How come $32.5 = 31.5$?

Question

Below is a visual proof (!) that $32.5 = 31.5$. How could that be?

Answer 1

It's an optical illusion - neither the first nor second set of blocks actually describes a triangle. The diagonal edge of the first is slightly concave and that of the second is slightly convex.

To see clearly, look at the gradients of the hypotenuses of the red and blue triangles - they're not 'similar'.

gradient of blue triangle hypotenuse = 2/5
gradient of red triangle hypotenuse = 3/8

Since these gradients are different, combining them in the ways shown in the diagram does not produce an overall straight (diagonal) line.

Answer 2

Since this question has been bumped up, and some people find pictures helpful, may as well supply illustrations of the explanation.

The area of the red triangle is $\frac12(8)(3) = 12$ and that of the blue triangle is $\frac12(5)(2) = 5$. The area of the yellow and green regions are clearly 7 and 8 respectively, so the total coloured area is 12 + 5 + 7 + 8 = 32 which is less than 32.5, the area of a 13×5 triangle. Indeed, in the figure below you can see that the hypotenuses of the red and blue triangles "dip" below the actual hypotenuse of the large triangle; this accounts for the 0.5 difference in area.

Similarly, in the second figure, the hypotenuse of the red and blue triangles are above the actual hypotenuse, which accounts for the 0.5 difference in area again.

The difference between the two figures is a thin parallelogram with vertices at the endpoints of the hypotenuses in both figures (coloured pink below), which has area exactly 1. (If it looks much less, that's the nature of the optical illusion.)

Also, there are some links to similar figures at Wikipedia's article on this missing square puzzle.

Answer 3

Surprisingly, the general case remains to be explained.

Consider the recursion $F(n+2) = F(n+1) + F(n)$, with $F(1) = F(2) = 1$
producing the Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 ...

Let the blue triangle have horizontal and vertical sides of length $F(n+1)$ and $F(n-1)$, respectively.
Let the red triangle have horizontal and vertical sides of length $F(n+2)$ and $F(n)$, respectively.
So that
in the upper figure the two triangles create a rectangle with an area of $F(n)F(n+1)$
and
in the lower figure the two triangles create a rectangle with an area of $F(n-1)F(n+2)$

By the Fibonacci identity $F(n-1)F(n+2) - F(n)F(n+1) = \pm1$
we can see that the two rectangles are bound to differ in area by exactly one unit.

As n increases the gradient/slope of each triangle tends to the square of the Golden Ratio $ \phi$,
one from above and one from below.
Thus, as $\phi^2$ = 0.3819660112...
2/5 = 0.4 > $\phi^2$
3/8 = 0.375 < $\phi^2$
5/13 = 0.3846... > $\phi^2$
etc.

Answer 4

Overlay the two triangles to see the difference. This is known as the missing square puzzle.

Answer 5

The red and blue triangles are not similar (the ratios of the sides are 3/8 = 0.375 and 2/5 = 0.4 respectively), so the "hypotenuse" of your big triangle is not a straight line.

Answer 6

No, this is not a proof of your statement. If you look very closely, you will see that the hypotenuse of the triangles aren't straight. You can also verify this algebraically by calculating the internal angles of the triangle using trigonometry.

The other way to demonstrate the non-straightness of the hypotenuse is in the fact that the red and blue triangles aren't similar, which can be seen by the difference in the ratios of their width and height. For the blue triangle, this is 2/5 and for the red triangle this is 3/8. As they are both right angled triangles, these ratios would be the same for similar triangles and the fact that they aren't means the interior angles are different.