Golden Ratio in Geometry

The Golden Ratio (Golden Mean, Golden Section) is defined as φ = (√5 + 1) / 2. The classical shape based on φ is the golden rectangle where φ appears alongside the perfect (unit) square:

The golden rectangle has dimensions 1×φ such that removing the unit square one is left with the rectangle (φ - 1)×1 similar to the original rectangle. Indeed, the most fundamental property of the golden ratio is

All these equations (and an extra one) can be summarized in the following diagram [Olsen, p. 54]:

The golden rectangle is easily constructed from a square as shown in the diagram below:

Changing the order of operations, the problem of inscribing a square into a semicircle has a golden rectangle as a biproduct. (The problem is easily solved using homothety, the same way as inscribing a square into a triangle.)

The Golden Ratio and its inverse are roots of quadratic equations. Perhaps surprisingly, the roots of all other quadratic equations also relate to the Golden ratio (N. Lord, Golden Bounds for the Roots of Quadratic Equations, The Math Gazette, v 91, n 522, Nov. 2007, p. 549.) Indeed, let r be a root of the quadratic equation ax² + bx + c = 0. Then the quadratic formula gives

from which

The equality is of course achieved for the Golden Ratio.

The golden ratio pops up in several geometric configurations, sometimes quite unexpectedly. It makes a stellar appearance in the pentagonal star:

(For a more descriptive treatment check S. Brodie's construction of the pentagon and the derivation of cos(36°) = (1 + √5)/4.) In passing, the isosceles 72^o-36^o-72^o and 36^o-108^o-36^o triangles are known as the golden triangles because, as you see in the diagram, the bisector of a base angle in the acute one cuts off a smaller 72^o-36^o-72^o triangle leaving a 36^o-108^o-36^o one. A trisector of the apex angle of the latter divides it again into two golden triangles. In both triangles, the ratio of a big side to a small one is of course φ, what else?

As we shall see below, the golden ratio crops up in various circumstances, often quite unexpectedly. The greatest surprise for me was to learn that a pentagon need not necessarily be regular to house all the occurrences of the golden ratio as if it were.

For the details check another page.

The golden ratio is related to the ubiquitous 3-4-5 triangle [Huntley, pp. 43-44].

Let ABC be such a triangle with BC = 3, AC = 4 and AB = 5. Let O be the foot of the angle bisector at B. Draw a circle with center O and radius CO. Extend BO to meet the circle at Q and let P be the other point of intersection of BO with the circle. Then PQ / BP = φ.

Indeed, BO being an angle bisector, O divides AC in the ratio of the sides AB : BC:

From here, AO = 5/2 and CO = 3/2. Thus the circle's radius r is 3/2. By the Power of a Point Theorem,

In other words,

From which, BO = 3√5/2. We thus find BP = 3(√5 - 1)/2. And finally,

(Incidently, the circle is tangent to the hypotenuse AB.)

Another way of linking the 3-4-5 triangle to the golden ratio has been discovered by Gabries Bosia while pondering over the knight's move in chess. The latter is naturally associated with a 1:2:√5 triangle. Both appear in the following diagram:

An intriguing showing of φ in an equilateral triangle was observed by George Odom, a resident of the Hudson River Psychiatric Center, in the early 1980s [Roberts, p. 10]. Upon hearing it from Odom, the late H. M. S. Coxeter submitted it as a problem to the American Mathematical Monthly (Problem E3007, 1983). The problem has also been reproduced by J. F. Rigby in [Pritchard, p. 294] and mentioned by K. Hofstetter in a recent article.

Let L and M be the midpoints of the sides AB and AC of an equilateral triangle ABC. Let X, Y be the intersections of LM extended with the circumcircle of ΔABC. Then LM / MY = φ.

Indeed, if 2a is the side length of ΔABC, then AM = MC = LM = a and XL = MY = b. By the Intersecting Chords Theorem,

In other words,

Denoting a/b = x, we see that

which is (2), so that indeed x = φ. A derivation based on the presence of similar triangles was posted by Jan van de Craats as a solution to Coxeter's problem and included by R. Nelsen in his collection Proofs Without Words II.

In the above mentioned article, K. Hofstetter, offered another elegant way of obtaining the Golden Ratio.

It will be convenient to denote S(R) the circle with center S through point R. For the construction, let A and B be two points. Circles A(B) and B(A) intersect in C and D and cross the line AB in points E and F. Circles E(B) and F(A) intersect in X and Y, as in the diagram. Because of the symmetry, points X, D, C, Y are collinear. The fact is CX / CD = φ.

Assume for simplicity that AB = 2. Then CD = 2√3, CX = √15 + √3, so that

Also, CD / DX = φ. Finally, observe that points E and F lie on C(D). It follows that the whole construction can be accomplished with compass only.

In a subsequent paper, Hofstetter gave a 5-step algorithm for dividing a segment in golden ratio:

Draw A(B) and B(A) and let C and D be their points of intersection. Draw C(A) and let it intersect A(B) in E and CD in F. Draw E(F). This intersects the line AB in points G and G' such that AB:AG = φ and AG':AB = φ.

