Alive and Well in Possum Trot

I first wrote this post in December of 2014. During my recent withdrawal from trying to post updates daily, I have had time to read and be re-inspired, and as usual, geometry brought me out of my seclusion. So I will be extending some beautiful thoughts on some ties between paper-folding, geometry and even some extensions to advanced math. To all of those who sent letters with good wishes and expressions of concern for my health (I've never felt as good as now). thank you.

Anyone who has ever gone searching through the internet trying to find the history of some mathematical object or problem knows that more often than not, you will find lots of nice diversions long before you ever find what you went looking for, and sometimes those diversions sidetrack the whole search and become a new adventure of their own.

During such a search I discovered that  you can find a lot of arithmetic just folding a paper.  While searching for something else, I came across a nice paper fold to find the thirds of a sheet.  And along the way, some other beautiful geometry and algebra exercises.

But first, to find and mark 1/3 of a square.  I began with a single square sheet of grid paper (shown at top) and bisected the top edge with a fold to mark the 1/2 edge, then folded the adjacent corner up to that point.  The rest is sweet algebra.

For explanation, I sketched out the fold on geogebra and copied below:

The distance CE (labled x) is 1/2, as is the 1-x in this case.  CF and FE together must add up to one, so we can call them y and 1-y, and so we arrive at $ x^2 + y^2 = (1-y)^2$
Expanding we can get $x^2 + y^2 = 1 - 2y + y^2 $   and simplify that down to  $2y= 1 - x^2$ or more simply, just $ y = \frac{1-x^2}{2}$ 
Since we established that x= 1/2 the unit square, we can now complete that CF or y = 3/8 and since FE is 1-y we know it is 5/8.  None of which produces the 1/3 of a square I set out to find, but be patient, my children, there is another right triangle visible in the field (EDG) and since angle FED is a corner of our square, we know that they must be similar triangles with CF corresponding to ED and x corresponding to DG, or $\frac{3/8}{1/2} = \frac{1/2}{DG}$  Which gives us the DG = 2/3 (and we can get 1/3 by either bisecting DG with a fold, or simply unfolding the crease to get GB = 1/3.

I'll let you do the similar correspondence to get the  length of hypotenuse EG which turns out to be 5/6, meaning that you have all the sixths in one paper fold, GI= 1/6, GB= 1/3, CE=1/2,  DG = 2/3,  EG = 5/6.

Of course with all that information, we have to ask one final question; how long is the crease FH ?

Of course a more interesting question would be, what value of x for the length CE would make the crease FH the longest, and how long would that be? This is a topic I will return to soon with a couple of nice solutions from paper folding up through calculus. I hope you are all even 1/100th as excited as I am.