Saturday, February 22, 2014

An animated puzzle

Swiped from Jake VanderPlas:


TC said...

The puzzle, I take it, is how can you have four pieces which are identical be configured in such a way that although they cover the exact same area in the larger triangle come up with a missing square in one. Simply by arranging them in a different order they fail to cover one square. Hmmm

bjkeefe said...

Yes, that's the question.

TC said...

A very clever illusion. In either position of the colored areas, the blank area to the upper left is not half the total area of the 5 X 13 block. The boundary between the triangles and the upper left blank space is not a straight line, even though it appears to be. It is a broken line with the bulge of color extending more into the blank space when the green triangle is down, and less when the green is up. That's where the area for the "extra" blank square comes from.

When the green triangle is down look where the top edge of the green crosses the grid behind it. Then look where the line crosses the grid when the green triangle is up. It changes slightly. Since it appears to run from corner to corner it seems like the same line , but not so much.

bjkeefe said...

I see the artifacts, and that was my first guess when I first saw the puzzle. But I don't think that's it. I think it is a difference in area in rectangles due to the larger (or smaller) difference between the length of the sides.

If you remember your introductory calculus, you might remember the problem of maximizing the area of a rectangle of fixed perimeter. It turns out that a square covers/enclosed the most area for a given perimeter, and that the more the rectangle gets skinnier, the less area it covers/encloses.

If you hover your mouse over the image, you'll see a suggestion I made largely with you in mind.

toma said...

I think the two triangular pieces are shaped differently. They're not similar. The green one is a little steeper in incline (the slope, or rise over run, of the top line is 2/5 or .400) where the blue one is a little flatter (3/8 or .375). So as TC said, the true (hinted at) dividing line running from the bottom left corner to the top right corner of the graph is not being displayed by the puzzle. The slope of that imaginary line is 5/13, or .385, in between the slopes of the triangular pieces. When the blue piece is on the left, the 'dividing line' displayed by the pieces together runs below the true dividing line. When the green piece shifts to the left, the 'dividing line' displayed by the pieces together runs above the true dividing line. It's like plucking a guitar string and watching it vibrate to either side of its original place. With the green piece to the left, and the dividing line on the high side, there's more space underneath than there was before. Great puzzle.