Pegs Problem F2 Stitching Rhombuses II

A square frame has sides which have been divided into 2, 3 and 4 parts. Four strings have been stretched across the frame to produce a central rhombus. What fraction of the squares area is taken up by the rhombus? Generalize.

Chris Pritchard, Mathematics in School

18 Mathematics in School, J a n u a ry 2 0 1 5 The M A website w w w .m - a .o r g .u k

Pegs Problem F3 Folding onto the Diagonal

ABCD is a square. In the first figure, B and D have been folded onto AC to meet at G. In the second, they have been folded to B' and D'. What fraction of the square is taken up by the kite AECF and by the parallelogram ALCM?

Chris Pritchard, Mathematics in School

Mathematics in School, J a n u a r y 2 0 1 5 The M A website w w w . m - a . o r g . u k 19

Solutions to Pegs Problems Set Fby Chris Pritchard

Problem FI

Cut each square in half horizontally:

The points of intersection of the strings are equally spaced on this horizontal line, so that the central triangles area each time is determined by the fraction of the line segment on which its base sits. The sequence of fractions is:1 1 1 J. 13 5 7 9 ........... 2k-1

And since the triangles occupy the same area in the semi-square as the rhombuses do in the full square, this sequence and general formula apply to the problem as stated.

Problem F2

There are various ways of tackling this problem. One possibility is a dissection approach. The first figure

can be dissected and rearranged as below.

The yellow rhombus occupies 1/3 of the square.

Similarly, for the second figure

we can rearrange thus:

to show that the rhombus occupies 1/2 the square.

Alternatively, proceed as in FI:

Again the points of intersection of the strings are equally spaced on this horizontal line, so that the yellow triangles area is simply the fraction of the line segment occupied by its base. This time the sequence is:

20 Mathematics in School, J a n u a r y 2 0 1 5 The M A website w w w . m - a . o r g . u k

1 2 3 4 n- 134 5 6 ........... n+T"

This sequence and general formula apply equally to the rhombuses in squares problem.

Problem F3

This is a single problem in two guises because whether the folds onto the diagonal AC are in the first manner or the second the area produced is the same. It does, however, highlight the fact that there is a connection between kites and parallelograms.

C

Let DF = FG = GC = 1. Then FC = 72, by Pythagoras Theorem and the side of the square is 1 + 72.

Area AECF = i x 2V2 (l + >/2) = V2 (l + V5).

Area ABCD = (l + 7 2 )\

Proportion of square taken up by the kite is

72 V 2(l-72)-----j= = t-----prr----= 2 -7 2 or about 59%.1 + V2 (l + 72) (1-72)

Problem F4

Let the side of the regular pentagon have unit length and let the diagonals have length x:

M

By Ptolemys Theorem:

JL.KM=KL.JM+JK.LMx2 = x +1

x2 - x - l = 0 x =

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