an odd PI from dividing circle diameters?

May 2010
27
5
Question:
I want to algebraically find the diameter of a larger circle that has contact with two smaller circles and is bound not to grow larger than x=y (see attached image)

It seems for some reason that if I take 6*(smaller circle)/(larger circle), I get PI. (see attached image)

If I get PI, why do I get PI? Is that normal?

I have serious trouble solving the problem algebraic so I made a script that tries to find the solution. I'm not sure why, but the script is not very precis. I can barely squeeze out the result 3.141

I don't see how I can solve this algebraically when the larger circles diameter is unknown and also position is unknow? how?

I was inspired by the NowIsForevers post here at mathhelpforum:
"A fractal made with circles?"
http://www.mathhelpforum.com/math-help/geometry/145229-fractal-made-circles.html

here is the script (as3)
PHP:
circl.x =48.0;
circl.y =0;

var h1:Number = 48.0;
var k1:Number = 48.0;

var stepsize:Number = 1.0;

var R1 = h1;
var R2;
var radiusSum;
var dist;
var dir = true;

this.addEventListener(Event.ENTER_FRAME,iter);  

function circleSize() { 
    circl.height = circl.width = circl.x*Math.sqrt(2);
}  

function iter(event:Event)
{
	circl.x +=stepsize;
	circleSize();
	checkR1R2();
}

function checkR1R2()
{
	R2 = circl.height/2;
	radiusSum = R1 + R2;
	var val = (h1-circl.x)*(h1-circl.x)+(k1)*(k1)
	
	dist = Math.sqrt(val);
	
	if(dist<radiusSum)
	{
		updateStepsize(true)
	}else{
		updateStepsize(false)
	}
}

function updateStepsize(val)
{
	if(val==dir)
	{
		
	}else{
		stepsize *=-1;
		var rand = Math.random() /1000;
		stepsize /=(1.0001+rand);

		trace(stepsize);
	}
	dir=val;
	traceQuote();
}

function traceQuote()
{
	var bigcirl = circl.height;
	var smallcircl = 2*(R1+R1*Math.sqrt(2));
	trace("q: " + 6*smallcircl/bigcirl);
}
 

Attachments

May 2010
27
5
It was close but no cigar... probably nested in some way, I don't know?

===========================
Assumption that there exists radiuses such that:

\(\displaystyle
6\frac{r_2}{r_3} = \pi
\)

\(\displaystyle
r_2 = r_1 + \sqrt{2}
\)

We set the smaller radius to one
\(\displaystyle
r_1 = 1
\)

then:
\(\displaystyle
6\frac{\ 1 + \sqrt{2} }{r_3} = \pi
\)
\(\displaystyle
r_3 = 6\frac{\ 1 + \sqrt{2} }{\pi}
\)

The larger radius is set such that:
\(\displaystyle
r_3 = \sqrt{\left (\frac{x_2}{2}\right)^2 + \left (\frac{x_2}{2}\right)^2}
\)

\(\displaystyle
x_2 =\sqrt{2\left(\frac{6\left( \left( 1 + \sqrt{2} \right)\right)}{\pi}\right)^2}
\)

Coordinates:
\(\displaystyle
r_1\left[x_1,y_1 \right] =\left[r_1, r_1 \right]
\)

\(\displaystyle
r_3\left[x_2,y_2 \right] =\left[x_2, 0 \right]
\)

And the distance d between the coordinates are:\\*
\(\displaystyle
d = \sqrt{\left (x_1 - x_2\right)^2 + \left (y_1 - y_2\right)^2}
\)

\(\displaystyle
r_1 + r_3 = \sqrt{\left (1 - \sqrt{2\left(\frac{6\left( \left( 1 + \sqrt{2} \right)\right)}{\pi}\right)^2}\right)^2 + \left (1 - 0\right)^2}
\)


\(\displaystyle
r_1 + r_3 = 5.6105053746002749516707090913621151836980050237612053686454...
\)
\(\displaystyle
r_3 = 4.6105053746002749516707090913621151836980050237612053686454...
\)
\(\displaystyle
r_2 = 2.4142135623730950488016887242096980785696718753769480731766...
\)

since:
\(\displaystyle
6*\frac{ r_2}{r_3} = 3.1417990431242965565708535190608565953968151035714525865439...
\)

\(\displaystyle
6\frac{r_2}{r_3} \not = \pi
\)
 
Last edited: