# Pinboard geometry.

Using pinboards, your students can engage in some interesting geometrical discoveries as proposed by Bolt (1982). When pinboards are not available for everyone, paper with dots in a square array can be used. First of all, ask your students to construct a table containing the number of squares and the corresponding number of pins in each N x N square array (N = 1, 2, 3...). They should note that an N x N square array has [(N + 1).sup.2] pins.

Next consider, in particular, N = 2 which has nine pins and four one-unit squares as shown in Figure 1.

Ask your students to discover the total number of triangles that can be formed using any three of the pins as vertices. This exercise requires them to use a systematic counting method to ensure that no triangle has been missed. In fact, it is best to use two different systematic methods as a check on their answer.

One method is to label the dots alphabetically from A to I, and then count the number of triangles in each group that contains a particular letter as a vertex. Each triangle counted must have its three letters in alphabetical order to avoid repeats. For example, the triangles in the group for the pin D are DEG, DEH, DEI, DFG, DFH, DFI, DGH, DGI, DHI, yielding a subtotal 9. Of course, three pins in a straight line (collinear) do not form a triangle. Here are the sub-totals for each letter respectively using this method.
```
A    B    C    D   E   F   Total

25   20   13   9   6   3   76
```

Another method is to group the triangles according to their side lengths and then count the number in each group. This method helps students to identify types of triangles and to calculate distances within the array.
```
Type                                                     Number

(l, 1, [square root of (2)]) right-angled                  16
(l, 2, [square root of (5)]) right-angled                  16
(l, [square root of (2)], [square root of (5)]) obtuse     16
(l, [square root of (5)], [square root of (8)]) obtuse     8
([square root of (2)], [square root of (5)],               4
[square root of (5)]) isosceles
(2, 2, [square root of (8)]) right-angled                  4
(2, [square root of (2)], [square root of (2)])            S
isosceles
(2, [square root of (5)], [square root of (5)])            4
isosceles
Total                                                      76
```

Note that for N = 1 the total number of triangles is only 4, hence the number for N = 3 will be very large. Some of your brighter students might like to try for this, while the others are still working on N = 2 by the two methods suggested.

Using the same 3 x 3 array of pins, ask your students to find the number of different shapes that can be formed whose area is one half of a square unit. A shape is different if it cannot exactly cover another shape after it is rotated or flipped over. From the previous exercise it is seen that there are only two shapes; a small right-angled triangle within one square with side lengths (1, 1, [square root of (2)]) and an obtuse-angled triangle with side lengths (l, [square root of (2)], [square root of (5)]). Using these two as basic shapes, plus the right-angled triangle (l, [square root of (2)], [square root of (5)]), ask your students to make up a table showing the number of different shapes that can be formed with areas 1, 1[1/2], 2, [1/2] 2, 3, 3 [1/2], 4 respectively. Remember that all vertices must be at pins. There should be quite a few n-sided polygons discovered during this process.

The above activity is much more fundamental and significant than the usual exercises of finding areas of rectangles using length times breadth, as it is related to tiling patterns or tessellations. The exercise can clearly be extended to a 4 x 4 array of pins.

Consider next a 5 x 5 array of pins (i.e., N = 4). Ask your students to make as many different shapes as they can which have a perimeter of 10 units. Two are shown in Figure 2.

They should eventually find all six different shapes and to note the areas contained in each one.

Use the same-sized pinboard (5 x 5) to discover all the polygonal shapes that have one pin only inside the boundary lines of the polygon. Make up a table showing the area (A) of each polygon and the number of pins (b) on its boundary. Deduce a formula connecting these two.

Next find all the shapes with 12 boundary pins exactly and 0, 1, 2, 3, 4, 5, 6, 7, 8 ... pins inside. Is there a limit to the number of interior pins when the boundary has 12 pins? What determines this limit?

Ask your students to discover a formula relating the area (A) of the polygon shapes that have 12 pins on the boundary to the number (i) of interior points.

There is a relation, known as Pict's theorem, which gives the area (A) of any polygon created on a pinboard in terms of the number of boundary points (b) and the number of interior pins (i). Using the results that have been obtained so far, and any other experiments with square array pinboards of different sizes, ask your students to try to discover Pict's formula, which is

A = i + 1/2 b - 1

Happy discoveries!

Reference

Bolt, B. (1982). Mathematical activities. Cambridge: Cambridge University Press.
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