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Lately, we've had plenty of puzzles based on the regular pentagon and its geometric properties. So I propose one that literally brings it all together.

two isosceles triangles, one with two gold side-lengths and one blue side-length, and the other with two blue side-lengths and one gold side-length

Use eleven copies of the larger (left) piece above and seven copies of the smaller (right) piece (these numbers being featured prominently in the dice game referred to in the title) to assemble a regular pentagon with no overlapping of triangle areas. The triangle sides should also define the pentagram formed from the diagonals. Blue segments are all the same length and gold segments are longer by, of course, the golden ratio.

You should be able to discern several golden-ratio relationships among the line segments from the resulting configuration.

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Step 1:

Construct the shapes to scale; a golden triangle with unit base (knowing that two of the angles are 72°), and a golden gnomon with unit legs.

A graphing program showing the construction of the golden triangle and the golden gnomon. A unit base along the x-axis (from the origin to point B) has two legs at 72° angles joining above a point C — forming the golden triangle. A circle about the origin passes through C, and about 1.618 on the axes. About point B, a unit circle is drawn intersecting the other circle at a point D — this forms the golden gnomon.

Step 2:

Screenshot, crop, print (2 pages, 9 per page), cutout.

Step 3:

Assemble the pieces; a small pentagon, then a pentagram, then a larger (inverted) pentagon.

Small pieces of paper with the graph above printed on them, cut into triangles and gnomons and assembled into a pentagon with a smaller inverted pentagon in the middle, with points radiating out forming a pentagram.

Results:

The small pentagon has side length $1$, and diagonal $\varphi$.
The large pentagon has side length $\varphi+1$, and diagonal $2\varphi+1$.

As well as the 11 unit-base golden triangles (with $\varphi$ legs), there are similar golden triangles with base $\varphi$ and legs $\varphi+1$, and with base $\varphi+1$ and legs $2\varphi+1$.

Similarly, well as the 7 unit-leg golden gnomons (with $\varphi$ base), there are golden gnomons with legs $\varphi$ and base $\varphi+1$, and with legs $\varphi+1$ and base $2\varphi+1$.

By similar pentagons,* triangles, or gnomons:

$$\frac1\varphi = \frac\varphi{\varphi+1} = \frac{\varphi+1}{2\varphi+1}$$

$$\boxed{\varphi = \frac{\varphi+1}\varphi} = \frac{2\varphi+1}{\varphi+1}\,\textrm{.}$$

* The medium sized pentagon (sides $\varphi$, diagonal $\varphi+1$) doesn’t appear in the assembly, but can be formed with 4 triangles and 3 gnomons.

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    $\begingroup$ lol, I recognised you before I saw the user card from the gameboy plushie :P $\endgroup$ Commented 5 hours ago
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    $\begingroup$ @Otakuwu Yay, BMO! $\endgroup$ Commented 5 hours ago
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Answer:

desired tiling

The ratio of lengths red : yellow : green : blue is 1 : φ : φ2 : φ3, where φ is the golden ratio.

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In this version the segment colors are preserved, which makes the golden ratio relationships described in other answers pop out readily.

enter image description here

For instance, the sides of the outer pentagon measure $1+\varphi$ (one blue segment and one gold segment) and the connections between the outer and inner vertices measure $\varphi$; by definition of the golden ratio we then have $(1+\varphi)/\varphi=\varphi$.

Lucas numbers, the fourth dimension, and new materials

The pentagon division shown here is actually one out of a sequence of such divisions into the two types of triangles indicated here. This sequence is discussed in this MSE answer. The number of smaller triangles and the number of larger triangles are consecutive Lucas numbers (which may appear in either order) defined by

$L_0=2,L_1=1, L_{n+2}=L_n+L_{n+1}.$

The total number of triangles is the next higher Lucas number; for instance (in this puzzle) $L_4=7$ small triangles. $L_5=11$ large ones, and $L_6=18$ total.

If we go to finer and finer divisions in this sequence, the generated triangle vertices fill in a ten-fold quasilattice. This is actually a four-dimensional integer lattice projected into the plane in a manner that maps the tenfold lattice symmetry in the higher dimension into a cyclic symmetry in the plane. This quasilattice and its associated higher dimensionality play an important role in the study of quasicrystalline materials.

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