Haberdasher-Puzzle

[Werner_E]

Since @ttokoro  recently once again tried to take us into the world of riddles and puzzles (-> Find BC:CD - PTC Community), I thought I would follow up with a problem that I dealt with some time ago.

Its the 123 year old Haberdasher problem, posed originally by the famous puzzle inventor Henry Dudeney. Actually there are various different publishing dates around - 1902, 1903, 1907 ...

The task is to divide an equilateral triangle of side length 1 with three straight, but not necessarily continuous, cuts into four parts so that they can then be put together to form a square.

Dudeney provided the solution only in the form of a drawing, but without specifying the exact dimensions.

Let D and E be the midpoints of AC and AB. The first cut runs from E to a point F on AB.

The other two cuts run from D and G perpedicular to EF. G is a point an AB and it can easily be seen that AG = AF+0.5 (or equivalent GB = 0.5-AF).

The four pieces now can be assembled to form a rectangle:

Depending on the choice of F, different rectangles may result.

The task now is to determine the value for AF so that the rectangle is a perfect square:

 

There are solutions around where AF=0.25 which actually is a good approximation, but the rectangle formed that way IS NOT a perfect square.

The correct value for AF is slightly larger, about 0.2545076167.

So what I would be interested in is a (as simple and short as possible) way to determine the exact symbolic value for AF so that the four pieces finally form a perfect square.

If necessary I can post the exact solution for reference but my method to arrive at it is a bit laborious and I am hoping for an easier way.

[AlfredFlaßhaar]

A solution is offered at https://mathworld.wolfram.com/HaberdashersProblem.html as a line segment s2. It supposedly originates from Stan Wagon. I don't know the solution, but it must be a very tedious process involving area balance. The areas could be calculated using the area formula for quadrilaterals (side lengths and interior angles).

[Werner_E]

Thanks for the tip. I had already found this treatise at Wolfram, but unfortunately no derivation is given.
The expression given there agrees with my calculation. It's not that complicated, but I was still hoping for something simpler 😉
My calculation expressed the coordinates of all points that are not fixed as a function of AF and then calculated the value of AF for which the lengths of the two smaller cut lines, which are perpendicular to the first cut EF, are exactly half the side of the square. The side of the square can easily be specified because of the equality of area of the triangle and the square.
Here is the corresponding part from the calculation (the sheet is otherwise quite a mess 😉 ). The identifiers are different from the ones I gave in the drawing above.
x = AF, P3 = D and P7 = G.

We could do the calculations by hand but with point coordinates like

its was quite convenient to have Mathcad at hand 🙂

There is also a nice geometric constructive solution circulating on the net, which you could of course also simply “recalculate”. But I don't think that would be any less laborious.

E.g. Haberdasher's problem puzzle plan | CRAFTSMANSPACE

Version 2 here is the 'wrong' 0.25 approximation but version 1 (I think this was originally published by Dudeney himself) seems to be correct.

[Werner_E · Accepted solution]

I guess there is nothing more to come here.

Therefore I close the thread by showing my (not very well structured) file with the calculation of the distance AF (in the file these are the points P0 and P1).

[ttokoro]

Thanks your worksheet.

I like paper cutting Puzzles also.

Prime 11 can show your sheet without animation.