For a ${\displaystyle K}$consisting of points in R^2, define the function B such that ${\displaystyle B(K)}$as the Union of ${\displaystyle K}$and all points which can be produced in the following way. For each set of points A, B, C, & D from ${\displaystyle K}$all different so that no three of A, B, C & D are co-linear. E is the point (if it exists) where ABE are colinear and CDE are co-linear.

If ${\displaystyle K_{0}}$= the vertices of a regular Pentagon centered at 0,0 with one vertex at (1,0), does there exist N such that ${\displaystyle B^{N}(K_{0})}$includes any point of the form (0, y)? (extending the question to any N-gon, with N odd) Naraht (talk) 05:16, 31 December 2024 (UTC)[reply]

I think you meant to write ${\displaystyle B^{N}(K_{0}).}$ --Lambiam 07:55, 31 December 2024 (UTC)[reply]

Changed to use the Math.Naraht (talk) 14:37, 31 December 2024 (UTC)[reply]

I'm not 100% sure I understand the problem, but try this: Label the vertices of the original pentagon, starting with (1, 0), as A, B, C, D, E. You can construct a second point on the x-axis as the intersection of BD and CE; call this A'. Similarly construct B', C', D', E', to get another, smaller, regular pentagon centered at the origin and with the opposite orientation from the the original pentagon. All the lines AA', BB', CC', DD', EE' intersect at the origin, so you can construct (0, 0) as the intersection of any pair of these lines. The question didn't say y could not be 0, so the answer is yes, with N=2.

There is some theory developed on "straightedge only construction", in particular the Poncelet–Steiner theorem, which states any construction possible with a compass and straightedge can be constructed with a straightedge alone if you are given a single circle with its center. In this case you're given a finite set of points instead of a circle, and I don't know if there is much theory developed for that. --RDBury (talk) 13:12, 1 January 2025 (UTC)[reply]

Here is an easy way to describe the construction of pentagon A'B'C'D'E'. The diagonals of pentagon ABCDE form a pentagram. The smaller pentagon is obtained by removing the five pointy protrusions of this pentagram.  --Lambiam 16:53, 1 January 2025 (UTC)[reply]

Here is one point other than the origin (in red)

If there is one such point, there must be an infinite number of them. catslash (talk) 22:52, 2 January 2025 (UTC)[reply]

Just to be clear, the black points are the original pentagon K, the green points are in B(K), and the red point is the desired point in B²(K); the origin is not shown. It would be nice to find some algebraic criterion for a point to be constructible in this way, similar to the way points constructible with a compass and straightedge are characterized by their degree over Q. --RDBury (talk) 01:45, 3 January 2025 (UTC)[reply]

Once you have a second one (such as the reflection of the red point wrt the x-axis), you have all intersections of the y-axis with the non-vertical lines through pairs of distinct points from ${\displaystyle \textstyle {\bigcup _{n}B^{n}(K_{0})}.}$ --Lambiam 16:22, 3 January 2025 (UTC)[reply]

RDBury *headslap* on (0,0) Any idea on y<>0? (← comment from Naraht)

See the above construction by catslash.  --Lambiam 16:12, 3 January 2025 (UTC)[reply]

The red point is at ${\displaystyle x=0}$, ${\displaystyle y=-{\sqrt {\frac {5-{\sqrt {5}}}{10}}}}$. catslash (talk) 16:18, 3 January 2025 (UTC)[reply]