Angle trisection

With two other clas­sical prob­lems, doubling the cube and squaring the circle, this problem came from Ancient Greece and was occu­pying human minds for centuries. By the way, even ancient math­e­mati­cians guessed that this prob­lems can not be solved using only compass and straight­edge, which was proved later. Attempts to extend tools lead to the first studies of conic sections, compli­cated curves and building inter­esting devices.

Consider a linkage being a paral­lel­o­gram with two fixed hinges. You remember from school that oppo­site angles of a paral­lel­o­gram are equal. This is true for every paral­lel­o­gram, for any state our mech­a­nism as well.

Are you sure that for any?

Our system has a special point when all edges are anti-paral­lel­o­gram.

### Sir Alfred Bray Kempe 1849—1922

Kempe was an english lawyer but a born math­e­mati­cian. In 1879 he publishes his solu­tion of the four colour problem. The royal math­e­mat­ical society imme­di­ately elected him as a fellow and later he becomes a knight for his contri­bu­tion to math­e­matics. One believed in the proof of Kempe till 1890 when Percy Heawood published a paper aston­ishing the math­e­mat­ical world: it shows incor­rect the reasoning of Kempe. However, some ideas of his proof were correct and in a century were used in the computer solu­tion of the problem.

This special feature of the mech­a­nism was not consid­ered in the reasoning of Alfred Kempe who proved in 1876 a theorem telling that there is a mech­a­nism that can only falsify your signa­ture and do nothing else. To be more precise, every bounded part of a plan alge­braic curve is a hinge trajec­tory for some plane linkage. The theorem itself is true, but the mistake in Kempe's proof was found only in 1984 and corrected by the end of the XX century.

Anti-paral­lel­o­gram inherits from paral­lel­o­gram the equality of oppo­site sides. It turns out that there is a constraint on the angles to: they are pair­wise equal!

Attach a smaller paral­lel­o­gram angles at the red hinge are also equal.

Stretching the directing lines we get a linkage one can use to draw a bisector of any given angle.