The disappearing area problem is a popular and baffling puzzle, widely available on the internet as Macromedia Flash, animated GIF, and Java animations. The puzzle shows a jigsaw of sorts made of two triangles and two polygons. If the pieces of the puzzle are rearranged, the resulting shape has a missing square, visible below at (1,8). This doesn't seem possible — where in the world does that missing square come from?
Note that the puzzle cannot be reproduced perfectly in ASCII, so these are approximations. A clean copy can be found by performing a Google image search for missing square puzzle.
7|aaaaaaa\ 7|__ cc|a\
8|________\ 8 __|cc|aa\
0|bb|__ cc|d\ 0|bb|__|aaaa\
1 2 3 4 5 1 2 3 4 5
The disappearing area problem is made possible by a very small deviation in angles between the three triangles involved — a deviation too small to be readily noticeable by eyeballing it and intentionally obscured by the thick border lines which are typically used to draw the problem.
Essentially what we have here is two triangles and two polygons that, when combined, appear to form a large triangle but in reality do not. We can prove this any number of ways, from simple area calculations to trigonometry. Taking the first triangle arrangement only:
The Pythagorean Theorem:
The hypotenuse of a right triangle can be found by the Pythagorean Theorem. The hypotenuses of the two component triangles should add up to the hypotenuse of the full size "triangle". Taking √(a2 + b2):
√(52 + 132) = 13.928388
However the two component triangles don't add up!
√(82 + 32) + √(52 + 22) = 8.544004 + 5.385165 = 13.929169
Too large by 0.000781. It is clear the large triangle is not a valid combination of the component parts, although we see how closely they do match! This is why they seem to match up at a visual inspection. It defies intuition that such a small mismatch can account for a whole 1 unit square missing though.
The area of a triangle is ½ base times height. The full size "triangle" should have a base of 5 and a height of 13. Therefore:
(5×13)/2 = 32.5
However the pieces that make it up are:
(2×5)/2 + (3×8)/2 + (3×5) = 32.0
There's a missing half a unit here. Conversely, the rearranged triangle has half an extra unit. The two missing halves from the two rearranged triangles are what cause the missing 1 unit square to mysteriously disappear.
Intuitively, the top angle made by the hypotenuse and the 8 side, and hypotenuse and the 13 side should be exactly the same angle. By the law of similar triangles in fact, all the angles of all three triangles should be exactly the same. But let's take one that's immediately obvious:
sin-1(5/13) = 22.619865 degrees
sin-1(3/8) = 22.024313 degrees
Again, we see a mismatch of 0.595552 between the component triangles and the larger triangle they are supposed to build. Just over half a degree is very hard to see by visual inspection however, and a thick border line covers the mismatch up nicely.
Any drawing program, or even graph paper, will quickly identify the discrepancy. However if using graph paper it will help to make the triangles as large as possible, and you may have to zoom in to see the mismatch in the drawing program. Building the three triangles, 13x5, 8x3, and 5x2 and putting them together as shown will show that the hypotenuse of the 13x5 triangle is outside the "hypotenuse" of the 8x3 and 5x2 triangles put together. The three hypotenuses form a triangle with sides 13.928388, 8.544004, and 5.385165 where they "should" actually combine to form just one straight line. Again, the thick border line hides this. Nevertheless, the connecting points between the two component triangles are not a straight line, there is a very hard-to-see corner where they meet.
Area of the mismatched section:
The area of this triangle must be 0.5 — the discrepancy we discovered in the Area proof. To find it we use Hero's formula for the area of a triangle, taking the three hypotenuses as the sides of the triangle.
√(s(s-a)(s-b)(s-c)) = 0.500000
So we see that this is just an illusion, smoke and mirrors of a particularly insidious type, especially to those people who may find math and geometry somewhat difficult to understand to begin with. Although the component parts almost match up to build the larger triangle, there is just enough of a mismatch to account for that missing square in the rearranged triangle. However it is spread over such a wide area that it becomes very thin, and easily hidden with a thickly drawn border line.
Wow! How do I make my own?
These triangles are all based on Fibonacci Numbers. If Fk is element k in the Fibonacci Sequence, the large triangle will have sides Fk+1 and Fk-1. The medium triangle will have sides Fk and Fk-2. The small triangle will have sides Fk-1 and Fk-3.
It is a property of Fibonacci Numbers that if you make a square with sides Fk then a rectangle with sides Fk-1 and Fk+1 will have an area which is off by exactly ±1 unit. This example uses F6 which is 8 (1 1 2 3 5 8 13). The large triangle is half of the rectangle with sides F5 and F7 which are 5 and 13. However the component parts are actually half of the original F6 square!
1| |1 2 3 4 5|1
2| | |2
3| |___ _____|3
4| | | |1
5| | | |2
6| | | |3
7| | | |4
1 2 3 1 2 1 2 3
We see that this 8×8 square is made of the component parts of two of the triangles. Two 8×3 triangles make the 8×3 rectangle, etc. This makes a square with area 64, which is of course twice what we found in the Area proof. Putting two of the large triangles together however gives us a rectangle with the expected off-by-one area, 13×5 = 65. Cutting the problem in half to make triangles instead of squares and rectangles simply disguises the fact that the resulting parts don't add up. Using the squares and rectangles would immediately show the missing square, but in the triangles the missing half square is harder to see.
Taking the next Fibonacci number would result in a large triangle with sides 21×8 "made" from two component triangles with sides 13×5 and 8×3. The larger k becomes, the harder the missing piece is to see because the area is always off by exactly one unit. With a larger k, that one unit is a smaller fraction of the whole and therefore less obvious.