cosh(y) + cosh(x)sinh(y),

so we get these multiple-angle identities:

cosh(2x) = cosh²(x) + sinh²(x) = 2cosh²(x) - 1
sinh(2x) = 2sinh(x)cosh(x)

cosh(3x) = cosh(2x+x) = (2cosh²(x) - 1) cosh(x) + (2sinh(x)cosh(x)) sinh(x)
= (2cosh²(x) - 1 + 2sinh²(x)) cosh(x) = 4cosh³(x) - 3cosh(x)
sinh(3x) = sinh(2x+x) = 2sinh(x)cosh(x)cosh(x) + (2cosh²(x) - 1)sinh(x)
= (4cosh²(x) - 1) sinh(x)

cosh(6x) = cosh(2(3x)) = 2cosh²(3x) - 1
= 2 (4cosh³(x) - 3cosh(x))² - 1 = 32cosh⁶(x) - 48cosh⁴(x) + 18cosh²(x) - 1
sinh(6x) = sinh(2(3x)) = 2sinh(3x)cosh(3x)
= 2 (4cosh²(x) - 1) sinh(x) (4cosh³(x) - 3cosh(x)) = 2 (16cosh⁵(x) - 16cosh³(x) + 3cosh(x)) sinh(x)

cosh(7x) = cosh(6x+x) = cosh(6x)cosh(x) + sinh(6x)sinh(x)
= (32cosh⁶(x) - 48cosh⁴(x) + 18cosh²(x) - 1) cosh(x) + 2 (16cosh⁵(x) - 16cosh³(x) + 3cosh(x)) (cosh²(x) - 1)
= 32cosh⁷(x) - 48cosh⁵(x) + 18cosh³(x) - cosh(x) + 2 (16cosh⁷(x) - 32cosh⁵(x) + 19cosh³(x) - 3cosh(x))
= 64cosh⁷(x) - 112cosh⁵(x) + 56cosh³(x) - 7cosh(x)

Also, since φ² = φ+1, the other powers of φ involve the Fibonacci numbers:

φ³ = 2φ + 1
φ⁴ = 3φ + 2
φ⁵ = 5φ + 3
φ⁶ = 8φ + 5
φ⁷ = 13φ + 8

Putting all these pieces together, we get

cosh(7s/2) = 64(φ/√2)⁷ - 112(φ/√2)⁵ + 56(φ/√2)³ - 7(φ/√2)
= 64(13φ+8)/(8√2) - 112(5φ+3)/(4√2) + 56(2φ+1)/(2√2) - 7(φ)/(√2)
= (8(13φ+8) - 28(5φ+3) + 28(2φ+1) - 7(φ)) / √2
= (13φ + 8) / √2 = φ⁷/√2.
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
Tetronian
 
Posts: 538
Joined: Tue Sep 18, 2018 4:10 am

Re: Hyperbolic tiling, 7:1

Postby Marek14 » Wed Feb 08, 2023 5:04 pm

That means that cosh(2s/2) (for double edge) is 2*cosh²(s/2)-1 = (φ+1)-1 = φ, which works (that's the edge of {5,6} and cos(π/5)/sin(π/6) is φ). So this could be used for direct verification whenever we have regular tilings with integral edge ratios. There are a few like this I know of, all with 1:2 ratio:

{5,4} - {3,10}/{5,6}/{10,5} - this example
{6,4} - {inf,6} - √3/√2 -> 2*3/2-1 = 2, works
{7,4} - {3,14}/{14,7}
{3,8}/{8,4} - {8,8}
{9,4} - {6,9}
{10,4} - {5,10}
{12,4} - {4,12}
{18,4} - {3,18}/{18,9}
Marek14
Pentonian
 
Posts: 1191
Joined: Sat Jul 16, 2005 6:40 pm

Re: Hyperbolic tiling, 7:1

Postby mr_e_man » Wed Feb 08, 2023 5:46 pm

mr_e_man wrote:4*2arcsin(φ⁻¹) + 2arcsin(φ⁻¹/√2) + 2arcsin(φ⁻⁷/√2) = 360°

Dividing by 2 and subtracting:

arcsin(φ⁻¹/√2) + arcsin(φ⁻⁷/√2) ≟ 180° - 4arcsin(φ⁻¹)

The arcsine of any positive number is between 0° and 90°, so the left side of this equation is between 0° and 180°.
Also, since φ⁻¹ < 1/√2, it follows that arcsin(φ⁻¹) < 45°, so the right side is also between 0° and 180°.
The cosine function is invertible on this interval, so it can be applied without changing the truth of the equation:

cos(arcsin(φ⁻¹/√2) + arcsin(φ⁻⁷/√2)) ≟ cos(180° - 4arcsin(φ⁻¹))

Here we can use the addition law, cos(x+y)=cos(x)cos(y)-sin(x)sin(y), along with cos(arcsin(x))=√(1-x²), to expand the left side:

cos(arcsin(φ⁻¹/√2) + arcsin(φ⁻⁷/√2))
= √(1 - (φ⁻¹/√2)²) √(1 - (φ⁻⁷/√2)²) - (φ⁻¹/√2) (φ⁻⁷/√2)
= √(1 - φ⁻²/2) √(1 - φ⁻¹⁴/2) - φ⁻⁸/2
= √(1 - φ⁻²/2 - φ⁻¹⁴/2 + φ⁻¹⁶/4) - φ⁻⁸/2

And here's the right side:

cos(180° - 4arcsin(φ⁻¹))
= -cos(4arcsin(φ⁻¹))
= 1 - 2cos²(2arcsin(φ⁻¹))
= 1 - 2(1 - 2sin²(arcsin(φ⁻¹)))²
= 1 - 2(1 - 2φ⁻²)² = 1 - 2(2φ - 3)²
= 16φ - 25

So the equation becomes:

√(1 - φ⁻²/2 - φ⁻¹⁴/2 + φ⁻¹⁶/4) - φ⁻⁸/2 ≟ 16φ - 25

Multiplying by 2φ⁸ and adding 1:

√(4φ¹⁶ - 2φ¹⁴ - 2φ² + 1) ≟ 32φ⁹ - 50φ⁸ + 1
√(4(987φ+610) - 2(377φ+233) - 2(φ+1) + 1) ≟ 32(34φ+21) - 50(21φ+13) + 1
√(3192φ + 1973) ≟ 38φ + 23

Both sides are obviously positive, so squaring is invertible and thus doesn't change the truth of the equation:

3192φ + 1973 ≟ (38φ + 23)²
3192φ + 1973 ≟ (38² + 2*38*23)φ + (38² + 23²)

At last we see that it is true.
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
Tetronian
 
Posts: 538
Joined: Tue Sep 18, 2018 4:10 am

Re: Hyperbolic tiling, 7:1

Postby Marek14 » Wed Feb 08, 2023 10:13 pm

Incredible! :)
Marek14
Pentonian
 
Posts: 1191
Joined: Sat Jul 16, 2005 6:40 pm


Return to Other Geometry

Who is online

Users browsing this forum: No registered users and 2 guests

cron