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.)

Tran Quang Hung has prepared a graphical illustration below where the ratio of the blue to a red segment is Golden.

golden ratio in a circle and a 1x2 rectangle

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 occurrence 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:

5-step construction of the golden ratio, #2

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 third paper is an additional 5-step division of a segment into the golden section.

5-step construction of the golden ratio, #3

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 $G.$ $AG:BG = \phi .$

The diagram contains several additional occurrences of the Golden Ratio. The details have been placed in a separate page.

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

5-step construction of the golden ratio, #4, rusty compass

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 4th 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 equilateral 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 diagram. $\displaystyle\frac{OE}{EM}=\phi .$

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 equilateral triangle.

golden ratio by Tran Quang Hung, construction #2

If the circumcircle $(FGH)$ 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:/p>

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 several 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.

Here is an invention of Elliot McGucken.

Golden Ratio With Two Equal Circles And a Line

Details can be found in a separate file.

Elliot came up with another construction:

Golden Ratio With three polygons, source

Details can be found in a separate file.

Following in Elliot's footsteps, Tran Quang Hung offered a different construction:

Golden Ratio With Two Unequal Circles And a Line

Details can be found in a separate file.

Tran Quang Hung and Elliot McGucken have independently came up with the following construction:

Golden Ratio In a 3x3 Square

Details can be found in a separate file.

Tran Quang Hung then found another occurrence of the Golden Ratio in a $3\times 3\;$ square

Golden  Ratio In a 3x3 Square II

Details can be found in a separate file.

Tran Quang Hung found the Golden Ratio in three tangent circles:

Golden Ratio In Three Tangent Circles

Here's beautiful self-explanatory diagram by John Arioni:

Golden Ratio Poster

The above suggests another construction:

Golden Ratio Next to Poster

Tran Quang Hung has devised another configuration of a $1\times 3\;$ rectangle and a circle that produce The Golden Ratio. Check the ratio of the blue segment to a red one:

Gloden ratio with a circlee and a 1x3 rectangle

There is a not immediately obvious relation between the case of $1\times 2\;$ and $1\times 3\;$ rectangles.

Nguyen Thanh Dung has devised the most straightforward construction of the Golden Ratio:

Golden Ratio In a 2x2 Square: Without And Within, diagram 1

This led first to an equivalent representation formed around a $2\times 2\;$ square:

Golden Ratio In a 2x2 Square: Without And Within, diagram 2

And, upon some deliberation, to the following one:

Golden Ratio In a 2x2 Square: Without And Within, diagram 3

Nguyen Thanh Dung has also come up with two related constructions:

Golden  Ratio With Two Unequal Circles And a Line II, Construction 1

and

Golden  Ratio With Two Unequal Circles And a Line II, Construction 2

The latter may be confused with that of George Odom but is quite distinct.

The following construction of the Golden Ratio has been posted by Tran Quang Hung at the CutTheKnotMath facebook page.

Golden Ratio in Equilateral and Right Isosceles Triangles, Construction 1

Golden Ratio has popped up in the Odom-like constructions involving a square and a regular hexagon:

Golden Ratio in Equilateral triangle, square and regular hexagon

For details visit trigonography.com.

Dũng Thanh Nguyên has posted the following diagram at the CutTheKnotMath facebook page:

Golden Ratio Astride Equilateral Triangle

As there's more to it, the proof and further exploration have been placed on a separate page.

Tran Quang Hung came up with an overlooked occurrence of the Golden Ratio in a regular pentagon:

Another Golden Ratio in Regular Pentagon

Then, $\displaystyle\frac{ON}{QN}=\frac{PR}{QR}=\varphi.\;$ The details are in a separate page.

Tran Quang Hung has found the Golden Ratio in a configuration of a regular pentagon and a regular hexagon:

Golden Ratio 5-gon & 6-gon

Then, $\displaystyle\frac{CQ}{BQ}=\frac{DR}{ER}=\varphi,\;$ the Golden Ratio. The details are in a separate page.

It did not take Tran Quang Hung long to find the Golden Ratio in an isosceles trapezoid with a $60^{\circ}\;$ degrees angle:

Golden Ratio in an Isosceles Trapezoid with a 60° Angle

Then $\displaystyle\frac{XB}{XA}=\frac{YC}{YB}=\frac{TA}{TD}=\frac{MC}{MA}=\varphi,\;$ the Golden Ratio. The details are in a separate page.

