Art Inspires GeometryAuthor(s): John SharpSource: Mathematics in School, Vol. 33, No. 1 (Jan., 2004), pp. 2-5Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30215645 .

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ART INSPIRES

GEOMETRY

by John Sharp

Mathematics and art have many links, some more obvious than others. At the beginning of the twentieth century, the influence of geometry especially was very widespread. It is not easy to describe the work of the Cubists in strict geometrical terms, but the work of the Dutch based De Stijl group of artists is much more approachable for its geometrical simplicity. The most famous De Stijl artist is Mondrian. Although he used geometry, he was also intuitive. The artist Theo van Doesburg (1883-1931) was perhaps the most influential of the group. I find his work manages to be simple yet surprisingly rich in both art and geometry and the geometry is much more obvious than in the work of Mondrian. The geometry also offers inspiration for investigation in many ways in the mathematical classroom as well as the ability to open the eyes of students to the fact that mathematics has a beauty of its own. The following set of problems arose from a pair of articles by Marion Walter and David Pimm in the journal For the Learning of Mathematics (Walter, 2001; Pimm, 2001).

The starting point is a painting which van Doesburg called Arithmetic Composition 1, and a simulation is shown in Figure 1. This was van Doesburg's last major work before his early death in 1931.

Fig. 1

I am going to pose a number of problems based on this painting and then give (most of) the solutions. There are many other possibilities, which is why it is such a rich subject. These problems are also aimed at making you think about posing others and in a mathematics lesson could be a method of stretching imagination by asking students for their own ideas on "What if?."; apart from the geometry and algebra, you can also ask about the aesthetics and delve into how much the visual aesthetics matches the mathematical beauty. They also offer a set of problems which are not based

on rote learning and solve practical problems of geometry that most people would not associate with artists. Because the problems raised are open-ended, it is an ideal topic for a mixed ability class. The Walter and Pimm articles offer more ideas with different approaches.

Problem 1 (unanswered)

Why did he give it the title Arithmetic Composition 1 when it is obviously based on a geometric progression? Neither Marion Walter nor David Pimm have an answer to this question. Any suggestions would be welcomed.

Problem 2

The composition is based on drawing a tilted square in the isosceles triangle formed by bisecting the square whose base is on the diagonal and whose other vertices are on the sides of the original square. Another horizontal square is created whose side is half the side of the original one and the process is repeated. Starting with a unit square, what is the side of the resultant square, and what proportion is this of the diagonal? What is the value of a in Figure 2?

A B

E

D F C Fig. 2

Problem 3

What is the relative area of two successive squares? Van Doesburg only paints four squares. Suppose the construction is carried on infinitely. What proportion of the original square is the area of the tilted squares? A similar problem is posed by the squares in Figure 3 if continued infinitely; in this case it is easy to see the solution instantly as a visual proof; and much easier to see if the construction lines are drawn in as shown at the top of the next column.

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Fig. 3

Problem 4

Suppose we consider a rectangle instead of a square, but with the next stage formed by halving each time. In Figure 4 and subsequent figures, let the sides of the rectangle be in the ratio 1:n and the vertex of the first tilted square be at a distance a from the bottom right vertex of the rectangle. Find a formula for a in terms of n. Calculate n for a few rectangles.

B

Fig. 4

Problem 5

Knowing the value of a in Figure 4, find a formula for the side EF of the first square. What are the ratios of DH, HG and GB as a formula? Consequently, what is the ratio GB:HG:DH?

Problem 6

Figure 4 shows the result for a rectangle whose sides are in the ratio 1:2, that is n = 2. If FH is extended then it misses the side of the next square by going to the right. What is the value of n for which the sides successive squares are on the same line as in Figure 5?

Fig. 5

Problem 7

Another approach to the position of the tilted square is to have the corner of the square H of Figure 4 be half-way along the diagonal, as shown in Figure 6. You can find the value of n using similar triangles, but the result of finding the ratio GB:HG:DH in Problem 5 gives a much easier solution. Figure 6 shows the result.

Fig. 6

Other Problems to Investigate

There are many other ways you can go in inventing your own problems. The ones above by no means exhaust the case for the square. The following are more suggestions. Some answers, but not solutions, are given below.

(a) Instead of creating a square on the diagonal, you could place a rectangle with sides in the same ratio as the

original. This leads to two possibilities: with the long or the short side on the diagonal. Figure 7 shows a rectangle with sides in the ratio 1:2 showing both versions of the rectangles. What is the area of overlap in terms of the general rectangle with sides in ratio 1:n?

Fig. 7

(b) Figure 8 shows the case where all the diagonals of the squares in a rectangle lie in a line. What is the ratio of the sides of the rectangle?

Fig. 8

(c) In the version using a square (like Figures 5 and 6), it is also possible to have a tilted polygon which is not a square, or even a circle. Figures 9-11 show some examples. Investigate the side of the polygon, or radius

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of the circle. Which polygons can be placed in more than one orientation? What is the result of extending Problem 3 to these constructions?

