What's the trick to this .gif image?

I’m sure everybody but me has seen this before, but here’s an animated gif that shows 8x8 == 5x13, that is, 64 == 65. Obviously there’s a trick to it, what’s the trick?

It looks to me like some “fudging” is going on when the various portions get moved around (I swear I saw some “blurring”).

Yup, it cheats.

Here ya go: http://www.jimloy.com/puzz/missing.htm

In the second array, there is a long, thin parallelogram between the other pieces, with an area of 1 square. It doesn’t show up well in low-resolution displays.

Oh feh. I was hoping it was something more cool than “use thick edges on the line so people don’t notice the gap.”

The slope of the diagonals on the two shapes is different: for the triangles, it’s 3/8, while on the quadrilaterals it’s 2/5. You can see that in the rearrangement, the main diagonal looks a bit “fat” due to these different angles; that’s where the extra square went.

“…comes from.”

My way to disprove this would be to add up the angles. You know the angles at the corners have to add up to 90 degrees and the angles in the center have to add up to 180 degrees.

I’ll leave the actual calculations to the reader, since this is another way of noticing the slope issue that Dr. Strangelove points out.

It’s similar to this one:

http://en.wikipedia.org/wiki/Missing_square_puzzle

Note that that one doesn’t rely on thick lines to cover an obvious gap, since the fudged border is the outer edge of the figure. It points out another factor - the eye just isn’t very good at noticing small differences in slope, and will easily see two adjoining line segments as a straight line when they aren’t.

There are numerous variations on this, some of them artistic, and Sam Loyd had a lot of those.

In the link quoted here, see the further link to the “Get Off The Earth” puzzle (near the bottom of the page) for a well-known one.

The area of the parallelogram in the middle of the rectangle would have to be 0 for the diagram to be accurate. Which would mean that the magnitude of the cross product of (5, 2, 0) and (8, 3, 0) would be 0. Calculating however, gives us

(5, 2, 0) × (8, 3, 0) = (2 ⋅ 0 − 0 ⋅ 3)i + (0 ⋅ 8 − 5 ⋅ 0)j + (5 ⋅ 3 − 2 ⋅ 8)k

which works out to

(5, 2, 0) × (8, 3, 0) = 0i + 0j − 1k

and then to

(5, 2, 0) × (8, 3, 0) = − k.

The magnitude of that, of course, is just 1, which, of course, is also the area of the hidden parallelogram.

It would be cool to go the angle route and calculate the sines and arctangents exactly, and then see everything cancel out in the end, but I think that would be a giant mess. I might do it later.