Mathematics and the Christmas Tree (A Classroom Experience)Author(s): Judita CofmanSource: Mathematics in School, Vol. 9, No. 4 (Sep., 1980), pp. 22-23Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30213582 .

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four fours? Seventeen was sometimes used in place of 15. This was presumably because 16 + 1 gave the appropriate integers. Are there any solutions to the 1 5 - problem which do not produce a solution to the 17 - problem with a change in sign?

After completion we discovered that the integers can be generated using logarithms anyway - and using only three fours at that! It can be extended to accommodate four fours.

However, we had restricted ourselves to arithmetical operations only.

4 1. 4-4+- 4

4 4 2. -+-

3. 4

+/4-4 4

4. 4+4-+4- 4

5. /4+4+ 4

6. 4+4-4+V4

7. 4+V4+ 4

4

8. 4+4+4-4

9. 4+4+4

10. 4x4-4-4 10. 4x4-4-V44

4! 4 11.

V4 4

12. 4x4-/4-V4

4! 4 13.4 +-

V4 4

14. 4+4+4-V4

4 15. 4x4-- 4

16. 4+4+4+4

17. 4x4+4 4

17. 4 4 + - 4 4! 18.- + 4+ 44

4 19. 4!-4-

- 4 4 4

20. 4+4 .4 .4

4 21. 4!-4+

- 4

22. 4!-4+4-V4

23. 4!-/4+ 4 4

24. 4x4+4+4

25. 4!+V4- 4

26. 4!+4-4+V4

4 27. 4!+/4+

28. 4!+4+4-4

4 29. 4!+4+- 4

30. 4! + -/4+ V4+ V4

.4 4 31. -4 74 -4

32. 4x4+4x4

1/4 x x/4 33. 4!+

.4'

34. 4!+4+4+V4

44 35. 4!+- 4

36. 4!+4+4+4

4 37. 4!+4+4

.4'

4 38. 4 x 4

_-V4 .4

39. V4x4!- .4'

40. 44-V4-V4

.4 4 41. +. 74. 4'

42. 4x 4+ 4 4

43. 44-- 4

44. 44+4-4

4 45. 44+x

46. 4! + 4! + V4- 4 4

47. 4!+4!-- 4

48. 4!+4!+4-4

49. 4!+4!+-

50. 4!+4!+4-V4

The list of numbers from 5 1 to 100 will be published in the next edition. Readers might like to design their own.

Mathematics and the

Christmas Tree

(a classroom

experience)

by Judita Cofman, Putney High School, London

From time to time I try to organise a session of free activities in class. The pupils (aged 11 to 16) are given a "model" and asked to write down their observations and comments, possibly accompanied by suggestions for future projects.

Our last model consisted of a set of congruent equilateral triangles arranged into an equilateral triangular shape as shown in Figure 1. The number of rows in the shape was unlimited. The triangles were numbered and the translated images of AA,OB, were shaded.

My aim was to find out: a) would the pupils be more attracted by geometric shapes or by properties of numbers included in Figure 1, and b) would there be any attempts to

22

Fig. 1

0

A, B1 39

2 A2. 3 8 B2 B~6

.5 7 9

10 1

12 113

14 15T 3A x,16x

18 20 22 24 17 19 21 23 '25

27 29 31 33 35 26 28 y30 32\334,1,36* 38 40 42 44 46 48

3\ /39 /41 \45 \//4\7 49

51 53 55 57 59 61 63

/ \.\

/\ /\/

52 4 6 58 r.0 62 64

\ I .

I \

I \

I i-,

i \\II\

I \ I\\ I/

\

\\1 An - -

, , \/X \ , /

\// \ ,/

\ - ,k

st - -)t - - - ,- -- -

apply knowledge on geometry to detect facts about numbers, or vice versa.

Here is a brief outline of the experiment. Pupils of all age groups could discover and describe

properties of number sequences like:

1,2,3,4,5, .

1,3,5,7,9, . . 1,4,9,16, . 1,2,5,10,17,

.. (the number of shaded triangles in each row);

. . (the number of triangles in each row); . (the last number in each row);

.. (the first number in each row); etc.

In U3 pupils were mostly concentrating on divisibility conditions by looking at numbers in triangles situated on lines parallel to the axis of symmetry OOk. One participant has noticed that AAoOB,,o contains a small triangle T in central position, and built up a sequence of concentric layers of triangles around T leading to new number sequences (Fig. 2).

