# Illusory area

Discussion in 'Physics & Math' started by noodler, Sep 28, 2009.

1. ### noodlerBannedBanned

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There is a way to divide up a square into parts, and reassemble the parts so there is an extra bit of area in the reassembly.

If the square has sides 8 units long, how do you rearrange the sides in effect, so that you can make an area equal to 8x8 + 1? This means sectioning the square a certain way, and translating and rotating the sections (in a certain way, which has a minimal # of such required movements).

This is simple multiplication and "triangulation" of the square, but why do you get the extra 1/64th, in terms of total area, and where does it come from or go to, when you decompose the second figure back into a square?

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3. ### PeteIt's not rocket surgeryModerator

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Like a hole in the middle?

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5. ### noodlerBannedBanned

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You don't make any holes, you section the square.
Start with an 8x8 area, divide it n times, and reassemble n + 1 sections so you now have an area = 8x8 + 1 'units' in magnitude.
Why does this change after a 'simple' rearrangement of the (ratios of the) sides? Hint: a rectangle is probably the simplest transformation of the square.

You start with (5+3)(5+3); after the sectioning/translating/rotating part, you have (5+3+5)(0+5).
In effect you move 5 units from one side of the square to the other, and reduce the same side by 3 units...?
But 13x5 = 65, which is 64 + 1 the required transformation of the area 'function'. Doing the inverse makes a unit vanish, but how come? Is it related to Pythagoras, or just "arithmetic geometry"?

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7. ### James RJust this guy, you know?Staff Member

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It's not clear what you're doing, but why don't you try it with a sheet of paper.

Take a sheet of paper size 8x8 inches, say, and rule it off into 1 inch squares. Then, try to cut it and rearrange the pieces so that you end up with 65 square inches of paper, when you started with 64 (and holes don't count!).

Let me know how you go with this.

8. ### PeteIt's not rocket surgeryModerator

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Common sense and good mathematics says that this can't be done.
I strongly suspect that you're making a mistake.

Last edited: Sep 28, 2009
9. ### kevinalmRegistered Senior Member

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That's a rather old bit. The trick is that a casual observer dosen't notice that the pieces don't precisely fit. There are thin 'slivers' of space between that add up to one square.

10. ### noodlerBannedBanned

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Well, someone went off on a bit of a tangent; dividing the square into smaller squares might get you somewhere, but the solution for those who can't see it is this:

Take one square sheet of paper, decide it has sides equal to 8 units. Divide the square into unequal [rectangular] parts in the ratio 3/5.

Now divide the unequal parts again, but divide both in half in the following way: the larger part (5x8) is sectioned across its longer sides, from a point 3 units along the first side to a point 5 units along the other side. The smaller part (3x8) is divided diagonally. You can now fit the right triangles to the quadrilateral sections, join these two right triangles along their hypotenuse and you have a rectangle with sides = 13x5.

Done, and done.

11. ### PeteIt's not rocket surgeryModerator

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That is similar to this illusion:

Four pieces are assembled in two different ways to make a 5x13 right triangle.
The first arrangement neatly completes the triangle, but the second arrangement leaves a hole. What goes on?

12. ### PeteIt's not rocket surgeryModerator

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The difficulty is that they fit together to make a convex quadrilateral, not a right triangle. What appears to be the hypotenuse of a 5x13 right triangle is not a single straight line, but two lines with a small angle between them.

When you fit the two "triangles" together to make the 13x5 rectangle, you'll find a small gap between in the shape of a long, thin parallelogram... with an area of 1 unit.

13. ### kevinalmRegistered Senior Member

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The 'hypotenuse' isn't a straight line.

14. ### noodlerBannedBanned

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But, when I make two rectangles with sides of (8,3) and (8,5) respectively, then divide them as above, I get a rectangle with sides of (8,13).

There are no gaps, the hypotenuse is a straight line. it would be curved if I didn't cut the paper straight, which I probably only managed approximately. However I am certain that I could cut a sheet of something with a straight line to an arbitrary precision. Mathematically the hypotenuse of a right triangle is a straight line so there is no gap between the two 8x13 triangles.

When you halve a rectangle with sides of 8 and 3, you get two right triangles with the same sides of course, the hypotenuse is a square root (of the sums of squares on the sides).

The slope exactly matches the slope you make in a 8x5 rectangle, by cutting it from a point 3 units to a point 5 units along the two opposite longer sides. The slopes match because the quadrilateral s have a 2:5 sub-triangle above a 3:5 rectangle, add one half of a 8x3 rectangle to get a 5:13 triangle.

Do the trig, see if you can show the angles are different so they aren't similar triangles, which will defeat my argument in one.

Last edited: Sep 28, 2009
15. ### PeteIt's not rocket surgeryModerator

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I found a picture of this puzzle (by Martin Gardner):

16. ### kevinalmRegistered Senior Member

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Count the squares to calculate the slope. 2/5 vs. 3/8, that's not straight.

17. ### PeteIt's not rocket surgeryModerator

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Not quite. Look at the gradient of each section.

The side of the quadrilateral goes across 5 and up 2, a gradient of 0.4, an angle of about 21.8 degrees to the horizontal.

The side of the triangle goes across 8 and up 3, a gradient of 0.375, an angle of about 20.6 degrees to the horizontal.

18. ### PeteIt's not rocket surgeryModerator

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Correction. The puzzle was not formulated by Martin Gardner (it seems that it is at least as old as 1858), but he discussed it in Mathematics, magic, and mystery (read it at Google books).

I love Martin Gardner's books. I remember reading one when I was maybe ten, and I still make hexaflexagons for fun sometimes.

19. ### noodlerBannedBanned

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Are you saying you can't draw a rectangle with sides of 5 and 13 units, then draw a line diagonally through it?

After not being able to do this, you can't divide the two right triangles and make all the sections into a square? It can't be done there will be a hole somewhere?

Look closely at that picture...

ed: I made a typo above, please read "5x13" where you see "8x13"

Last edited: Sep 28, 2009
20. ### James RJust this guy, you know?Staff Member

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Just try it with an actual sheet of paper. Look at Pete's diagram in post #12. When you assemble the pieces, you'll find there's a thin "hole" in the middle of the right-hand diagram, of area one square unit.

21. ### kevinalmRegistered Senior Member

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He had a column (or a least was a frequent contributor) in Scientific American late 70's early 80's. Enjoyed him myself. Got a compendium of his SA articles somewhere I got as a subscription premium. Have to dig it out again.

22. ### noodlerBannedBanned

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OK, start with a sheet 5x13 inches, or cm. See what happens when you make a square sheet which is 8x8 inches or cm.

23. ### PeteIt's not rocket surgeryModerator

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No.
We're saying that the diagonal through that rectangle does not correspond with the sides of the pieces cut from the 8x8 square.

Get some graph paper, and draw a line from (0,0) to (13,5). This is the diagonal of the 5x13 rectangle.

Look closely, and you will see that the line will pass below (5,2) and above (8,3). These are the corners of the sections cut from the square.