# 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 $Y=(y,0),$ $-1\lt x,y\lt 1.$ 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:

1. $x=0,\,y=2/3.$

The implied construction has been discussed previously.

2. $x=-1/2,\,y=7/8.$

The implied construction has been discussed previously, although the relation is less transparent than before.

3. $x=1/4,\,y=1/2.$

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:

1. Construct circle $C(O,B)\,$ centered at $O\,$ and passing through $B.$
2. Construct circle $C(B,O).\,$ The two circles intersect at $G,H.$
3. Join $GH;\,$ let $I\,$ be the intersection of $GH\,$ with $BO.$
4. Construct circle $C(B, I);\,$ let $D\,$ be its intersection with $C(O,B).$
5. Construct circle $C(G,B);\,$ this circle intersects $C(O,B)\,$ the second time in $C.$
6. Join $CD;\,$ its intersection with $BO\,$ is the sought point $M\,$ that divides $BO\,$ in Golden Ratio.