The Golden Ratio

Construct the golden ratio and a golden rectangle
Dividing a segment
(Continued fractions)
(Golden spirals )
Golden triangles
Regular pentagons

Construct the golden ratio and a golden rectangle

Construct the golden ratio

Construct a line L. Construct a line M perpendicular to L at a point A. Mark a segment AB on M of length 1, or call the length 1 if you don't care what units of length you use.
Mark a segment AC of length 1 along L. Mark a segment CD of length 1 along L, so that AD has length 2.
With compass point at B, mark off distance BD along line M in the direciton opposite A. Call the intersection point E.

The ratio of lengths AE/AD is called the golden ratio. This number is usually denoted by the greek letter tau, but I will use g, which is easier to type.

What is the length of BD? Hint: use the Pythagorean theorem.
What is the length of AE?
What is the golden ratio? Express it in two ways: with radicals, and as a decimal approximation.
Show that the golden ratio (in radical form) satisfies the polynomial equation x^2 - x - 1 = 0. (Note: the symbol ^ means "raised to the power"; so x^2 means x raised to the 2nd power, or x squared.)
Use algebra to rearrange the equation for g in the previous question to show that

g^2 = g + 1 (that is, g^2 can be replaced by g+1 whenever convenient)
1/g = g - 1 (that is, 1/g can be replaced by g-1 whenever convenient)

Find an expression for g^3 that involves only g and integers, and no higher powers of g.
Find an expression for g^4 that involves only g and integers, and no higher powers of g.
(Extension) Find an expression for g^n that involves only g and Fibonacci numbers, and no higher powers of g.

Construct a golden rectangle

With your compass point at E, construct a circle of radius AD. With your compass point at D, construct a circle of radius AE. The intersection point of the circles is the fourth vertex, F, of rectangle AEFD.
Make a square with side AD: With your compass point at A, mark point H at distance AD along AE. With your compass point at D, mark point G at distance AD along DF.

The rectangle ADFE, or any rectangle similar to it, is called a golden rectangle.

Still assuming that AD = 1, find EH.
Show that the ratio EF/EH is the golden ratio; that is, show that rectangle EFGH is also a golden rectangle. (You will have to remember how to rationalize the denominator.)

Dividing a segment

Suppose you have a segment AC, and a point B on the segment. This determines three lengths: the whole, AC, and two parts, AB and BC. You can form several ratios. Consider the ratio of the whole AC to the part AB, and the part AB to the other part BC. Can these ratios ever be equal? (online interactive diagram)

Find AC/AB and AB/BC when

B is the midpoint of AC
B is a third of the way along AC
B is a 3/4 of the way along AC

Here is a way to answer the question of whether the ratios can be equal, using algebra. Call the length AC = 1, and the length AB = x.

Find an expression in x for BC.
Write a proportion using these expressions that says AC/AB = AB/BC.
Solve for x. You will need to use the quadratic formula. One of the solutions will be negative; use the positive one.
Use this x to find the ratio AC/AB and simplify.

You should get the golden ratio.

Golden triangles

Definition (this is not standard terminology): A tall golden triangle is any isosceles triangle with two long sides and one short side, where the ratio of long to short is the golden ratio. A short golden triangle is any isosceles triangle with two short sides and one long side, where the ratio of long to short is the golden ratio.

The angles in the golden triangles

You will need to use:

The AA triangle similarity theorem.
Facts about isosceles triangles.
Angle sum in a triangle.
The golden ratio satisfies the equations g^2 - g - 1 = 0, g^2 = g+1, and 1/g = g-1.

Consider a tall golden triangle ABC, with base length AB = 1 and the other two sides, AC and BC, both of length g, the golden ratio. Call the measure of each base angle of this triangle m.

