Thread: an odd PI from dividing circle diameters?

1. an odd PI from dividing circle diameters?

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:
http://www.mathhelpforum.com/math-he...e-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 dist;
var dir = true;

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;
var val = (h1-circl.x)*(h1-circl.x)+(k1)*(k1)

dist = Math.sqrt(val);

{
}else{
}
}

{
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);
}[/PHP]

2. 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:\\*
$\displaystyle d = \sqrt{\left (x_1 - x_2\right)^2 + \left (y_1 - y_2\right)^2}$

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

$\displaystyle r_1 + r_3 = 5.610505374600274951670709091362115183698005023761 2053686454...$
$\displaystyle r_3 = 4.610505374600274951670709091362115183698005023761 2053686454...$
$\displaystyle r_2 = 2.414213562373095048801688724209698078569671875376 9480731766...$

since:
$\displaystyle 6*\frac{ r_2}{r_3} = 3.141799043124296556570853519060856595396815103571 4525865439...$

$\displaystyle 6\frac{r_2}{r_3} \not = \pi$