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If we allow two sets of congruent pieces, then the regular pentagon can divided into various numbers of parts according to the Lucas numbers. Usually these are defined by the recursion $L_1=1,L_2=3, L_{n+1}=L_n+L_{n-1}$ for $n\ge2$. If we run the recursion backwards, we find $L_0=2$ and $l_{-n}=(-1)^nL_n$. We observe the (positive) numbers from this sequence in the divisions below, where the two sets of congruent areas are represented by different colored dots. With an unbounded number of iterations the ratio of the numbers of pieces tends to $\phi=(1+\sqrt5)/2$, and the nodes fill in a tenfold symmetric quasilattice.

The rules for the divisions are as follows:

1. Start by drawing two diagonals of the pentagon. This produces two shapes of isosceles triangles, one acute and one obtuse. Each shape will be reproduced with smaller size and greater count of triangles by the subsequent divisions governed by Rules 2 and 3.

2. When the acute triangles are larger, subdivide them by bisecting one of the base angles. The base angle to be bisected is chosen as follows: rotate the figure so each acute triangle in turn has the apex on top, then select the base angle at the bottom left of the triangle.

3. When the obtuse triangles are larger, subdivide them by drawing one trisector of the apex angle. The trisector to be drawn is chosen by rotating the fugure so that each obtuse triangle in turn has its apex on top, then drawing the trisectir that os directed downwards and to the left.

If we allow two sets of congruent pieces, then the regular pentagon can divided into various numbers of parts according to the Lucas numbers. Usually these are defined by the recursion $L_1=1,L_2=3, L_{n+1}=L_n+L_{n-1}$ for $n\ge2$. If we run the recursion backwards, we find $L_0=2$ and $l_{-n}=(-1)^nL_n$. We observe the (positive) numbers from this sequence in the divisions below, where the two sets of congruent areas are represented by different colored dots. With an unbounded number of iterations the ratio of the numbers of pieces tends to $\phi=(1+\sqrt5)/2$, and the nodes fill in a tenfold symmetric quasilattice.

The rules for the divisions are as follows:

1. Start by drawing two diagonals of the pentagon. This produces two shapes of isosceles triangles, one acute and one obtuse. Each shape will be reproduced with smaller size and greater count of triangles by the subsequent divisions governed by Rules 2 and 3.

2. When the acute triangles are larger, subdivide them by bisecting one of the base angles. The base angle to be bisected is chosen as follows: rotate the figure so each acute triangle in turn has the apex on top, then select the base angle at the bottom left of the triangle.

3. When the obtuse triangles are larger, subdivide them by drawing one trisector of the apex angle. The trisector to be drawn is chosen by rotating the fugure so that each obtuse triangle in turn has its apex on top, then drawing the trisector that os directed downwards and to the right.

If we allow two sets of congruent pieces, then the regular pentagon can divided into various numbers of parts according to the Lucas numbers. Usually these are defined by the recursion $L_1=1,L_2=3, L_{n+1}=L_n+L_{n-1}$ for $n\ge2$. If we run the recursion backwards, we find $L_0=2$ and $l_{-n}=(-1)^nL_n$. We observe the (positive) numbers from this sequence in the divisions below, where the two sets of congruent areas are represented by different colored dots. With an unbounded number of iterations the ratio of the numbers of pieces tends to $\phi=(1+\sqrt5)/2$, and the nodes fill in a tenfold symmetric quasilattice.

If we allow two sets of congruent pieces, then the regular pentagon can divided into various numbers of parts according to the Lucas numbers. Usually these are defined by the recursion $L_1=1,L_2=3, L_{n+1}=L_n+L_{n-1}$ for $n\ge2$. If we run the recursion backwards, we find $L_0=2$ and $l_{-n}=(-1)^nL_n$. We observe the (positive) numbers from this sequence in the divisions below, where the two sets of congruent areas are represented by different colored dots. With an unbounded number of iterations the ratio of the numbers of pieces tends to $\phi=(1+\sqrt5)/2$, and the nodes fill in a tenfold symmetric quasilattice.

The rules for the divisions are as follows:

1. Start by drawing two diagonals of the pentagon. This produces two shapes of isosceles triangles, one acute and one obtuse. Each shape will be reproduced with smaller size and greater count of triangles by the subsequent divisions governed by Rules 2 and 3.

2. When the acute triangles are larger, subdivide them by bisecting one of the base angles. The base angle to be bisected is chosen as follows: rotate the figure so each acute triangle in turn has the apex on top, then select the base angle at the bottom left of the triangle.

3. When the obtuse triangles are larger, subdivide them by drawing one trisector of the apex angle. The trisector to be drawn is chosen by rotating the fugure so that each obtuse triangle in turn has its apex on top, then drawing the trisectir that os directed downwards and to the left.

If we allow two sets of congruent pieces, then the regular pentagon can divided into various numbers of parts according to the Lucas numbers. Usually these are defined by the recursion $L_1=1,L_2=3, L_{n+1}=L_n+L_{n-1}$ for $n\ge2$. If we run the recursion backwards, we find $L_0=2$ and $l_{-n}=(-1)^nL_n$. We observe the (positive) numbers from this sequence in the divisions below, where the two sets of congruent areas are represented by different colored dots. With an unbounded number of iterations the ratio of the numbers of pieces tends to $\phi=(1+\sqrt5)/2$, and the nodes fill in a tenfold symmetric quasilattice.

If we allow two sets of congruent pieces, then the regular pentagon can divided into various numbers of parts according to the Lucas numbers. Usually these are defined by the recursion $L_1=1,L_2=3, L_{n+1}=L_n+L_{n-1}$ for $n\ge2$. If we run the recursion backwards, we find $L_0=2$ and $l_{-n}=(-1)^nL_n$. We observe the (positive) numbers from this sequence in the divisions below, where the two sets of congruent areas are represented by different colored dots. With an unbounded number of iterations the ratio of the numbers of pieces tends to $\phi=(1+\sqrt5)/2$, and the nodes fill in a tenfold symmetric quasilattice.