Fig. 9

Fig. 10

Fig. 11

(d) Invent your own problem, mixing and matching ideas from other problems, for example in finding the areas of the infinite number of circles defined by Figure 11.

Solutions

The problems have been posed so that they can be copied and given out for a class project. Teachers, being busy, need the solutions too, although there may be more than one approach to the result.

Solution 2

Marking the congruent segments as, in Figure 12, helps in identifying similar triangles. It also shows that the side of the square GH is a third of the diagonal DB. By using similar triangles BDC and EFC, it is then easy to see that EC is also a third of

BC. Since DB = /2, then the side of the

tilted square EFHG is /2/3 and its area is 2/9 of the area of square ABCD.

Solution 3

From solution 2, we know that the area EFHG is 2/9 of the area of square ABCD. The next square has a side which is a half of square EFHG and the next square has a side a half of that one. Each area is thus a quarter of the previous one. This gives an infinite geometric progression with first term

A

Fig. 12

(a) of 2/9 and common ratio (r) of 1/4. The sum of an infinite geometric progression is:

a. 2/9 8 23 SO which evaluated

is1-1/4 27 33

The visual proof of Figure 3 shows that the black area (if the division continues to infinity) is one third of the original square and is a proof of the expression:

S2( 2n

(2) 3

Solution 4

From Figure 4, since triangles BCD and ECF are similar, FC = na and FE = GE =

(an2+1)a. Triangles ECF and BGA are also similar and so

(/n2+1)a 1-a na (an2+1)a

which when solved for a gives n

a=n2 + n.+

For the first few integers values of n this gives:

n a

1 (a square) 1/3 2 2/7 3 3/13

Solution 5

From Figure 4, if s is the side of the square EFHG, then

S - (n2+ l )n

n2+n+l

Also triangles BCD, BGE and FHD are all similar with the long to the short side in the ratio n:1. So it is quite easy to show that BG = s/n, HG = s and HF - ns. This gives the ratio GB:HG:DH as 1/n : 1: n = 1: n : n2

Solution 6

This solution requires some extra lines (Figure 13). FH is extended to meet line AB and PF is drawn to be perpendicular to AB at P

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M

Fig. 13

The strategy in solving this problem is to define MB in two different ways and equate them.

NQ is a half of BE because AN is a half of AB and so the rectangle is halved each time. So NG - (1-a)/2. Triangles MNQ and BCD are similar, so

MN 1 1 - a MN -and hence MN - NQ n 2n

NB is a half of AB and so equal to n/2. Thus:

1-a n MB = MN + NB-=I- + -

2n 2

Also MB- MP + PB. Now triangles MPF and BCD are similar and PB - FC = na, so

1 MB = -+ na.

n

Equating these two expressions for MB and using the expression relating a and n in Problem 4, gives the following equation for n:

n4-n3-2n-1 =0.

This requires an iterative method to solve for n and a can then be found using the expression in Problem 4. This gives n - 1.7943096... and a = 0.298362...

Solution 7

From Solution 5 (using the labelling of Figure 4), the ratio GB:HG:DH = 1: n : n2. Now we want H to be the midpoint of BD, so that GB + GH - HD. Consequently,

1+n-n2.

This is a quadratic whose solution (n > 1) is the Golden Section (1+ a5)/2 or approximately 1.6180399...

Other problems - hints and some solutions

(a) In the case like Figure 7, I suggest you consider rectangles with ratios 1:n and 1: 1/n. Then you can use previously derived results. Think what this means in terms of rotations or reflections and scaling to get the two rectangles to overlap.

(b) The shape of the rectangle where the diagonals of the squares are all in one line as in Figure 8 gives a rectangle 1:n, where n is the root of the equation:

n4- n3- n2-2n - 1 -0.

The solution to this equation is 2.065994892..., but note also how similar the equation is to the one for the solution of Problem 6.

(c) For the circles case, the circle is the incircle of a right- angled isosceles triangle. So if the outer square is of unit length, the radius of the circle is

tan 22.5' n22 = 0.2928932... {2

References Pimm, David 2001 'Some Notes on Theo van Doesburg (1883-1931) and his

Arithmetic Composition 1', For the Learning of Mathematics, 21, July, pp.31-36.

Sinclair, Nathalie 2002 'Reconstructing a Painting with Geometric Eyes', For the Learning of Mathematics, 22, November, pp.19-22.

Walter, Marion 2001 'Looking at a Painting with a Mathematical Eye', For the Learning of Mathematics, 21, July, pp.26-30.

Keywords: Theo van Doesburg; Art and Mathematics; Puzzles from Art.

Author John Sharp, 20 The Glebe, Watford WD25 OLR. e-mail: sliceforms@compuserve.com

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