What are the lengths that can be constructed with straightedge but without compass? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T12:30:57Z http://mathoverflow.net/feeds/question/61059 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61059/what-are-the-lengths-that-can-be-constructed-with-straightedge-but-without-compas What are the lengths that can be constructed with straightedge but without compass? Colin Tan 2011-04-08T13:16:50Z 2013-04-25T20:40:14Z <p>Most field theory textbooks will describe the field of constructible numbers, i.e. complex numbers corresponding to points in the Euclidean plane that can be constructed via straightedge and compass. This field is the smallest field of characteristic 0 that is closed under square root (i.e. is Pythagorean) and is closed under conjugation.</p> <p>I'm interested in know: What is the field of numbers that can be constructed if we disallow compass and use only straightedge? </p> <p>I have not checked this up, but it seems that this question led Hilbert to formulate his 17th problem, particularly the version involving polynomials with rational coefficients (rather than the real coefficients which Artin proved). I'm also interested in knowing more about this history too. </p> http://mathoverflow.net/questions/61059/what-are-the-lengths-that-can-be-constructed-with-straightedge-but-without-compas/61068#61068 Answer by quid for What are the lengths that can be constructed with straightedge but without compass? quid 2011-04-08T14:05:44Z 2011-04-09T01:06:03Z <p>In view of the many comments, I will make a (I hope correct) summary of these comments in CW mode; everybody please feel free to edit:</p> <ol> <li><p>If one starts with a completely 'blank sheet of paper' it seems that one can do almost nothing with a straightedge alone.</p></li> <li><p>However, as mentioned by François Brunault given certain 'initial constellations' one can construct some additional interesting points using a staightedege alone (see <a href="http://fr.wikipedia.org/wiki/Construction_a_la_regle_seule" rel="nofollow">here</a> (in French)).</p></li> <li><p>Daniel Briggs suggested to 'add' just one circle with known center (the unit circle). If one does this, then by the <a href="http://en.wikipedia.org/wiki/Poncelet-Steiner_theorem" rel="nofollow">Poncelet-Steiner</a> theorem (mentioned by François Brunault) one can already construct everything one can construct with straightedge and compass.</p></li> <li><p>L. Spice mentioned that by the <a href="http://en.wikipedia.org/wiki/Mohr-Mascheroni_theorem" rel="nofollow">Mohr-Mascheroni theorem</a> the converse situation (only a compass no straightedge) allows also to construct everything one can construct with straightedge and compass.</p></li> <li><p>The book <em>Leçons sur les constructions géométriques</em> by Lebesgue is entirely devoted to the question of geometric constructions using various instruments. The table of contents of this book (in French, again) is available <a href="http://www.gabay.com/sources/Liste_Fiche.asp?CV=148" rel="nofollow">here</a>.</p></li> </ol> http://mathoverflow.net/questions/61059/what-are-the-lengths-that-can-be-constructed-with-straightedge-but-without-compas/61069#61069 Answer by thei for What are the lengths that can be constructed with straightedge but without compass? thei 2011-04-08T14:10:42Z 2011-04-08T14:10:42Z <p>It depends on what you regard as a starting point.</p> <p>For ruler and compass, we start with the points 0 and 1.</p> <p>For an unmarked ruler, this is not a good start, because an unmarked ruler is good at conserving cross-ratios, but if you start with two (or three including $\infty$) points, there is no cross-ratio yet to conserve.</p> <p>With a marked ruler, this problem disappears because you obviously get all the integers and then are able to construct parallel lines by building a complete quadrilateral over three equidistant points. So you already get all the rational numbers.</p> http://mathoverflow.net/questions/61059/what-are-the-lengths-that-can-be-constructed-with-straightedge-but-without-compas/61074#61074 Answer by Gerald Edgar for What are the lengths that can be constructed with straightedge but without compass? Gerald Edgar 2011-04-08T15:29:18Z 2011-04-08T15:29:18Z <p>Bogomolny's site "Cut the Knot" has lots of interesting math... <a href="http://www.cut-the-knot.org/impossible/straightedge.shtml" rel="nofollow">http://www.cut-the-knot.org/impossible/straightedge.shtml</a></p> http://mathoverflow.net/questions/61059/what-are-the-lengths-that-can-be-constructed-with-straightedge-but-without-compas/61078#61078 Answer by Dave Richeson for What are the lengths that can be constructed with straightedge but without compass? Dave Richeson 2011-04-08T16:19:02Z 2011-04-08T16:19:02Z <p>The short answer is that nothing is constructible. As is standard, we begin with the two points $(0,0)$ and $(1,0)$. Then we can draw a line between them, and that's it. We can't draw any more lines, and hence we can't construct any new points. The Euclidean rules say that we are only allowed to draw a new line if we are joining two already-constructed points, and a point can only be constructed if it is the intersection of two lines (or, irrelevant to this discussion, two circles or a line and a circle).</p> <p>However, suppose you begin with a finite collection of points $(x_1,y_1),\ldots, (x_n,y_n)$. Let $C$ be the set of points constructible from this set using only a straightedge (unmarked). If a point $(x,y)$ is in $C$, then $x$ and $y$ can be formed from $x_1,\ldots,x_n,y_1,\ldots,y_n$ using the operations $+$, $-$, $\cdot$, and $\div$ (since new points are created as intersections between lines). However the converse of this is not true (as the two-point example shows). I suppose you could say more about what $C$ looks like, but it would probably be messy.</p>