Golden Ratio in Geometry

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

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

(1)

$\phi : 1 = 1 : (\phi - 1).$

To see this, we reduce (1) to

(2)

$\phi ^{2} - \phi - 1 = 0,$

with two solutions $(1 \pm \sqrt{5})/2$ of which one is positive and the other negative. The positive one is exactly $\phi$ while the negative is $-1/\phi.$ The reciprocal $\phi ' = 1/\phi $ is sometimes also termed the golden ratio, but more often is referred to as its conjugate. Of course also

(3)

$\phi = 1 + \phi '.$

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 $ax^{2} + bx + c = 0.$ Then the quadratic formula gives

$\displaystyle r = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a},$

from which

$\begin{align} |r| &\le (|b| + \sqrt{|b|^{2} + 4|a|\cdot|c|}) / 2|a|\\ &\le (1 + \sqrt{1 + 4}) / 2 \cdot \max\{|a|, |b|, |c|\} / |a|\\ &= \phi \cdot \max\{|a|, |b|, |c|\} / |a|. \end{align}$

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 $\cos(36^{\circ}) = (1 + \sqrt{5})/4.)$ In passing, the isosceles $72^{\circ}-36^{\circ}-72^{\circ}$ and $36^{\circ}-108^{\circ}-36^{\circ}$ 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 $72^{\circ}-36^{\circ}-72^{\circ}$ triangle leaving a $36^{\circ}-108^{\circ}-36^{\circ}$ 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 $\phi,$ what else? But this is not the only occurence of the golden ratio in regular pentagon.

A regular pentagon can be formed by folding and tightening a narrow band of paper:

trefoil knot - out of flattened and tightened piece of paper

(This is the basis for my logo.)

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 $PQ / BP = \phi .$ 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 $1:2:\sqrt{5}$ triangle. Both appear in the following diagram:

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

1-2-sqrt(5) triangle in the 3-4-5 triangle

An intriguing showing of $\phi$ 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 $\Delta ABC.$ Then $LM / MY = \phi .$

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

$MX\cdot MY = AM\cdot MC.$

In other words,

$(a + b)\cdot b = a\cdot a.$

Denoting $a/b = x,$ we see that

$1 + x = x^{2},$

which is (2), so that indeed $x = \phi.$ 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 $B(E)$ and $A(F)$ intersect in $X$ and $Y,$ as in the diagram. Because of the symmetry, points $X,$ $D,$ $C,$ $Y$ are collinear. The fact is $CX / CD = \phi .$ (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 $AB:AG = \phi$ and $AG':AB = \phi.$

For a proof, suppose $AB$ has unit length. Then $CD = \sqrt{3}$ and $EG = EF = \sqrt{2}.$ Let $H$ be the orthogonal projection of $E$ on the line $AB.$ Since $HA = 1/2,$ and $HG^{2} = EG^{2} - EH^{2} = 2 - 3/4 = 5/4,$ we have $AG = HG - HA = (\sqrt{5} - 1)/2.$ 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 $G.$ $AG:BG = \phi .$

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

$IG/GH = DI/FH = (\sqrt{3})/2 / (\sqrt{15})/4 = 2/\sqrt{5}.$

It follows that $IG = 2\cdot IH/(\sqrt{5} + 2) = (\sqrt{5} - 2)/2,$ and $AH = 1/2 + IG = (\sqrt{5} - 1)/2.$ 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 $RS = a \lt b.$ The line $MN$ parallel to the bases is of length $\sqrt{(a^{2} + b^{2})/2}.$ 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 $PSW,$

$\displaystyle\frac{SM}{SP}=\frac{MV}{PW}=\frac{\sqrt{\frac{a^{2}+b^{2}}{2}}-a}{b-a}$

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

In the construction, $BC = 3\cdot AD,$ $CE = AD,$ $CE \perp BC,$ $BF = EF = FH,$ $FH \perp BE,$ $BI = BH$, $IJ\parallel AB,$ and $GJ\parallel BC.$ $AG/BG = \phi .$

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

$(1 + \phi ):1 = (1 + 2\phi ):\phi ,$

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 $b,$ $a \lt b.$ The straight line through the inflection points meets the graph in two other points with abscissas $x_{L}$ and $x_{R}$ say: $x_{L} \lt a \lt b \lt x_{R}.$

