
View Full Version : Illusory area
noodler 092709, 07:32 PM 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?
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.
Like a hole in the middle?
noodler 092709, 08:05 PM 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"?
James R 092709, 09:53 PM 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.
You start with (5+3)(5+3); after the sectioning/translating/rotating part, you have (5+3+5)(0+5).
Common sense and good mathematics says that this can't be done.
I strongly suspect that you're making a mistake.
kevinalm 092709, 10:00 PM 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"?
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.
noodler 092709, 10:02 PM 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.
That is similar to this illusion:
http://www.opticalillusions.info/Illusions_images/Missingsquare.png
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?
You can now fit the right triangles to the quadrilateral sections, join these two right triangles...
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.
kevinalm 092709, 10:17 PM The 'hypotenuse' isn't a straight line.
noodler 092709, 10:21 PM 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 subtriangle 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.
I found a picture of this puzzle (by Martin Gardner):
http://www.mathematik.unibielefeld.de/~sillke/pictures/puzzles/fakedissect.gif
kevinalm 092709, 10:26 PM 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.
Count the squares to calculate the slope. 2/5 vs. 3/8, that's not straight.
There are no gaps, the hypotenuse is a straight line.
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.
I found a picture of this puzzle (by Martin Gardner):
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 (http://books.google.com.au/books?id=PUCdaD4l8C&lpg=PA135&ots=rubpx2FdJ&dq=martin%20gardner%20square%20rectangle%2064%2065&pg=PA132#v=onepage&q=&f=false)).
I love Martin Gardner's books. I remember reading one when I was maybe ten, and I still make hexaflexagons for fun sometimes.
noodler 092709, 10:36 PM 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"
James R 092709, 10:41 PM 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 righthand diagram, of area one square unit.
kevinalm 092709, 10:45 PM 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 (http://books.google.com.au/books?id=PUCdaD4l8C&lpg=PA135&ots=rubpx2FdJ&dq=martin%20gardner%20square%20rectangle%2064%2065&pg=PA132#v=onepage&q=&f=false)).
I love Martin Gardner's books. I remember reading one when I was maybe ten, and I still make hexaflexagons for fun sometimes.
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.
noodler 092709, 10:46 PM OK, start with a sheet 5x13 inches, or cm. See what happens when you make a square sheet which is 8x8 inches or cm.
Follow the pattern http://www.mathematik.unibielefeld.de/~sillke/pictures/puzzles/fakedissect.gif
Are you saying you can't draw a rectangle with sides of 5 and 13 units, then draw a line diagonally through it?
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.
kevinalm 092709, 10:51 PM Or in other words we're saying that the diagonal through the final arrangement isn't a line, it's a _very_ thin parallelogram, if that's any clearer.
OK, start with a sheet 5x13 inches, or cm. See what happens when you make a square sheet which is 8x8 inches or cm.
The sections cut from the rectangle will be slightly bigger than they should be to assemble the square, and you will have a small overlap.
For example, consider the triangle which should have sides 8 and 3.
The triangle you cut from the 5x13 rectangle must have sides in the same proportion of 5 to 13, right?
So, if you cut it to make one side 8, the other side will be slightly more than 3.
(3 and a thirteenth, to be precise.)
That is similar to this illusion:
http://www.opticalillusions.info/Illusions_images/Missingsquare.png
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?
Noodler, look at this puzzle.
Can you confirm for yourself that the two triangles are actually quadrilaterals?
That the hypotenuse of the top triangle is actually bulging slightly downward, while the hypotenuse of the bottom triangle is bulging slightly upward?
Look closely at where the lines cross the grid corners. Draw it yourself on some graph paper. Really  go out and buy some if you have to. It's really a wonderful thing to learn and understand how this illusion works.
noodler 092709, 11:08 PM ??
The diagonal of the rhs figure, is all inside the square on the lhs.
It only means you need to cut vertically, to the diagonal from the "X" sides of the 13,5 rectangle, a little further. The vertical cuts in the inner rectangle are on the outside of part of the square.
