Visual Methods

21 February 2008 – 6:23 pm by Phil O'Donnell

A well-constructed drawing can help to understand or stimulate a mathematical idea. The following puzzles explore some of the visual techniques that can be employed.

We begin by considering representations of the integers, and use the principle that if you count the same thing in two different ways you reach the same total.

We can represent the sum of the first n consecutive integers,

1 + 2 + 3 + … + n

as the quantity of dots in the following figure:


fig1_small.png

By considering an upside down copy of these dots, and by adjoining this copy to the original we are left with a rectangle of n by n+1 dots. The next figure demonstrates this for the case n = 5.

fig2_small.png

So, twice the sum of the first n integers is n(n+1), and therefore:

1 + 2 + 3 + … + n = n(n+1)/2

This is the nth of the so called triangle numbers, normally written Tn, which tend to crop up frequently in combinatorics.

Problem 1: What is 1 + 3 + 5 + … + (2n - 1), the sum of the first n odd numbers?

Next, we consider the dissection and rearrangement of two-dimensional shapes.

A beautiful example is as follows. Given a right angled triangle with sides a, b and c (for the hypotenuse), we observe that we can arrange four copies of it within a square of side a+b in two different ways:


fig3_small.png
The shaded regions in both squares must be equal in area since all we have done is to rearrange the positions of the triangles within the larger square. The shaded region in the first square has area c2, and the shaded regions in the second square have areas a2 and b2. This gives us the famous Pythagorean theorem:a2 + b2 = c2An interesting problem is to cut a cross made up of five squares into four equal parts such that they can be reassembled into a square. The solution is as follows:

fig4_small.png
But how would one arrive at such a solution in the first place?
A common way to approach such problems is to first tile the plane with the shape to be dissected and then tile the plane again with the shape that is our objective.If we can overlay these two tilings in such a way that they match then the required dissection becomes clear. To see what this means we examine the above problem. The cross tiles the plane as follows:

fig5_small.png
By joining the centres of each cross together, we get an overlaid square tiling of the plane:

fig6_small.png
This gives the dissection into four equal parts, but observe that we can slide the two tilings about to get countless other dissections, although not necessarily into equal parts:

fig7_small.png

Problem 2: Cut the following shape, a pentomino, into three pieces (not all equal) that can be rearranged into a square.


fig8_small.png

Then, by considering the more general shape made up of two squares, prove the Pythagorean theorem!

fig9_small.png

Finally, we look at how infinite series can be represented visually.

By starting with a square of side 1, and repeatedly cutting off half of what remains we can arrive at the following infintite dissection:


fig10_small.png
The areas of each piece cut are the geometric series 1/2, 1/4, 1/8, 1/16, … whose nth term is (1/2)n. The dissection demonstrates that the total area of these pieces is 1. That is:1/2 + 1/4 + 1/8 + 1/16 + … = 1If instead we were to repeatedly divide into thirds, we might arrive at the following disection:

fig11_small.png
By shading in pieces corresponding to the geometric series 1/3, 1/9, 1/27, 1/81, … whose nth term is (1/3)n, we see that the shaded and non shaded regions are identical interlocking spirals:

fig12_small.png
The area of each region is therefore 1/2, and so:1/3 + 1/9 + 1/27 + 1/81 + … = 1/2

Problem 3: The following figure is constructed from a square in to which we fit a circle into which we fit a square into which we fit a circle … and so on. What portion of the original square is shaded?


fig13_small.png

This is a Big Ideas blog post. As you may know, Big Ideas is a series of events in London. We also podcast, blog and get up to other things.


  1. 5 Responses to “Visual Methods”

  2. Dig the pictures. What did you use to draw them?

    Have to say I think most of us have been taught to solve the infinite sum algebraically:

    x = 1/3 + 1/9 + ….

    So:
    x = 1/3 + 1/3(1/3 + 1/9 + …)

    which means that:
    x= 1/3 + x/3

    Rearrange to get x = 1/2.

    That approach seems pretty natural to me so don’t see much benefit from a geometric approach, esp since it’s obvious that you can reuse the algebraic technique whilst the rather clever geometric observation feels like it’s a bit of a one-off.

    For your last example (circles/squares) are you looking for someone to try to work out the solution?

    I suppose you can observe (geometrically) that the second largest square has half the area of the outer by rotating it by 45 degrees. Is that the geometric trick you are looking for or is there something more fiendishly clever you can do?

    I’d use a bit of algebra and a bit of geometry to solve.

    Let:
    s = shaded portion of outer square
    = unshaded portion of second largest square
    u = unshaded portion of outer square
    = shaded portion of second largest square

    Then obviously:
    s = 1-u

    Since we know the ratios of the area is 1/2 already, we get:

    s = 1/2 + 1/2 * shaded portion of second largest square
    = 1/2 + u/2

    The first 1/2 comes from the observation that the outer green stripe occupies half of the outer square as the second outer square is half the size.

    Since u = 1 - s we have:

    s = 1/2 + (1-s)/2

    Solve to get:
    s = 2/3

    … which feels plausible to me.

    T

    By Toby on Feb 22, 2008

  3. Toby, you wrote:

    “x= 1/3 + x/3

    Rearrange to get x = 1/2.”

    The quantity represented by the symbol “x” is an infinite series which may not converge to a finite value. The standard algebraic rules for manipulation of symbols representing finite numbers do not necessarily apply to symbols representing infinite quantities. Before you could execute this “rearrange” action, you would first need to demonstrate to me that such actions are legitimate in dealing with infinite quantities. In other words, I can’t get from here:

    “x= 1/3 + x/3″

    to here:

    “Rearrange to get x = 1/2.”

    without first doing some deep pure mathematics.

    By Peter on Mar 3, 2008

  4. Great comment

    By kpss Matematik on May 29, 2008

  5. well, Toby’s logic would presume the infinite series converges. The answer =1/2 would reinforce that it does, right? (This is definitely NOT my area of xpertise). A converging infinite series “number” is still just a definable “number”.

    By Bill on Jun 22, 2008

  1. 1 Trackback(s)

  2. Feb 22, 2008: Carnival of Mathematics 1000 « JD2718

Post a Comment