Construct a point D on side BC so that AD = 1. This divides the triangle ABC into two smaller triangles, ABD and ACD.
Prove that angle ABD = angle ADB = m.
Prove that triangle ABD is similar to triangle ABC.
Use proportions and the golden ratio to prove that BD = g-1.
Conclude that CD = 1.
Show that angle CAD = angle ACD.
Find an expression for angle ACD in terms of angle ADB, then an expression for angle ACD in terms of m.
Find and equation for the sum of the angles in triangle ABC in terms of m.
Solve for m.
Find the other angles in a tall golden triangle.

Now you know all the side lengths (up to a factor) and all the angles of any tall golden triangle.

Prove that ACD is a short golden triangle: that it has two equal short sides, and the ratio of long to short is the golden ratio.
Find the number of degrees in each of the angles of a short golden triangle.

Corollary: You have also proved that a tall golden triangle can be divided into a smaller tall golden triangle and a short golden triangle.

Prove that a short golden triangle can be divided into a smaller short golden triangle and a tall golden triangle. (And explain how to divide it.)

How to construct a regular pentagon.

Construct a tall golden triangle (1,g,g). Construct one short golden triangle (1,1,g) on each of the long sides of the tall triangle, with the sides of length g matching. The outline of this figure is a pentagon.

Prove that the pentagon is equilateral.
Prove that the pentagon is equiangular.

A more direct construction of a regular pentagon.

Given a side AB, construct a regular pentagon ABCDE.

For convenience, assume AB = 1.

Extend segment AB to ray AB. Find the midpoint M of AB. Construct a perpendicular at A.
Construct circle c1 with center A and radius 1. Call the intersection point with the perpendicular F.
With compass point at M, mark off distance MF at G on ray AB. Call the length of AG g. (What is g?)
Construct circles c2 and c3 with centers A and B and radius g. Call D the intersection of c2 and c3, and E the intersection of c1 and c3.
Construct circle c4 with center B and radius 1. Call C the intersection of c2 and c4.

ABCDE is a regular pentagon. (Why? This is a homework question.)

Individual homework

Go through the more direct construction of a regular pentagon and prove that it really does make a regular pentagon. You may use the results of the group work on the first construction of a regular pentagon.
Construction of circumscribed circle around a regular pentagon, and regular decagon. For extension points, prove the parts in square brackets.

Start with a regular pentagon. (Use a whole sheet of paper. You may use a protractor to save time.) Mark the vertices ABCDE, in order.
Construct the perpendicular bisectors of any two sides. Call the intersection point X.
Construct the circle with center X and going through any vertex of the pentagon. [The circle goes through all the vertices of the pentagon.]
Find the other point Y where line AX intersects the circle. It should be between C and D.
Open your compass to distance CY and repeatedly mark this distance around the circle, until you get back to C. If you're not careful or are using a dull pencil, you might not get back to C. [All the marked points on the circle are the vertices of a regular 10-gon.]

Adapt the ideas from the construction of the 10-gon to give a straightedge and compass construction of a regular 12-gon. (What figure should you start with?) How could you construct a regular 24-gon? (You don't have to do it--just describe briefly.)
Again start with a regular 5-gon ABCDE, and construct the circumscribed circle as above.

Construct (with straightedge and compass) an equilateral triangle AFG inscribed in the same circle. Easiest way: Do the construction described in the first assignment of the course, and use the circle around the pentagon as your original circle. Keep drawing circles with centers on the original circle until you get back to the first one. This should give you the vertices of a regular hexagon. Use A and every other vertex of the hexagon to get an equilateral triangle. Mark the other vertices F and G so that the order of points around the circle is ABFCDGE.
Mark off the distance CF around the circle as before. You should hit all the points ABFCDGE, and several others.
Connect all the vertices you just marked off with line segments in order around the circle. What have you just constructed?
What does this have to do with the least common multiple of 3 and 5? Generalize the construction to least common multiples of any two integers greater than 2. Don't go into great detail, just a general statement and recipe.

(Extension) Invent a procedure using only straightedge and compass for dividing a given segment into two pieces whose lengths satisfy the golden ratio. You are not given a segment or length 1, and you may not use any ruler markings.