Ahem, I guess this means the extra "unit" is all in the hypotenuse, and you cant get this to fold up exactly inside a square because of Pythagoras, I thought you could allow for the difference because of the relation between odd and even products. since the product of two odd numbers is odd, it had to do with making an even "geometric product", but there's a remainder after all.
But is there a way to divide a square other than this "illusory" way so it does approximate a rectangle with oddinteger length sides? What about nonEuclidean spaces like the sphere and so on? And why is 8 the LCM factor so you don't deal with fractions?
It must mean that the 1/64th extra unit is only illusory if you have to translate along straight edges or flat planes.
The diagonal of the rhs figure, is all inside the square on the lhs.
It only means you need to cut vertically, to the diagonal from the "X" sides of the 13,5 rectangle, a little further. The vertical cuts in the inner rectangle are on the outside of part of the square.
Um What?
The slopes match because the quadrilateral s have a 2:5 subtriangle above a 3:5 rectangle, add one half of a 8x3 rectangle to get a 5:13 triangle.
The slopes don't match.
The three triangles in question are:
5:13 (half the 5x13 rectangle)
2:5 (from the quadrilateral)
3:8 (half the 3x8 rectangle)
These are not similar triangles.
noodler 092809, 12:55 AM Yes, ok, thanks.
i was thinking about how the illusion appears to make an evensided figure into one with oddvalued sides. Anyways, it doesn't work in Euclidean space, but how would you make it work in a space with figures in it that have fractional sides?
Or how could you contract the edges of the mismatched sides, so the edges join, and you get a slight, I suppose, ridge or something, like a dihedral angle ?
noodler 092809, 01:55 AM Alrighty I think I got this; the article is about hyperbolic geometry, but in E the space of flat, intersecting planes, triangles have straight lines for edges.
The leftover in the problem shows up in two ways; because the points of the intersecting quadrilaterals should be at (5,2) and (8,3) inside the rectangle, to be on the perimeter of the 5x8 rectangle in the square.
If you make three cuts instead of one straight across from corner to corner of the 5x13, the "quadrilaterals' will have 5 sides, joining these will leave a convex void space along the intersection; the same deal with the hypotenuse of each "triangle" which is now a quadrilateral and a join means another void space. That's two void inner products from a sectioning function, both the result of an angular difference in the diagonal of a rectangle, taken directly, or in sections (the middle section will intercept the other sides near the corners).
With a straight diagonal there will be a mismatch in the sides of the two rectangles for the square, the larger will be slightly more than 8 units long, the smaller will be more than 3 units wide.
noodler 092809, 05:43 AM Damn, I'm imagining things. The three cuts diagonally across the 13x5 rectangle go from (0,0) to (5,2), then (5,2) to (8,3) then to (13,5). The larger subrectangle for the square will match up, each half of it is a 5x5 square with a 2:5 triangle removed.
The remaining triangles from the RHS rectangle, will be concave along their hypotenuse, or if you make the inner cut go from (5,2) to (8,3) you put all the void space in the triangles that make up the smaller rectangle in the square. If the inner cut extends beyond this range there should be two void spaces. The question is then, why does a figure with 65 square units, become a figure with 64 units, some of which is missing?
noodler 092809, 06:32 PM Oops I think I got that wrong, the RHS will have a zigzag through it that means the "triangles" will have slightly convex sides, sliding them together means there will be an overlap in the smaller rectangle of the square.
But if you extend the "zag" or middle part of the cut the resulting larger rectangle will have the correct length sides but part of the middle missing. The missing part will be in the overlap of the other two sections in the smaller rectangle. From the square to the rectangle you expand the inner space, from the rectangle to the square you have to fold the extra real unit up.
It all sort of becomes apparent when you draw the lines, through the central point of the 13x5 rectangle. You can see that an inner section from (5,2) to (8,3) means the "triangles" have a bent hypotenuse, the slope of this section is 1/3, the outer "zigs" are 2/5 so sliding the sides together will mean an excess, which is where the extra unit is.
So no real paradox in Euclidean space then.
Well done!
Kudos to you, noodler, for thinking it through so thoroughly.
