Some additional notes on a paper folding problem, and it's solutions

 A few days ago I re-posted a 2014 paper folding exercise, and closed with a unanswered problem about the minimization of a crease. Today I would like to answer the general question as posed and solved about a century ago by British Puzzleist Henry E Dudeney in his 1926 book Puzzles and Solutions; “Fold a [rectangular] page, so that the bottom outside corner touches the inside edge and the crease is the shortest possible. That is about as simple a question as we could put, but it will puzzle a good many readers to discover just where to make that fold.”

If we make the bottom of the rectangular sheet AB, then Dudeney's solution calls for folding the sheet to form a crease that is the perpendicular bisector of AB at forming ACB, then again bisecting AC to form point D. Now he departs paper folding for a compass construction to make a semi-circle with diameter equal to DB (see image below. Point E is selected where this circle crosses the bisector of AB, and the crease, Dudeney says, is the line from D through E to the crease.

 Dudeney offered no proof of his solution.  

Martin Gardner in his July 1959 Scientific American column.  After describing some paper folding techniques for math including how to fold a paper to form multiple creases to give the envelope of a parabola, he writes, "Closely related to this folding procedure is an interesting problem in elementary calculus.".  Gardner then proves that the fold DF shown by Dudeney (without mentioning Dudeney, which is strange as he re-published Dudeney's book mentioned above only two years sooner)  is indeed, the minimum length crease.  I will provide an example solution from the calculus now, but point out that if you accept that the minimum crease happens 1/4 of the way along AB, an excellent geometry problem is to find the length of L from straight geometry and algebra.  

We begin with the Pythagorean Theorem to get L2=x2+(y+z)2.$L^2=x^2+(y+z{)}^2.$

We need to get rid of some variables, so we look apply the same thm. to the triangle AGE, to get x2=y2+(1−x)2$x^2=y^2+(1-x{)}^2$.   to get y2=2x−1$y^2=2x-1$ .

now we look to the similar triangles AEG and FEH.  Using the ratios FHEH:GAEA$\frac{FH}{EH}:\frac{GA}{EA}$ we can arrive at z=1−xy$z=\frac{1-x}{y}$ 

Inserting this into the original equation for z2$z^2$ we can arrive at z2=1−x22x−1$z^2=\frac{1-x^2}{2x-1}$

And now we can rewrite the original L^2 equation with substituting in all variables for x to get the rather ugly but workable, L2=x2+(2x−1)+2(1−x)+(1−x22x−1)$L^2=x^2+(2x-1)+2(1-x)+(\frac{1-x^2}{2x-1})$  

at this point the geometry algebra solution for L^2 can be found by substitution of x= 3/4 (if we believe Dudeney).  

The better Algebra student can graph this relation on his Graphing Calculator or computer software and see that the minimum value does appear at x=.75 and yield the approximate value of L^2

Otherwise some more simplification can produce the more manageable L2=2x32x−1$L^2=\frac{2x^3}{2x-1}$.  

From here the Calculus student can just apply the derivative and set it equal to zero.

2x2(4x−3)(1−2x)2$\frac{2x^2(4x-3)}{(1-2x{)}^2}$ =0  

This gives the two possible solutions of x=0 and x=3/4.  Refusing the solution that the shortest crease length is 0, we arrive at x=3/4 , and from evaluating this in the last L^2 simplification gives us a length of L= 33√4$\frac{3\sqrt{3}}{4}$ 

As an added joy, while researching this I came across a great little source of fun ideas all related to that most simple geometric object, the equilateral triangle.  Stay tuned, be surprised, maybe even  learn a new term, I did.