Then

$\displaystyle x_{L}=a\frac{1+\sqrt{5}}{2} +b\frac{1-\sqrt{5}}{2}\\ \displaystyle x_{R}=b\frac{1+\sqrt{5}}{2} +a\frac{1-\sqrt{5}}{2}.$

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 $\phi$ and two semicircles of diameter $\phi ^{-1}.$

three equal circles in a semicircle

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 $\phi.$

The only time the sides of a right triangle are in geometric progression is when the triangle is similar to the one with sides $1,$\sqrt{\phi },$ $\phi$ [Charming Proofs, p. 74]. These triangles are often referred to as Kepler's triangles.

In general, if the sides of a triangle are in geometric progression, the factor of the progression necessarily falls into the range $(1/\phi , \phi ).$

The area of annulus defined by the radii, $a$ and $b \lt a$ is equal to the area of an ellipse with major and minor axis equal to $a$ and $b,$ respectively, is when $a/b = \phi $ [Charming Proofs, p. 37].

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

$\displaystyle\frac{1}{\sqrt{R}} = \frac{1}{\sqrt{R_{1}}} + \frac{1}{\sqrt{R_{2}}}.$

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, $R_{1} = a$ and $R_{2} = b.)$

golden ratio in the chain of tangent circles

He then showed that the coordinates $x_{n}$ of the centers of the circles on the line converge to a limit $x_{\infty}$ that satisfies

$x_{\infty } - x_{1} : x_{2} - x_{\infty } = \phi \sqrt{a} : \sqrt{b}.$

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 $\phi:$

golden ratio from two equal circles - a wonderful window

(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, $A_{2}$ divides $A_{1}A_{3}$ in the golden ratio, namely, $A_{1}A_{3} / A_{1}A_{2} = \phi .$

(For a proof check a separate page.)

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

$\cos\theta = \tan\theta .$

(Such an angle certainly exists because graphs of the functions $y = \cos(x)$ and $y = \tan(x)$ intersect in the interval $[0, \pi /2).$ The equation is equivalent to $\cos^{2}\theta = \sin\theta ,$ or

$x^{2} + x - 1 = 0,$

with $x = \sin\theta .$ Since $|\sin\theta | \le 1,$ we have a unique solution $x = (\sqrt{5} - 1) / 2 = 1/\phi .$ It follows there is a right triangle with hypotenuse $\phi ,$ one leg $1$ and the other $\sqrt{\phi }.$ If $\theta$ is the angle opposite the leg $1,$ then $\cos\theta = \sqrt{\phi } / \phi ,$ while $\tan\theta = 1 / \sqrt{\phi },$ 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 $60^{\circ}$ centered at $B$ passing through $A,$ $C$ intersects side $MN$ by golden ratio.

For a proof, check a separate page.

Here's another neusis construction with a proof on a separate page:

The base of the rectangle equals $1/\phi .$

John Molokach came up with a construction of the golden ratio by first constructing a golden rhombus, that is, a rhombus in which the ratio of the diagonals (long to short) equals $\phi .$

Golden ratio via the golden rhombus

In one sweep, John additionally obtains $1/\phi ,$ $1/\phi ^{2},$ and $\phi ^{2}.$ I placed the derivation on a separate page.

Bùi Quang Tuån found a simple construction: if $\displaystyle\frac{OM}{CM}=3,$ then $\displaystyle\frac{MN}{NO}=\phi .$

Golden ratio by Bui Quang Tuan

I placed a proof in a separate file.

And here's one more construction from Bùi Quang Tuån. if $\displaystyle CM=\frac{BC}{2}$ then $\displaystyle\frac{CN}{NO}=\phi .$

Golden ratio by Bui Quang Tuan, #2 - construction

I placed a proof in a separate file. Bùi also came up with a simple unified approach that underlies several at first sight independent construction of the Golden Ratio.

Bùi Quang Tuån has turned in another construction that is made transparent by the following disgram. $\displaystyle\frac{OE}{EM}=\phi .$

golden ratio by Bùi Quang Tuån, construction

I placed a proof in a separate file.

Tran Quang Hung has contributed eight constructions of the Golden Ratio.

