# Dissection puzzle from the Gardner's book. How to define points without tools?

The figure shows a fairly well-known puzzle from the book by Martin Gardner. You need to cut the regular hexagonal star into pieces and fold the square.

Question: how to define points (marked in red) without using additional tools?

I am looking for an origami technique.

• What makes you think that Martin Gardner defined those points without any additional tools? Oct 12 '20 at 11:16
• I haven't done the calculation, but I believe my comments on this question also applies here. Oct 12 '20 at 11:25
• I'm pretty sure all of Martin Gardner's dissection puzzle solutions are constructible with compass and straightedge (and therefore paper folding too). Oct 12 '20 at 11:30
• Just checked and the puzzles include regular 7 and 9-gons, where compass and straightedge may fail, but paper folding will still be possible. Oct 12 '20 at 13:43

Not sure this is the most elegant way of doing it but it does the job:

Make a crease at the midline (in particular, marking the midpoint M of the base) and cut off the top and bottom triangle.
Align one of the triangles with its base on the midline and its midline on the base as in the picture. Its tip marks the point N. Make a vertical crease through the midpoint B of N and corner A.
Insert the two triangles on the left as indicated in the picture and mark the long vertical line trough their tips. Fold through B such that M falls on that vertical line at new point M'. Mark the horizontal midline and fold through A such that M' falls onto the horizontal midline. (Note: The fold is meets the top of the shape very close to but not exactly at the vertical midline.) This fold is one of the two we require. Fold through corner C and perpendicular to this fold to obtain the other one.

Let's assume the small triangles have sides of length 1. Then the square will have area $$3\sqrt{3}$$ and its side will be $$\sqrt{3\sqrt{3}}$$. We can check that the distance $$AM'$$ is half that length. Indeed, by construction we have $$AM'^2=BM^2-AB^2$$. These have lengths $$AB=\frac {3-\sqrt{3}}{4}$$ and $$BM=\frac{3+\sqrt{3}}{4}$$ such that $$AM^2=\frac{3\sqrt{3}}{4}$$ as claimed.

No solution – an experiment

This began before the "origami" restriction was edited into the question.

I asked myself: does it matter where point 2 is, so long as it forms a right angle?
First I cut off half the paper and used one edge as a ruler to draw line AB.
I then positioned the right-angled corner of the paper so that its sides touch C and D.
And drew the lines CE and DE to its corner, then extended DE to F.

The lines weres delicate and thin and did not copy/resize well, so I re-drew with a ruler.
I then cut it along the green lines (not AB) and reformed the parts as shown.

The result isn't a square but a rectangle, so I have not solved the puzzle.

Are the two top corners square?
Looking at the top right corner, the two sliver angles come from points C and D.
The other parts of C and D always sum to 90°, so the slivers always sum to 30°.