# Golden Ratio in Geometry

The φ = (√5 + 1) / 2. The classical shape based on φ is the

*Golden Ratio*(*Golden Mean*,*Golden Section*) is defined as*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

(1) | φ : 1 = 1 : (φ - 1). |

(2) | φ^{2} - φ - 1 = 0, |

(3) | φ = 1 + φ'. |

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 asquare 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, ax² + bx + c = 0. Then the quadratic formula gives

__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
r = (-b ± √b² - 4ac) / 2a,

from which

|r| | ≤ (|b| + √|b|² + 4|a|·|c|) / 2|a| |

≤ (1 + √1 + 4) / 2 · max{|a|, |b|, |c|} / |a| | |

= φ · max{|a|, |b|, |c|} / |a|. |

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 ofcos(36°) = (1 + √5)/4.) In passing, the isosceles 72 and 36 triangles are known as the36 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.^{o}-36

^{o}-72

^{o}

^{o}-108

^{o}-36

^{o}

*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^{o}-108

^{o}-36

^{o}

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. For a proof, see

*Golden Ratio And the Egyptian Triangle*.
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:

José Antônio Fabiano Mendes from Rio de Janeiro, Brazil, has observed additional appearances of the1:2:√5 triangle. The inradius of the 3-4-5 triangle is 1 and the distance between the incenter and the circumcenter is √5/2:

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

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

MX·MY = AM·MC.

In other words,

(a + b)·b = a·a.

Denoting a/b = x, we see that

1 + x = x

^{2},
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*.
(Linda Fahlberg-Stojanovska made a camcast that follows Odom's construction by paper folding a circle. The resulting pentagon is very nearly regular, but not quite.)

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. (The proof has been placed on a separate page.)

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 andAG':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 HG we haveAG = HG - HA = (√5 - 1)/2. This shows that G divides AB in the golden section.

^{2}= EG

^{2}- EH

^{2}= 2 - 3/4 = 5/4,

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.

IG/GH = DI/FH = (√3)/2 / (√15)/4 = 2/√5.

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. (For details of the proof check a separate page

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.

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

(1 + φ):1 = (1 + 2φ):φ,

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

Assume the graph of a 4a < b. The straight line through the inflection points meets the graph in two other points with abscissas xx

^{th}degree polynomial has inflection points with abscissas a and b,_{L}and x_{R}say:_{L}< a < b < x

_{R}.

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}.

As can be easily verified, when three equal circles touch each other in sequence, and a larger semicircle, the ratio of the radius of the latter to the diameter of the small circles is φ.

The only time the sides of a right triangle are in geometric progression is when the triangle is similar to the one with sides 1, √φ, φ [Charming Proofs, p. 74]. These triangles are often referred to as

*Kepler's triangles*.
The area of annulus defined by the radii, a and b < a is equal to the area of ellipse with major and minor axis equal to a and b, respectively, is when a/b = φ [Charming Proofs, p. 37].

The relation between the radii of three pairwise tangent circles that are also tangent to a line is well known

1/√R = 1/√R

_{1}+ 1/√R_{2}.
Giovanni Lucca has recently observed that the relation is actually the one that defines the Fibonacci sequence, if applied to a chain of circles standing on a straight line and tangent to their immediate members. (In the diagram below, R and R

_{1}= a

_{2}= b.)

He then showed that the coordinates x

_{n}of the centers of the circles on the line converge to a limit x_{∞}that satisfies
x

_{∞}- x_{1}: x_{2}- x_{∞}= φ√a : √b.
Jerzy Kocik (

*Math Magazine*, 83 (2010) 384-390) enjoys a window in his house with an abundance of Golden Ratio and its powers. Start with two small central circles of unit diameter. The radius R of the two circles on their left and right, given that a pair of congruent circles (dotted) is simultaneously tangent to all the other circles, is exactly φ:
(The diagram also sports multiple appearances of the golden rectangle and Kepler's triangle.)

In a 2011 article, M. Bataille, offered an elegant way of constructing the Golden Ratio via an equlateral triangle and a square.

Given an equilateral triangle ABC, erect a square BCDE externally on the side BC. Construct the circle, center C, passing through E, to intersect the line AB at F. Then, B divides AF in the golden ratio. (For a proof check a separate page.)

Also, in a 2011 article, Jo Niemeyer, offered a beautiful way of constructing the Golden Ratio with three equal segments, their midpoints and a pair of perpendicular lines.

