The Art Of Mathematical Origami

         Over the holidays I came across a puzzle book entitled ‘Can You Solve These?’  Naturally I had the instant urge to prove that indeed I could solve every puzzle that appeared in the book.  As indeed I did, and in fact the next few posts on this blog will be regarding some of the techniques I used and some nice little proofs I came up with along the way.

         This particular post concerns a geometric puzzle in which the puzzler is required to determine whether a particular arrangement of paper is reachable by means of a finite number of folds (without cutting or tearing).  The first step in my solution was to establish that to fold one point of the paper onto another, a fold must be made that is the perpendicular bisector of the line joining the two points.  It makes sense intuitively that the point to be translated and the destination be equidistant from the fold line, and it seems quite acceptable that this fold always be at right angles to the segment between these points, but this result can be proved rigourously.

        Firstly, we state that we wish to translate point A on a sheet of arbitrarily sized paper onto point a on the same paper.  To do this, we make a fold from points B and C on the edges of the paper.  Now we know that AB will end up lying on aB, and AC will lie on aC, and therefore |AB| = |aB| and |AC| = |aC|.  Marking these in we see that ABaC is a kite, and therefore the vertical diagonal, BC is the perpendicular bisector of the horizontal one, Aa.

Fig. 25 Translating points on a page by folding.

          The next step in my investigation was to extend the method to a translation of two points simultaneously.  I soon established a means for doing so.  Beginning with any sheet of paper with two points, A and B marked on it, and two others with the same distance as the previous, a and b, which are not necessarily on the same angle.

Fig. 26 Translating a set of points

          We now make a fold, CD, in order to translate A such that AB is parallel to ab, A’B.  If B were below this fold line, the process would be more difficult.  Thankfully, in my case B was well above, but in this example B lies on the fold lie, which is permissible, since B needn’t move from its original position.  Now, since A’B || ab, the distance apart is equal and a line which is perpendicular to one is necessarily perpendicular to the other, and therefore, the fold that translates A onto a, C’D’, also translates B onto b (excuse the diagram, the points aren't placed perfectly parallel and equal in distance).  This was all I needed to complete the proof!