Eric Broug in his book *Islamic Geometric Patterns* gives
straightedge and compass construction of some simpler patterns.
It is clear his techniques will provide constructions for many
Islamic patterns.


Looking at formal constructibility, the Wikipedia pages gives Gauss' result that 
7, 9, 11, 13, 14, 18... etc sided polygons are not constructible. Hence the pattern


<http://tilingsearch.org/HTML/data160/J43C.html>


is not constructible since it contains a regular
9-pointed star polygon.


I have over 800 Islamic patterns on my web site but
I use a computer and trigonometry to produce my images. It seems that
about 40 Islamic patterns on my site are not constructible.


Given an Islamic pattern that is not excluded from construction by Gauss'
result, it is almost certainly constructible if the following is true:



> 
> Given two points on the plane, a polygon ($n$ sides) can be constructed with the two points
>  as an edge, provided $n$ is not equal to 7, 9, 11, 13, 14, 18... etc.
> 
> 
> 


This result would allow patterns to be built up piece-by-piece.


EDIT, Will Jagy: from his profile page, the OP's website, in this address preset to display the tilings in a slideshow format on a web browser, is at


<http://www.tilingsearch.org/>



What Islamic tiling patterns are constructible?
Yes, that is right. It seems all you are missing is this: given a number of sides $n$ such that the regular polygon of $n$ sides is constructible (by the results of Gauss and Wantzel), how to force the edge length to be a fixed length, call it $L$?
$$ $$
All you need to do is construct the regular $n$-gon. Draw the perpendicular bisector of any edge, which is then a line that passes through the center of the regular polygon, and stop the line at that center so that it is actually a half-line, a "ray." Call this ray $r.$ The ray $r$ starts in the interior of one central triangle of the regular polygon, call the radii acting as the two othe edges ot that triangle $e\_1$ and $e\_2$ 
$$ $$
Now, parallel to $r,$ construct the parallel line at distance $$ \frac{L}{2}$$
from $r.$ At some point $P,$ the ray $r$ intersects either $e\_1$ or $e\_2.$ The point $P$ is a vertex of the polygon you want. To find the rest of the polygon just draw a circle with center at the center of the polygon and passing through $P,$ then extend the radii passing through the vertices of your original polygon until they reach the circle. 


