Hyperbolic tiling, 7:1

[Marek14]

I wanted to share this beautiful tiling that was recently found, based on a equality that I discovered:

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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}.

[ICN5D]

I like #6 and #7. It's crazy to see those triangles get warped into hooks and spikes like that.

[mr_e_man]

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.)

[Marek14]

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)

[mr_e_man]

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.

[Marek14]

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}

[mr_e_man]

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.

[Marek14]

Incredible!