Just in time for the festive 2016 season Tran Quang Hung found another instance of the Golden Ratio:

Golden Ratio in Pentagon And Two Squares

The details are in a separate page.

And yet another 2016 discovery by Tran Quang Hung in the nick of time, on December 31:

Golden Ratio in Pentagon And Three Triangles

The details are in a separate page.

Tran Quang Hung made already several finds in 2017. Here's the first one: $\displaystyle\frac{AS}{SX}=\frac{CS}{BS}=\varphi\,$ in a mutually beneficial relationship.

Golden Ratio in a Mutually Beneficial Relationship, illustration

The details are in a separate page.

Tran Quang Hung keeps adding to his 2017 discoveries. Reflect a regular pentagon $ABCDE\,$ in each of its sides. Join the vertices of the so obtained pentagons skipping two at a time.

Star, Six Pentagons and Golden Ratio

The resulting star cuts the sides of the pentagon $ABCDE\,$ in the Golden Ratio. The diagram sports additional occurrences of the Golden Ratio and other secrets.

John Arioni came up with a construction of the Golden Ratio that includes $\sqrt{2},\,$ and $\pi/3:$

golden ratio by John Arioni, ratio

The details are in a separate page.

Elliot McGucken has discovered the Golden ratio in a configuration of two and three equilateral triangles, two of which are congruent:

Elliot McGucken + Leo Giugiuc + Alexander Bogomolny

The details are in a separate page.

Tran Quang Hung's frequent encounters with Golden Ratio become an anticipated event. Here's he found the Golden Ratio in a configuration of three regular pentagons

Golden Ratio in Three Regular Pentagons, illustration

This configuration proved to conceal other instances of the glorious number:

Golden Ratio in Three Regular Pentagons, extra occurrence

The details are in a separate page.

By changing configuration Tran Quang Hung's discovered another occurrence of the Golden Ratio. The configuration is in fact rich with pairs of segments in the Golden Ratio.

Golden Ratio in Three Regular Pentagons, illustration

The details are in a separate page.

Replacing three regular pentagons with five equilateral triangles, Tran Quang Hung's discovered another occurrence of the Golden Ratio.

Golden Ratio in Wu Xing, illustration

The details are in a separate page.

And another one by Tran Quang Hung:

3-4-5, Golden Ratio, illustration

The details are in a separate file.

John Molokach has published a construction that embeds into that of John Arioni:

Golden Ratio In Three Circles And Common Secant #2

The details can be found in a separate file.

It was possible to perceive the presence of the Golden Ratio in Problem 22 from the 2014 American Mathematical Competition:

problem 22 from 2014 AMC, solution

The details can be found in a separate file.

As was observed by Ercole Suppa (Italy), there's Golden Ratio in square:

golden ratio in square

and that led to other occurrences:

golden ratio in square, extra #2

The details can be found in a separate file.

Kadir Altintas posted a problem at the Peru Geometrico facebook group

A Modified Sangaku, illustration

The details can be found in a separate file.

Kadir Altintas posted an additional problem at the Peru Geometrico facebook group

Golden Ratio in  Equilateral Triangle on the Shoulders of George Odom

The problem reduces to George Odom's construction. The details can be found in a separate file.

Spt Ân (Vietnam) has been posted at the Peru Geometrico facebook group.

Golden Ratio in Two Squares, Or, Perhaps in Three, source

We relate this to the configuration of three squares discussed earlier. The details can be found in a separate file.

Thanos Kalogerakis came up with a construction, using three circles, which can be reduced to two:

Golden Ratio in Circles,problem 1,illustration

This can be nicely modified into a construction with four circles:

Golden Ratio in Circles,problem 2,illustration

The details can be found in a separate file.

Kadir Alkintas found the Golden Ratio in a special isosceles triangle but it came out squared. Ercole Suppa responded to the challenge with a construction of his own:

Golden  Ratio in Isosceles Triangle II,illustration

The details can be found in a separate file.