Geometric transformations mapping some of the polygons

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~OT~i~L~br~i

15 3 4

5

14 13 2 6 6/7

12 10 9

8 11

Fig. 2

one onto another were studied. It was observed by one pupil from L, that for any k= 2,3, .. the number preceding k2 can be written in the form (k-1 )(k + 1 ).

In Ur the main contribution asserted that:

"The last number in the k-th row of AAkOBk must be k2, since k2: 1 is the ratio of the areas of the similar triangles AAkOBk and AA,OB,, their linear dimensions being in ratio k: 1."

This was a useful comment leading to investigations on similarity and perspectivity of shapes.

In Ls (first division) the following "mini-research" was undertaken:

1) It was noticed that for any pair of positive integers k and n, with k < n, the number z of the small triangles in trapezium AkAnBnBk in Figure 1 can be expressed as a difference of two square numbers, namely

z= n2- k2.

2) The special case in 1), when z is a square itself, stimulated search for Pythagorean triads. Two families were found by looking at trapeziums of smallest heights 1 and 2:

a) For n=k+ 1 is z= 2k+ 1. Thus for all odd square numbers z= x2> 1 there is a Pythagorean triad

x2- 1 X2+ 1 X, , ;

2 2

b) For n=k+ 2 is z= 4k+ 4. This implies that for all even square numbers z= x2 >4 there is a Pythagorean triad

x2-4 x2+4 X, - S

4 1 4

The results in a) and b) were brought into connection with the previously learnt formula (a+ b)2= a2+ 2ab+ b2. 3) There was an unexpected turn at work when one of the girls tried to visualise the variety of the Pythagorean triangles corresponding to the above triads and their multiples. This resulted in enlargements of parabolas, as follows:

The girl's idea was to draw Pythagorean triangles in a plane, all with a common vertex O and with one of their sides, adjacent to the right angle, on a common half-ray Ox. Where would the third vertices of these triangles lie? To answer the question, set up a cartesian co-ordinate system in the plane with origin O and x-axis containing Ox. Look first at triads from a). Denote by y the number x(x2- 1) in the triad. Plot some of the points (x,y) with co-ordinates from triad a): M,(3, 1(32-1)), N,(5, J(52- 1)), K,(7, (72- 1)), . . . . Denote the orthogonal projections of these points onto the x-axis by M,, N,, K,, ... respectively. The the triangles AOM,M,, AONN,, AOKK, . . . are Pythagorean; their third vertices M,, N,, K, ... lie on the parabola

P1 with equation y= (x2 - 1 ) (Fig. 3).

The next step is to enlarge every triangle obtained by scale factor 2 from centre O. The images M2, N2, K,,

.... of the

points M,, N,, K, ..., respectively are on the lines OM,, ON,, OK,, . . . , and have co-ordinates of the form

2x, 2 x2- 1 (2x)2-4 2

,2 2 - x 4 -- "

Fig. 3 I //

I l/ li N2

N,

M2

i

-4// I

/ I I SI

N I /

/1/

/I/

It did not take much time to deduce that M2, N2, K2 ....

are also on a parabola, say P2, with equation y= x(x2- 4). In general, if the scale factor of the enlargement is a positive integer t, then the third vertices of the Pythagorean triangles

from triad a) are on a parabola Pt with equation y= 1-x(x2-

t). 2t

Similarly, the third vertices of the triangles from triad b) are on 1

parabolas with equations y= -(x2- 4t2) for t= 1,2,3, .... 4t

4) The final problem was: Is it possible to construct a model, analogous to Figure 1, in the three-dimensional space? The suggestion was, to construct a regular tetrahedron from a set of congruent "small" regular tetrahedrons. This would mean, that any regular tetrahedron could be decomposed, first of all, into 23= 8 regular tetrahedrons. However this construction fails, and the anology with Figure 1 breaks down: Although one gets four regular tetrahedrons by connecting the midpoints of the edges in a regular tetrahedron T (each of them having an edge in common with T), the remaining part of T is an octahedron (Fig. 4).

Fig. 4

The outcome of our experiment I considered satisfactory. No doubt, I have learnt a great deal about the mentality of my pupils, their approach and attitude to work. Hopefully they have profited as well. In any case, some of them seemed to enjoy the activities. A little girl from U3 has remarked in her report: "A nice sort of Mathematics: it reminds me of a Christmas tree" (Fig. 5).A

Fig. 5 S334

6 ... 8

... 20 22

2

24 ...... 29 0

31

34 36 38 40 42

5 7 49 5

51 4556 58 20

6 62 64

5

66 68 70 7

-5i 8

23

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