For a proof, suppose AB has unit length. Then CD = √3 and EG = EF = √2. Let H be the orthogonal projection of E on the line AB. Since HA = 1/2, and HG² = EG² - EH² = 2 - 3/4 = 5/4, we have AG = HG - HA = (√5 - 1)/2. This shows that G divides AB in the golden section.

In the third paper in the series, Hofstetter mentions having been alerted to the fact that the above proof had been discovered previously by E. Lemoine (1902) and L. Reusch (1904). The essence of the paper is an additional 5-step division of a segment into the golden section.

The construction is this. For a given segment AB of length 1, form circles A(B) and B(A). Let C and D be the intersections of A(B) and B(A). Extend AB beyond A to the intersection E with A(B). Draw E(B) and let F be the intersection of E(B) and B(A) farther from D. DF intersects AB in G. AG:BG = φ.

To see why this is so, let I be the intersection of CD and AB and H the foot of perpendicular from F to AB.

It follows that IG = 2·IH/(√5 + 2) = (√5 - 2)/2, and AH = 1/2 + IG = (√5 - 1)/2. This shows that G divides AB in the golden section.

In a 2005 paper, Hofstetter offers a similar construction with a rusty compass whose opening can be set only once.

Draw A(B) and B(A) and find C and D at their intersection. Let M be the midpoint of AB found at the intersection of AB and CD. Construct C(M, AB), a circle with center M and radius AB. Let it intersect B(A) in F and another point, F being the farthest from D. Define G as the intersection of AB and DG. G then is the sought point.

The proof is straightforward.

First, BF = FM; so that the projection of F to AB (denoted K) is the midpoint of BM: KM = BK. Right triangles GMD and GKF are similar which gives a proportion:

It is convenient to assume AB = 4. Then FM = BF = 4, AM = BM = 2, KM = BK = 1. Solving right triangles we obtain KF = √15 and DM = √12 such that GM/GK = 2/√5. Since GM + GK = KM = 1, we derive

Further, AG = AM + GM = 2(√5 - 1) and BG = GK + BK = 6 - 2√5. Now by direct verification,

The golden ratio has been sighted in a trapezoid [Tong]. In the diagram, the bottom base PQ has length b and the top base RS = a < b. The line MN parallel to the bases is of length √(a² + b²)/2. This quantity that is known as the quadratic mean (or the root-mean-square) fits between a and b as any other mean would.

From the similarity of triangles MSV and PSW,

if b = 3a. The construction of the golden ratio based on this is depicted below

In the construction, BC = 3·AD, CE = AD, CE BC, BF = EF = FH, FH BE, BI = BH, IJ||AB, and GJ||BC. AG/BG = φ.

The golden ratio helps resolve Curry's paradox [Gazalé, p. 133]:

In order for the pieces to fit the rectangle tightly, from the similarity of two upper triangles we should have the proportion

which is equivalent to (2). If we change units from 1 to 5, we get the dissection below:

Assume the graph of a 4^th degree polynomial has inflection points with abscissas a and b, a < b. The straight line through the inflection points meets the graph in two other points with abscissas x_L and x_R say: x_L < a < b < x_R.

Then

One solution to the problem of bisecting the Yin-Yang symbol by straightedge and compass is shown in the diagram below:

The dashed line is formed by two semicircles of diameter φ and two semicircles of diameter φ^-1.

References

1. M. J. Gazalé, Gnomon: From Pharaohs to Fractals, Princeton University Press, 1999
2. K. Hofstetter, A Simple Construction of the Golden Section, Forum Geometricorum, v 2 (2002), pp. 65-66
3. K. Hofstetter, A 5-step Division of a Segment in the Golden Section, Forum Geometricorum, v 3 (2003), pp. 205-206
4. K. Hofstetter, Another 5-step Division of a Segment in the Golden Section, Forum Geometricorum, v 4 (2004), pp. 21-22
5. K. Hofstetter, Division of a Segment in the Golden Section with Ruler and Rusty Compass, Forum Geometricorum, v 5 (2005), pp. 135-136
6. H. E. Huntley, The Divine Proportion, Dover, 1970
7. R. B. Nelsen, Proofs Without Words II, MAA, 2000
8. S. Olsen, The Golden Section: Nature's Greatest Secret, Walker & Company, 2006
9. C. Pritchard (ed.), The Changing Shape of Geometry, Cambridge University Press, 2003
10. S. Roberts, King of Infinite Space, Walker & Company, 2006
11. J. Tong and S. Kung, A Simple Construction of the Golden Ratio, Forum Geometricorum, Volume 7 (2007) 31–32

Fibonacci Numbers

1. Ceva's Theorem: A Matter of Appreciation
2. When the Counting Gets Tough, the Tough Count on Mathematics
3. I. Sharygin's Problem of Criminal Ministers
4. Single Pile Games
5. Take-Away Games
6. Number 8 Is Interesting
7. Curry's Paradox
8. A Problem in Checker-Jumping
9. Fibonacci's Quickies

Golden Ratio

1. Golden Ratio in Geometry
2. Golden Ratio in an Irregular Pentagon
3. Golden Ratio in a Irregular Pentagon II
4. Inflection Points of Fourth Degree Polynomials
5. Wythoff's Nim
6. Inscribing a regular pentagon in a circle – and proving it
7. Cosine of 36 degrees
8. Continued Fractions