Probably there is no more appropriate place to house the Golden Ratio than the famous Yin-Yang diagram. It was John Arioni who made the discovery:

Golden Ratio in Yin-Yang, problem

The details can be found in a separate file.

An unexpected fact has been brought to may attention by Vincent Pantaloni and to his by Coln Wright. Colin came across the diagram below at the site of Mathematics Education Innovation.

Golden ratio by two tangents

References

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

Golden Ratio

  1. Golden Ratio in Geometry
  2. Golden Ratio in Regular Pentagon
  3. Golden Ratio in an Irregular Pentagon
  4. Golden Ratio in a Irregular Pentagon II
  5. Inflection Points of Fourth Degree Polynomials
  6. Wythoff's Nim
  7. Inscribing a regular pentagon in a circle - and proving it
  8. Cosine of 36 degrees
  9. Continued Fractions
  10. Golden Window
  11. Golden Ratio and the Egyptian Triangle
  12. Golden Ratio by Compass Only
  13. Golden Ratio with a Rusty Compass
  14. From Equilateral Triangle and Square to Golden Ratio
  15. Golden Ratio and Midpoints
  16. Golden Section in Two Equilateral Triangles
  17. Golden Section in Two Equilateral Triangles, II
  18. Golden Ratio is Irrational
  19. Triangles with Sides in Geometric Progression
  20. Golden Ratio in Hexagon
  21. Golden Ratio in Equilateral Triangles
  22. Golden Ratio in Square
  23. Golden Ratio via van Obel's Theorem
  24. Golden Ratio in Circle - in Droves
  25. From 3 to Golden Ratio in Semicircle
  26. Another Golden Ratio in Semicircle
  27. Golden Ratio in Two Squares
  28. Golden Ratio in Two Equilateral Triangles
  29. Golden Ratio As a Mathematical Morsel
  30. Golden Ratio in Inscribed Equilateral Triangles
  31. Golden Ratio in a Rhombus
  32. Golden Ratio in Five Steps
  33. Between a Cross and a Square
  34. Four Golden Circles
  35. Golden Ratio in Mixtilinear Circles
  36. Golden Ratio With Two Equal Circles And a Line
  37. Golden Ratio in a Chain of Polygons, So to Speak
  38. Golden Ratio With Two Unequal Circles And a Line
  39. Golden Ratio In a 3x3 Square
  40. Golden Ratio In a 3x3 Square II
  41. Golden Ratio In Three Tangent Circles
  42. Golden Ratio In Right Isosceles Triangle
  43. Golden Ratio Poster
  44. Golden Ratio Next to the Poster
  45. Golden Ratio In Rectangles
  46. Golden Ratio In a 2x2 Square: Without And Within
  47. Golden Ratio With Two Unequal Circles And a Line II
  48. Golden Ratio in Equilateral and Right Isosceles Triangles
  49. Golden Ratio in a Butterfly Astride an Equilateral Triangle
  50. The Golden Pentacross
  51. 5-Step Construction of the Golden Ratio, One of Many
  52. Golden Ratio in 5-gon and 6-gon
  53. Golden Ratio in an Isosceles Trapezoid with a 60 degrees Angle
  54. Golden Ratio in Pentagon And Two Squares
  55. Golden Ratio in Pentagon And Three Triangles
  56. Golden Ratio in a Mutually Beneficial Relationship
  57. Star, Six Pentagons and Golden Ratio
  58. Rotating Square in Search of the Golden Ratio
  59. Cultivating Regular Pentagons
  60. Golden Ratio in an Isosceles Trapezoid with a 60 degrees Angle II
  61. More of Gloden Ratio in Equilateral Triangles
  62. Golden Ratio in Three Regular Pentagons
  63. Golden Ratio in Three Regular Pentagons II
  64. Golden Ratio in Wu Xing
  65. Golden Ratio In Three Circles And Common Secant
  66. Flat Probabilities on a Sphere
  67. Golden Ratio in Square And Circles
  68. Golden Ratio in Square
  69. Golden Ratio in Two Squares, Or, Perhaps in Three
  70. Golden Ratio in Isosceles Triangle
  71. Golden Ratio in Circles
  72. Golden Ratio in Isosceles Triangle II
  73. Golden Ratio in Yin-Yang

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