Mathematical Exercises in Paper Folding: Part VIAuthor(s): James BruntonSource: Mathematics in School, Vol. 3, No. 4 (Jul., 1974), pp. 26-27Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30211237 .

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M1HEMICAL EXERCISESIN Paper

Fold Ing Part VI

by James Brunton, Redland College, Bristol

23 An Isometric Grid If one is careful with the preliminary folds there is no need to use the specially shaped rectangle referred to in my article on rectangles, an equilateral triangle can be constructed in a piece of paper of any shape. The steps are shown in Fig. 1 where the first fold, Fig. la is the perpendicular bisector of the base. The second fold is best done by keeping the thumb-nail on the left-hand

comer and stretching the base edge of the paper with the right hand while folding. From Fig. 1c it can be seen that AABB' if completed would be equilateral, and that LB'AB (60') is bisected by the fold. A similar fold with corner A will start the process of constructing the isometric grid, which can easily be continued with parallel folds, by bisecting the angles at the sides of the paper and by folding the sides to the middle (Fig. 2). Smaller triangles still can be formed by folding the width of the paper into eight equal strips.

An interesting consequence of the properties of the equilateral triangle is that the paper can be easily folded into six strips. In Fig. 1c, if/LACB is bisected by folding, the base edge of the paper is trisected (Fig. 3); if A is now folded to D and B to the consequent crease, the paper is in three equal strips (Fig. 4) and folding B to D and so on, will give six equal strips. Since folding into two, four, eight and so on parts is elementary, (these are binary numbers) and folding in half is clearly a binary operation, this ability to fold into three as well will enable one to fold into a number of equal strips which is of the form 2m3n where m and n are any whole numbers including zero.

24 The Golden Section There are several ways of adapting compass and straight- edge constructions which will depend on what one has

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in mind to do. For example, to construct a regular pentagon on a given base or to inscribe a regular pentagon in a given circle, the necessary proportions can be produced by the following paper-fold. The

paper should be rectangular rather than square, (A4 will do very well) and the width AB should be twice the required base or radius. The procedure is as follows: Fold the perpendicular bisector of AB, through C and bisect LB to meet the side of the paper at D. Fold A onto

D, so that two squares are formed in the end of

the paper, each with a side equal to required base (or radius) for the construction. Next, fold the bottom

edge of the paper, carefully along the diagonal AF of the rectangle AEFB so that the new crease exactly meets the corner A (Fig. 8).

Now it will be found that on a base equal to BC, diagonals of length CH will form a regular pentagon, Fig. 9; with a radius equal to BC (or FG) chords of

length FH will form the regular inscribed pentagon (see Fig. 10). For, reference to Fig. 8 will show that the bisector AJ of the LFAE divides FE in the ratio AF:AE i.e. -/5:1 and this establishes that the slope of AH is that of the diagonal of a Golden Rectangle. The rest follows.

An interesting adaptation of H. W. Richmond's con- struction for an inscribed regular pentagon, as quoted in Coxeter's Introduction to Geometry (p. 27) was suggested to me some years ago by A. R. Pargeter. Fold two perpendicular diameters, and fold the perpen- dicular bisector of one of the radii (Fig. 11). Carefully fold the bisector of LBEC as in Fig. 12, to find the point F in AB. A chord through F, at right angles to AB will meet the circle at P, and P2, where BPI and BP2

are sides of the desired pentagon (Fig. 13). The other two points on the circle are found by folding the per- pendicular bisectors of BPI and BP2. These five points folded onto the centre will produce a folded pentagon (Fig. 14).

25 2n-gon, starting from n-gon It is easy to see that when the corners of an equilateral triangle are folded onto the centroid, a regular hexagon is formed (Fig. 15). To perform a similar modification to a square necessitates a few extra folds, as follows: (Fig. 16) The diagonals of the square are folded and one of the 45' angles is bisected. Next, the corner is folded to the intersection of diagonal and bisector as indicated in Fig. 16b. This is the exact amount that needs to be folded off at each corner to convert a square to a regular octagon. The work can be somewhat shortened for the whole figure, if the procedure out- lined in Fig. 17 is followed. Before the bisector is unfolded, the adjacent corner is folded down to meet it

(Fig. 17a) then the step as at Fig. 16b and finally the same two steps with the remaining corners will com-

plete the octagon (Fig. 18). These two examples of modifying n-gons and

2n-gons, actually follow a general pattern which can be summarised as follows:

1. Bisect an angle of the polygon, at one vertex (A) 2. Join the two adjacent vertices (B and C) 3. Bisect the angle ABC 4. Fold A to the intersection of these two angle

bisectors (D) (Fig. 19). PQ is a side of the desired

2n-gon. It is left to the reader to establish the result!

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27

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