say it with pictures

Say It with PicturesAuthor(s): Tandy ClausenSource: Mathematics in School, Vol. 21, No. 4 (Sep., 1992), pp. 17-19Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30214905 .

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Clore Gallery, Tate, London

SAY

IT

WITH

1PIC TUR ES

by Tandy Clausen The Jonathan Miller School, Berkshire

One approach to many of the standard investigations which are commonly offered by the arbiters of GCSE coursework assessment involves the systematic collection of numerical data from a sequence of diagrams. The pupil then tabulates this data, and searches for a number pattern from which a generalized formula may be conjectured. Mark schemes designed to measure the pupil's aptitude for this type of task often assume such an approach, and apportion marks accordingly.

But the approach through number patterns may not be the only one available. To take one simple example, given a sequence of patterns of square tiles:

I~I1 O O O

the pupil may be asked to predict the number of tiles in the nth member of the sequence. She may draw a few more patterns, count the tiles and produce a table:

n: No. of tiles:

1 2 3 4 5 6 1 5 13 25 41 61

From this table the pupil may (or may not) be able to spot a number pattern which leads to a generalized formula.

But a different approch is possible, and may lead to a deeper understanding of the problem. This bypasses the

Mathematics in School, September 1992

number patterns and instead focusses directly on the pattern of the tiles themselves. A consideration of the tile patterns for the early members of the sequence generates a sort of generalized diagram, rather than a number pattern:

C] O O O

O

A rearrangement of the tiles in the generalized diagram leads directly to the formula:

- -n --1 -1 --

JI T..

17



It is evident from the diagrams, without recourse to any number patterns, that the total number of tiles is equal to the sum of n2 and (n - 1)2.

In the example just described either approach - working through the number patterns or working directly from the diagrams - may equally well succeed. But developing a generalized diagram or model, and then pulling the formula straight out of that, sometimes has clear advantages. For example, the formula for the sum of the first n natural numbers, 1 + 2 + 3 + .....+n=n(n+1), may be found algebraically -but the technique is complicated and hard to remember. On the other hand, one simple well-known diagram:

Ta

i f----- n + l-----j

says it all. In the same way, the sum of the squares may be

represented by 3-D models as W. A. Miller describes (Mathematics in School, 20:4), and so may the sum of the cubes:

33

+

13/ \/ o

...+n3

3+

2

2

3

+ n

In each case, the formula for the sum of the powers of n can be pulled directly out of the model. The model itself justifies the formula with no need for a lot of tedious (and

18

to the pupil often largely meaningless) algebraic manipu- lation.

So working directly from a model or diagram is sometimes much simpler than using number patterns and algebra. It can also be safer as there is no chance that irrelevant number patterns will mislead one into assuming the truth of an invalid generalization. A classic example here is the the old-established problem:

Join up n points on the edge of a circle. How many regions are created?

P Pi P2 P, 2 P 4 P

(D ) P3 (5 P3

The number pattern which may be found from a table of results is very obvious ... and wrong. The answer is not 2n-1

But by using the diagram to consider what actually happens when you add a new point and a new set of lines to the circle, we can see that:

Every time a new line crosses an old line, it enters a new region which it splits in two. So if a new line crosses t old lines, it will create t + 1 new regions.

So the question now becomes:

"How many old lines does each new line cross?"

The pattern for the number of old lines crossed by new lines can be pulled directly out of a diagram. Consider what happens when we add point P, to the circle:

P2 P2

P3 P3

P3P

P, P, / \

\ ' , 4 P4

i~~~~ I . #,

aarpr

P6 P7 P6 P \ I ">

.", ,

",\!/,,'aI

6 new lines must be drawn coming of of P-,.

How will the new P, lines cross the P6 lines?

P1P, will cross 0 of the P6 lines. P2P7 will cross 1 of the P6 lines. P3P, will cross 2 of the P6 lines. P4P, will cross 3 of the P6 lines. PsP, will cross 4 of the P6 lines. P6P7 will cross 0 of the P6 lines.

So the new P, lines will cross 1 + 2 + 3 + 4 of the P6 lines. These numbers constitute the beginning of the second

diagonal of Pascal's triangle, so their sum, 10, is to be found 1 row down on the third diagonal of the triangle:




12

3 3 1

1 6 4 10

1 s 10 5 1

In the same way, the new P, lines will cross 1 + 2 + 3 of the P5 lines:

P,P, will cross 0 of the P5 lines. P2P7 will cross 1 of the P5 lines. P3P7 will cross 2 of the P5 lines. P4P, will cross 3 of the P5 lines. PsP, will cross 0 of the P5 lines. P6P7 will cross 0 of the P5 lines.

Again this sum, 6, appears in the third diagonal of Pascal's triangle:

1 /

1 3 1

1 4 F 6 1

Similarly, the New P, lines will cross 1 + 2 of the P4 lines, and 1 of the P3 lines. These two sums, 1+2 and 1, give the first two numbers in the third diagonal:

1 1

1 2 11 2 3 =1

So the total number of old lines crossed by the new P7 lines is the "sum of the sums" of the first 4 counting numbers -

(1 +2+3+4)+(1 +2+3)+(1 +2)+(1)

Each of these sums is a number in the third diagonal of Pascal's triangle, so their sum occurs 1 row down in the fourth diagonal of the triangle:

1 1

W220 1 2

3 3 3

1 4 4 1

1 5 10 10 5

6 15 = 15 6 1

So the total number of old lines crossed by the new P,7 lines is equal to the number on the fourth diagonal 6 rows down on Pascal's triangle. Generalizing, the total number of old lines crossed by the new P, lines is equal to the number which appears on the fourth diagonal n - 1 rows down on Pascal's triangle.


Now each of the new lines created by the point P, will create one more region than the number of old lines it crosses; because, as we observed earlier, "if a new line crosses t old lines it will create t+ 1 new regions". So if "the total number of old lines crossed by the new P, lines is equal to the number which appears on the fourth diagonal n- 1 rows down on Pascal's triangle", then the total number of new regions created when we add the point P, is n -1 more than this. For example, the point P, will create 20 (the number of old lines crossed by new P, lines) plus 6 (the number of new P, lines) new regions. These numbers appear on the second and fourth diagonals of Pascal's triangle:

1

1 2 1

1 3 3 1

1 4 6 4

1 5 10 10 5 1

1 15 15 6 1

So all together the points P1, P2, P3 ..... P, will create a number of regions which is equal to the sum of the numbers up to n - 1 on the second diagonal, plus the sum of the numbers up to n-1 on the fourth diagonal, plus 1 (the region which was in the circle to start with). These two sums again occur on Pascal's triangle, 1 row down on the third and the fifth diagonal respectively:

+ 3

1 4 6 1

1 10 5 1

o a 15

i 6 1

So to find the total number of regions created by joining up n points on the edge of a circle, using Pascal's triangle, take:

The number n rows down on the third diagonal plus the number n rows down on the fifth diagonal plus 1.

This formula may be written more succinctly as:

Regions= "C4 + "C2 + 1,

but this sort of algebraic code makes the mathematical ideas being represented inaccessible to people who have not been taught how the code works!

The formula for the number of regions created by joining up n points on a circle, like the formulas for the tile patterns and for the sums of the powers described earlier, was pulled directly out of the diagram. We had no recourse to any - in this case spurious and misleading - number patterns. The formulas develop out of the models, and the models justify the formulas. As Miller (op. cit.) observes, this approach to problem-solving enables students "to get a taste of the excitement of creative activity in mathematics". It is easy, it is effective - and it's fun!

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