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MATHSEARCH 2000
Senior Project Prepared by Geoff Ball and Humphrey Gastineau-Hills, University of Sydney.

Pedan Polygons and Polyhedra - page 2

Section A. Preliminary Exercises. 20 marks.

Definition: An isosceles trapezium is one whose parallel sides are unequal and whose non parallel sides are equal.

Exercise 1 (6 marks):
For an isosceles trapezium ABCD with AB DC and AB<CD, prove that:
(i) ADC = BCD. Hint: Draw AX parallel to BC to meet DC in X.
(ii) ABCD is a cyclic quadrilateral;
(iii) the diagonals of ABCD are equal.

Exercise 2 (6 marks):
Suppose an isosceles trapezium has:
(a) parallel sides of lengths a and b with a < b;
(b) non-parallel sides of lengths c and
(c) the diagonals of length d
Use Ptolemy's theorem to prove that d2 - c2 = ab. We will call this Pedan's formula.
Notice that there are restrictions on the terms in Pedan's formula.
Importantly, the lengths a, b, c and d are constrained by the triangle inequality (that the sum of the lengths of any two sides of a triangle is greater than the length of the third side. )

Exercise 3 (8 marks):
(i) Find the Pedan isosceles trapezium ABCD with BC=AD= 2 and for which p+ < 35.

(ii) Find the Pedan isosceles trapezium ABCD with BC=AD=4 and for which p+ < 35.

Section B. Investigative Exercises. 40 Marks.

B1. Pedan Polygons.

Generally, finding Pedan pentagons, hexagons and higher order polygons requires considerable insight and patience though one can get the flavour of such investigations using congruent triangles to form isosceles trapezia, for instance, which in turn can be organised into Pedan hexagons.

The following hexagon, known to famous mathematicians such as Leonard Euler, is a Pedan hexagon.

Suppose we begin with the triangle P1P2P3 in which:

(a) P1P2 P3 = 120°, and
(b) P1P2 = 3 and P2P3 = 5, respectively

Now, on the same side of P2P3 as P1 is, construct P2P3P4 P3P2P1.

Note the order of the corresponding vertices which ensures this triangle is unique.

Complete the convex hexagon P1P2P3P4P5P6 using the same congruent triangles with angles of 120° at P4 and P5 respectively.

We hypothesise that the hexagon so formed is a Pedan hexagon!

Exercise 4. (2 marks) Use the cosine rule to show that the congruent triangles are Pedan.

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