bigideas

Following my post about visual methods from a few weeks ago, here are some solutions to the puzzles. If you haven’t already, you might like to read the original post and try them for yourself before reading on.

Problem 1

Problem 1 asked you to find an expression for 1 + 3 + 5 + … + (2n – 1), the sum of the first n odd numbers. We can represent the sum as a triangle. The case n = 5 is shown in the following figure:

We observe that this can be split into two triangles and then reformed into an n×n square:

Thus,

1 + 3 + 5 + … + 2n-1 = n²

Another way of putting this is that the sum of the n^th triangular number and its predecessor is n²:

T_n + T_n-1 = n²

Observe also that by starting with an n×n square, we can see how it is made up of the sum of the first n odd numbers in a quite different way:

Problem 2

Problem 2 asked you to cut a pentomino into three pieces that can be rearranged into a square.

One way to do this is to tile the plane with the pentomino and overlay a grid of squares. By sliding this around we arrive at a three-piece dissection:

The same kind of tiling and dissection works for the more general shape made up of two squares. If these squares are of sides a and b, then we observe that one of the pieces is a right-angled triangle with sides a and b whose hypotenuse (of length c, say) forms one of the sides of the square after rearrangment. The area of the original shape is a² + b², and after rearrangement the area is c².

Thus,

a² + b² = c²

which means we’ve proved Pythagoras’s theorem.

Problem 3

The final problem asked you for the proportion of the given figure that was shaded.

First observe that we can rotate every other square within its bounding circle by 45 degrees without altering area of the shaded region. This gives us:

We examine the top left quarter and rearrange it such that it is made up of three equal series of squares, two of which are shaded:

We conclude that 2/3 of this quarter is shaded, and hence 2/3 of the original square is shaded (since the quarters are all the same).

As Peter points out in the comments to the original post, this is a slightly different kind of problem because it involves an infinite series. Toby’s solution (also in the comments), which uses a rather refined 18th Century sort of algebraic approach, relies on our being able to prove that, for example, we may divide out a factor from an infinite series, and that fact isn’t obvious. One nice thing about the visual approach is that it makes even something like this seem intuitively “right”.