In a square $ABCD$, with $M,N,P,Q$ the midpoints of the sides, circle on $AP$ as diameter cuts $FN$ in Golden Ratio.

golden ratio by Tran Quang Hung, construction #1

Details can be found in a separate file.

Tran's second discovery takes place in equlateral triangle.

golden ratio by Tran Quang Hung, construction #2

If the circumcricle $(FGH)$ of equilateral $\Delta FGH$ serves as the incircle of equilateral $\Delta ABC$ such that $F$ is on $AH,$ $G$ on $CF,$ and $H$ on $BG,$ then $F,G,H$ divide the corresponding segments $AH,CF,BG$ in Golden Ratio. Details can be found in a separate file.

Tran's third discovery relates to a hexagon:

golden ratio by Tran Quang Hung, construction #3

Details can be found in a separate file.

Tran's fourth construction builds on two squares $ABCD$ and $MNPC,$ with $M$ on the diagonal $BD$ (which causes $N$ to lie on $AB.)$ Let $Q$ be the intersection of $CD$ and $MP,$ extended.

golden ratio by Tran Quang Hung, construction #4

If $2DQ=CD$ then $AN/BN=\phi.$ Details can be found in a separate file.

His fifth construction starts with two equilateral triangles $ABC$ and $ABC'.$ Let $M$ be the midpoint of $BC,$ $N$ the midpoint of $AB;$ $P$ the intersection of $MN$ with the circumcircle $(ABC');$ $AP$ crosses $BC$ in $D.$

golden ratio by Tran Quang Hung, construction #5

Then $CD/BD=\phi,$ the Golden Section. Details can be found in a separate file.

Here's Tran's sixth construction. Square $MNPQ$ is inscribed into square $ABCD$ so that the lines joining their vertices intersect the sides of $MNPQ$ at the midpoints $X,Y,Z,W,$ as shown:

golden ratio by Tran Quang Hung, construction #6, statement

Then the vertices of $MNPQ$ divide the sides of $ABCD$ in the Golden Ratio, e.g. $\displaystyle\frac{BM}{AM}=\phi.$ Details can be found in a separate file.

Tran's seventh construction: Equilateral $\Delta DEF$ is inscribed into equilateral $\Delta ABC$ so that its extended midlines $MN, NP, MP$ pass through the vertices of $\Delta ABC,$ as shown:$

golden ratio by Tran Quang Hung, construction #7

Then, say, the vertices of the inner triangle divide the sides of the outer triangle in the Golden Ratio, e.g., $\displaystyle\frac{BF}{AF}=\phi.$ Details can be found in a separate file.

Tran's construction #8: $ABCD$ is a rhombus with $2AC=BD;$ $(O)$ the inscribed circle; $E$ and $F$ the points of intersection of $(O)$ with $BD.$

golden ratio by Tran Quang Hung, construction #8

Then $F$ divides $DE$ in the golden ratio. Details can be found in a separate file.

Not to be overdone, Quang Tuan Bui devised a 5-step construction of the golden ratio:

golden ratio by Quang Tuan Bui, 5-step construction

In the diagram $\displaystyle\frac{MP}{NP}=\phi.$ Details can be found in a separate file.

He also noticed an appearance of the golden ratio at the intersection of a cross and a square inscribed in the same circle:

golden ratio by Quang Tuan Bui, between a cross and a square

Not every cross will do but the one that consists of five equal squares. For details check a separate file

Quang Tuan Bui came up with a wonderful configuration with severa instances of the Golden Ratio:

four golden circles by Quang Tuan Bui

In the diagram, the sides and the diagonals of the parallelogram are in Golden Ratio; the sides are divided by the centers of the circle in Golden Ratio - but that's not all. For details check a separate file.

I've stopped getting surprised. Tran Quang Hung has found the Golden Ratio among mixtilinear circles in equilateral triangle.

golden ratio in mixtilinear circles by Tran Quang Hung

Details can be found in a separate file.

In 2015 Tran Quang Hung has found once more the Golden Ratio in a combination of a semicircle, a square, and a right isosceles triangle:

Tran Quang Hung (2015) - Golden Ratio construction, construction

Details can be found in a separate file.

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

Golden Ratio

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