Golden Ratio in Circle - in Droves
This is always very satisfying to discover a unique underlying reason for apparently independent mathematical facts, especially when it is simple and leads to new discoveries. An example of such discovery was posted on facebook by Bùi Quang Tuån. This is a beautifully simple and general approach to deriving golden ratio constructions.
Consider a unit circle with center $O=(0,0)$ and diameter $AB,$ with $A=(-1,0)$ and $B=(1,0).$ Let $X=(x,0)$ and $Y=(y,0),$ $0\lt x,y\lt 1.$ Define also two points on the circle $C=(x,\sqrt{1-x^{2}})$ and $D=(y,\sqrt{1-y^{2}}).$
Let point $M=(m,0)$ be the intersection of $AB$ and $CD$ so that
(1)
$\displaystyle m=\frac{x\sqrt{1 – y^2} + y\sqrt{1 – x^2}}{\sqrt{1 – x^2} + \sqrt{1 – y^2}},$
which could be obtained from similar triangles $MXC$ and $MYD,$ or with a two-point equation of line $CD.$
At this point Bùi Quang Tuån finds the condition for $M$ to divide $BO$ in Golden Ratio: $\displaystyle\frac{BM}{MO}=\phi.$ It could be verified directly from (1) or with wolframalpha that
(2)
$\displaystyle m=\frac{x\sqrt{1 – y^2} + y\sqrt{1 – x^2}}{\sqrt{1 – x^2} + \sqrt{1 – y^2}}$
implies the following relation between $x$ and $y$:
(3)
$\displaystyle x=\frac{2m-ym^{2}-y}{m^{2}-2my+1}.$
To insure $\displaystyle\frac{BM}{MO}$ we should take $m=\displaystyle\frac{1}{1+\phi},$ which yields a simple expression:
(4)
$\displaystyle x=\frac{3y-2}{2y-3}.$
Because of the symmetry between $x$ and $y$ it is also true that
(4')
$\displaystyle y=\frac{3x-2}{2x-3}.$
There are several interesting cases:
$x=0,\,y=2/3.$
The implied construction has been discussed previously.
$x=-1/2,\,y=7/8.$
The implied construction has been discussed previously, although the relation is less transparent than before.
$x=1/4,\,y=1/2.$
This is Kurt Hofstetter Another 5-step division of a segment in the golden section, Forum Geometricorum 4 (2004) 21–22, see also
Remark
The second example ($x=-1/2,\,y=7/8)$ leads to a six-step division of a given segment in the Golden Ratio. It reminds of an earlier Kurt Hofstetter division algorithm. Assume it's the $BO$ segment that has to be divided:
- Construct circle $C(O,B)$ centered at $O$ and passing through $B.$
- Construct circle $C(B,O)$. The two circles intersect at $G,$ $H.$
- Join $GH;$ let $I$ be the intersection of $GH$ with $BO.$
- Construct circle $C(B, I);$ let $D$ be its intersection with $C(O,B).$
- Construct circle $C(G,B);$ this circle intersects $C(O,B)$ the second time in $C.$
- Join $BC;$ its intersection with $BO$ is the sought point $M$ that divides $BO$ in Golden Ratio.
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
- Triangles with Sides in Geometric Progression
- Golden Ratio in Hexagon
- Golden Ratio in Equilateral Triangles
- Golden Ratio in Square
- Golden Ratio via van Obel's Theorem
- Golden Ratio in Circle - in Droves
- From 3 to Golden Ratio in Semicircle
- Another Golden Ratio in Semicircle
- Golden Ratio in Two Squares
- Golden Ratio in Two Equilateral Triangles
- Golden Ratio As a Mathematical Morsel
- Golden Ratio in Inscribed Equilateral Triangles
- Golden Ratio in a Rhombus
- Golden Ratio in Five Steps
- Between a Cross and a Square
- Four Golden Circles
- Golden Ratio in Mixtilinear Circles
- Golden Ratio in Isosceles Right Triangle, Square, and Semicircle
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