 |
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.
1 2 3
4 
|