Thursday, 11 August 2022

Mind Your Decisions - maths fail

Watch the first minute or so of his recent video, until he's explained the question, then press pause:

He rambles on for another five minutes explaining, rather too quickly, an unnecessarily complicated way of working it out. Try doing it my way (below) first to really torture yourself when you are watching it!

I left the following comment:

Far too complicated!

It's length 4 from one corner of a hexagon to the opposite corner.
'Vertical' height of each triangle in the hexagon = sq rt 3.
Put the square flat on the table, fold up the flaps and look at it from the side.
Point E/F is 1 'across' and has hypotenuse sq rt 3.
Acos 1/sq rt 3 = 54.74 deg (no need to know length of the other side).

Then rotate the square so that you are looking at it diagonally from one corner.
Distance from E/F to a point diagonally opposite (assuming there are four hexagons all folded up) = 4 x sq rt 2.
Distance diagonally across the square = 2 x sq rt 2.
So point E/F is sq rt 2 'across' and has hypotenuse 2 (side length of hexagon).
Acos sq rt 2/2 = 45 deg (no need to know length of the other side).

Took my daugher two minutes once I had explained the basic approach. OK, she's studying maths...


Chromatistes said...

He's starting to construct a truncated octahedron.

Mark Wadsworth said...

C, indeed, but I'm not clever enough to know it's angles off by heart.

Bayard said...

Part 2, if you look at the figure at 1':05" and assume it to be two dimensional, then you can quickly see that the distance A-E, shall we call it y, is half the distance A-C, (rt2/2)y. Now if you return the figure to three dimensions, bisect it along the line E-A-C and look at it from the side you can see that the distance between A and the point vertically below E is the same distance as A-E when the figure was three-dimensional. Since the triangle A-G-E is a right-angle triangle then the distance E-G is rt(y^2-(rt2/2)y^2) = rt((1/2)y^2) = rt2/2 = A-G. Therefore the triangle must be isoceles and the included angle must be 45 degrees, (180-90)/2. No tables required.

Bayard said...

Sorry, that should be A-B = y and half A-C =(rt2/2)y.

Mark Wadsworth said...

B, fair enough. With maths, what takes time is working out HOW to work it out, not actually working it out once you know HOW. Yours is easier to work out, but it would have taken longer to devise HOW to work it out.

Bayard said...

It depends how long it takes you to suspect that the triangle A-G-E is an isoceles triangle. In both our cases we knew that as soon as we saw the result. It would have taken me a lot longer if I hadn't seen the result first, true, but my initial idea was always to work out the dimensions first.

Mark Wadsworth said...

B, if I could have been bothered, I could have worked out height from the first part - hyp sq rt 3, width 1 = height sq rt 2. Which is clearly equal to where point E/F ends up 'across' in part 2.

But you had to use your calculator to do acos for part 1, so I thought, use consistent approach for part 2.