# The pentagon.

I recently came upon an article containing a number of pentagon problems for senior high school students (Geretschlager, 2014). I decided to share the background and some of the easier problems with you and your students. Consider a regular pentagon which is convex with five equal sides. Ask your students to construct one by drawing a circle with a compass and marking five points on its circumference that are equally spaced apart. This means, of course, that they will have to use a protractor to ensure that the angle subtended at the centre of the circle is 360/5 = 72 degrees. The points are then joined as in Figure 1 to form the regular pentagon.Note that this method of construction is very useful for many regular polygons (n-gons) since 360 has many factors. See if your students can list the useful ones for regular polygons.

Now ask your students to determine the size of the internal angles of a regular pentagon between any two adjacent sides.

Most of the properties of the regular pentagon have been known for centuries. Five-sided figures were symbolic in many cultures as far back as 5000 years ago in Mesopotamia. The best known pentagon in modern Western culture is the Defence building in the U.S.A.

When the diagonals are constructed they form a figure known as the star-pentagon or pentagram (Figure 2).

The Pythagoreans used the pentagram as a symbol of health, but other cultures have used it as a symbol of good, evil or magic. Many artists have used pentagonal forms in their work including Albrecht Durer, Leonardo da Vinci, Salvador Dali and Maurits Escher.

Ask your students to construct a regular pentagon including all its diagonals. See if they can determine all the angles included in the ten triangles and the small pentagon inside.

Next remove the two diagonals from the apex leaving Figure 3 to be investigated.

The triangles AEB and FEB are clearly congruent, because they have two base angles and the included side BE equal in each. Therefore if the length of the side of the pentagon is denoted by s, we have:

AB = AE = BF = EF = s = BC = CD = DE

Denoting the length of a diagonal of the pentagon by d, we have:

BE = BD = CE = d

The triangles EBF and DCF are similar, because they have common angles and their sides are in proportion. Triangle DCF is a scaled-down version of triangle EBF. Hence,

DF/BF = CD/BE = s/d

Since BF = s then DF = [s.sup.2]/d and since BD = BF + DF we have:

d = s + [s.sup.2]/d

Multiplying through by d/[s.sup.2] yields:

[[empty set].sup.2] = [empty set] + 1

... where [empty set] = d/s is the ratio of any diagonal to any side of the regular pentagon. The solutions of this quadratic equation are (1+[square root of 5])/2 and (1-[square root of 5])/2, but we ignore the negative answer as irrelevant to our diagram.

Thus,

d/s = [empty set] = (1+ [square root of 5])/2 = 1.618

This is the Golden Ratio. The Greek letter [empty set] is used to represent the Golden Ratio in memory of Phidias, an ancient Greek sculptor, who used this ratio in idealising human proportions. The Pythagoreans saw the connection between [empty set] and the regular pentagon. They realised that [empty set] was an irrational number and embodied it in the pentagram, which they used as their secret sign. The Golden Ratio is supposed to give the most pleasing rectangular proportions, and has been incorporated in many architectural buildings including the Parthenon in Athens.

There are eighteen problems considered in the reference given at the end of this Discovery article. They vary in difficulty, but there are two that junior secondary students should be able to attempt and solve.

Problem 1

An equilateral triangle is constructed on one side of a regular pentagon with its apex inside the pentagon. Join the apex to the other three vertices of the pentagon. Fill in all the angles.

Problem 2

The base side of a regular pentagon is extended equally on each side of its base vertices. An equilateral triangle is constructed on this extended line so that its apex is just above the top vertex of the pentagon and its sides intersect the other pairs of sides of the pentagon. Find all the angles inside the figure.

Happy discoveries!

Reference

Geretschlager, Robert (2014). Journal of the World Federation of National Mathematics Competitions, 27(2),7-33, AMT Publishing, Canberra.

Printer friendly Cite/link Email Feedback | |

Title Annotation: | discovery |
---|---|

Author: | de Mestre, Neville |

Publication: | Australian Mathematics Teacher |

Date: | Sep 22, 2015 |

Words: | 758 |

Previous Article: | Expectation and variation with a virtual die. |

Next Article: | Some ways to get a piece of Pi Day action. |

Topics: |