Three equal segments A

_{1}B_{1}, A_{2}B_{2}, A_{3}B_{3}are positioned in such a way that the endpoints B_{2}, B_{3}are the midpoints of A_{1}B_{1}, A_{2}B_{2}respectively, while the endpoints A_{1}, A_{2}, A_{3}are on a line perpendicular to A_{1}B_{1}.
In this arrangement, AA

_{2}divides A_{1}A_{3}in the golden ratio, namely,_{1}A

_{3}/ A

_{1}A

_{2}= φ.

(For a proof check a separate page.)

Following D. MacHale (The Mathematical Gazette, v. 92, n. 525, Nov 2008, 536-537), let θ be an acute angle that solves

cosθ = tanθ.

(Such an angle certainly exists because graphs of the functions y = cos(x) and y = tan(x) intersect in the interval [0, π/2). The equation is equivalent to cos²θ = sinθ, or

x² + x - 1 = 0,

with x = sinθ. Since |sinθ| ≤ 1, we have a unique solution x = (√5 - 1) / 2 = 1/φ. It follows there is a right triangle with hypotenuse φ, one leg 1 and the other √φ. If θ is the angle opposite the leg 1, thencosθ = √φ / φ, while tanθ = 1 / √φ, which are the same.

Quang Tuan Bui came up with a construction based on that of George Odom: ABC and AMN are two equilateral triangles where M is midpoint of side BC. Arc 60° centered at B passing through A, C intersects side MN by golden ratio.

For a proof, check a separate page.

### References

- C. Alsina, R. B. Nelsen,
*Charming Proofs*, MAA, 2010 - M. Bataille,
__Another Simple Construction of the Golden Section__,*Forum Geometricorum*,Volume 11 (2011) 55 - M. J. Gazalé,
*Gnomon: From Pharaohs to Fractals*, Princeton University Press, 1999 - K. Hofstetter,
__A Simple Construction of the Golden Section__,*Forum Geometricorum*, v 2 (2002), pp. 65-66 - K. Hofstetter,
__A 5-step Division of a Segment in the Golden Section__,*Forum Geometricorum*, v 3 (2003), pp. 205-206 - K. Hofstetter,
__Another 5-step Division of a Segment in the Golden Section__,*Forum Geometricorum*, v 4 (2004), pp. 21-22 - K. Hofstetter,
__Division of a Segment in the Golden Section with Ruler and Rusty Compass__,*Forum Geometricorum*, v 5 (2005), pp. 135-136 - H. E. Huntley,
*The Divine Proportion*, Dover, 1970 - R. B. Nelsen,
*Proofs Without Words II*, MAA, 2000 - Jo Niemeyer,
__A Simple Construction of the Golden Section__,*Forum Geometricorum*, Volume 11 (2011) 53 - S. Olsen,
*The Golden Section: Nature's Greatest Secret*, Walker & Company, 2006 - C. Pritchard (ed.),
*The Changing Shape of Geometry*, Cambridge University Press, 2003 - S. Roberts,
*King of Infinite Space*, Walker & Company, 2006 - J. Tong and S. Kung,
__A Simple Construction of the Golden Ratio__,*Forum Geometricorum*, Volume 7 (2007) 31-32

### Fibonacci Numbers

- Ceva's Theorem: A Matter of Appreciation
- When the Counting Gets Tough, the Tough Count on Mathematics
- I. Sharygin's Problem of Criminal Ministers
- Single Pile Games
- Take-Away Games
- Number 8 Is Interesting
- Curry's Paradox
- A Problem in Checker-Jumping
- Fibonacci's Quickies
- Fibonacci Numbers in Equilateral Triangle
- Binet's Formula by Inducion
- Binet's Formula via Generating Functions
- Generating Functions from Recurrences
- Cassini's Identity
- Fibonacci Idendtities with Matrices
- GCD of Fibonacci Numbers
- Binet's Formula with Cosines
- Lame's Theorem - First Application of Fibonacci Numbers

### Golden Ratio

- Golden Ratio in Geometry
- Golden Ratio in an Irregular Pentagon
- Golden Ratio in a Irregular Pentagon II
- Inflection Points of Fourth Degree Polynomials
- Wythoff's Nim
- Inscribing a regular pentagon in a circle - and proving it
- Cosine of 36 degrees
- Continued Fractions
- Golden Window
- Golden Ratio and the Egyptian Triangle
- Golden Ratio by Compass Only
- Golden Ratio with a Rusty Compass
- From Equilateral Triangle and Square to Golden Ratio
- Golden Ratio and Midpoints
- Golden Section in Two Equilateral Triangles
- Golden Section in Two Equilateral Triangles, II
- Golden Ratio is Irrational

## No hay comentarios:

## Publicar un comentario