Problem of Apollonius

(Difference between revisions)
 Revision as of 17:00, 3 November 2009 (edit)← Previous diff Revision as of 18:49, 3 November 2009 (edit) (undo)Next diff → Line 64: Line 64: Solving the first equation for x: Solving the first equation for x: - $a_2x=d_2-c_2r-b_2y \rightarrow x=\frac{d_2-c_2 r -b_2 y}{a_2}$. + $a_2x=d_2-c_2r-b_2y \rightarrow x=\frac{d_2-c_2 r -b_2 y}{a_2}$ Substituting that in to the second equation allows us to find y in terms of known values and r: Substituting that in to the second equation allows us to find y in terms of known values and r: Line 76: Line 76: *$y=\frac{a_2 d_3-a_3 d_2 +(a_2 c_3 -a_3 c_2)r}{a_2 b_3-a_3 b_2}$. *$y=\frac{a_2 d_3-a_3 d_2 +(a_2 c_3 -a_3 c_2)r}{a_2 b_3-a_3 b_2}$. - Now we can substitute back in to find x + Rather than substituting back in to find x, it is actually simpler to go through the same process we used to find y to get x in terms of known values and r. + First, we solve the first equation for y. + $b_2y=d_2-c_2r-a_2x \rightarrow y=\frac{d_2-c_2 r -a_2 x}{b_2}$ + + Plugging into the second equation gives us + + $a_3 x+b_3\left(\frac{d_2-c_2 r -a_2 x}{b_2}\right) +c_3 r =d_3$ + + $a_3 b_2 x+b_3 d_2 -b_3 c_2 r-a_2 b_3 x +b_2 c_3 r=b_2 d_3$ + + $x(a_3 b_2-a_2 b_3)=b_2 d_3-b_3 d_2 +(b_2 c_3-b_3 c_2) r$ + + *$x=\frac{b_2 d_3-b_3 d_2 +(b_2 c_3-b_3 c_2) r}{a_3 b_2-a_2 b_3}$.}} + + With $x=\frac{b_2 d_3-b_3 d_2 +(b_2 c_3-b_3 c_2) r}{a_3 b_2-a_2 b_3}$ and $y=\frac{a_2 d_3-a_3 d_2 +(a_2 c_3 -a_3 c_2)r}{a_2 b_3-a_3 b_2}$, we plug in values for the a's, b's and c's (which we get from the original information about the centers and radii of the three circles) to calculate x and y. Using our very first equations for the circles, we can then solve for r. + + {{Switch|link1=click to show example|link2=click to hide example|1=|2= + + I need an example! }} }} +

Revision as of 18:49, 3 November 2009

This an example of a fractal that can be created by repeatedly solving the Problem of Apollonius.

Basic Description

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The problem of Apollonius involves trying to find a circle that is tangent to three objects: points, lines, or circles in a plane. The most famous of these is the case involving three different circles in a plane, as seen in the picture to the left. The given three circles are in red, green, and blue, while the solution circle is in black.

Apollonius of Perga posed and solved this problem in his work called Tangencies. Sadly, Tangencies has been lost, and only a report of his work by Pappus of Alexandria is left. Since then, other mathematicians, such as Isaac Newton and Descartes, have been able to recreate his results and discover new ways of solving this interesting problem.

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The problem usually has eight different solution circles that exist that are tangent to the given three circles in a plane. The given circles must not be tangent to each other, overlapping, or contained within one another for all eight solutions to exist.

Given three points, the problem only has one solution. In the cases of one line and two points; two lines and one point; and one circle and two points, the problem has two solutions. Four solutions exist for the cases of three lines; one circle, one line, and one point; and two circles and one point. There are eight solutions for the cases of two circles and one line; and one circle and two lines, in addition to the three circle problem.

A More Mathematical Explanation

There are many different ways of solving the problem of Apollonius. The few that are easiest to understand include using an algebraic method or an inverse geometry method.

Algebraic Method

This method only uses math up to the level of understanding quadratic equations. We will proceed by setting up a system of quadratic equations and solving for the radius, r, of the unknown circle.

We start by labeling the center of each of the given circles $(x$1$,y$1$)$, $(x$2$,y$2$)$, and $(x$3$,y$3$)$. We will call the center of the unknown circle $(x,y)$. $r$1,$r$2, and $r$3 are the different radii of each of the given circles.

From this we are able to write our equations:

• $(x - x$1$)^2 + (y - y$1$)^2 = (r \pm r$1$)^2$
• $(x - x$2$)^2 + (y - y$2$)^2 = (r \pm r$2$)^2$
• $(x - x$3$)^2 + (y - y$3$)^2 = (r \pm r$3$)^2$

Next we are able to expand each of the equations to see better how they can relate to each other.

Expanding gives us:

• $(x^2+y^2-r^2)-2xx$1$-2yy$1$\pm2rr$1$+(x$1$^2+y$1$^2-r$1$^2)=0$
• $(x^2+y^2-r^2)-2xx$2$-2yy$2$\pm2rr$2$+(x$2$^2+y$2$^2-r$2$^2)=0$
• $(x^2+y^2-r^2)-2xx$3$-2yy$3$\pm2rr$3$+(x$3$^2+y$3$^2-r$3$^2)=0$

We can now look at the equations and see how we can subtract them from each other. So we will take the second and third equation minus the first equation.

Second minus first gives us:

• $2(x_1-x_2)x+2(y_1-y_2)y+2(\pm r_1 \pm r_2)r=(x_1^2+y_1^2-r_1^2)-(x_2^2+y_2^2-r_2^2)$

Third minus first gives us:

• $2(x_1-x_3)x+2(y_1-y_3)y+2(\pm r_1 \pm r_3)r=(x_1^2+y_1^2-r_1^2)-(x_3^2+y_3^2-r_3^2)$

For the sake of simplicity, we'll define some new variables. Let

$a_2=2(x_1-x_2)$; $b_2=2(x_1-x_2)$  ; $c_2=2(\pm r_1 \pm r_2)$ ; $d_2=(x_1^2+y_1^2-r_1^2)-(x_2^2+y_2^2-r_2^2)$

$a_3=2(x_1-x_3)$; $b_3=2(x_1-x_3)$  ; $c_3=2(\pm r_1 \pm r_3)$ ; $d_2=(x_1^2+y_1^2-r_1^2)-(x_3^2+y_3^2-r_3^2)$

Now our two equations can be written as

$a_2 x+b_2 y +c_2 r=d_2$
$a_3 x+b_3 y +c_3 r=d_3$

Since this is a simple linear system of equations, we can solve it for x and y in terms of r.

Solving the first equation for x:

$a_2x=d_2-c_2r-b_2y \rightarrow x=\frac{d_2-c_2 r -b_2 y}{a_2}$

Substituting that in to the second equation allows us to find y in terms of known values and r:

$a_3\left(\frac{d_2-c_2 r -b_2 y}{a_2} \right)+b_3 y +c_3 r =d_3$

$a_3 d_2-a_3c_2 r -a_3 b_2 y+a_2 b_3 y +a_2 c_3 r =d_3 a_2$

$y(a_2 b_3-a_3 b_2)=a_2 d_3-a_3 d_2 +(a_2 c_3 -a_3 c_2)r$

• $y=\frac{a_2 d_3-a_3 d_2 +(a_2 c_3 -a_3 c_2)r}{a_2 b_3-a_3 b_2}$.

Rather than substituting back in to find x, it is actually simpler to go through the same process we used to find y to get x in terms of known values and r.

First, we solve the first equation for y.

$b_2y=d_2-c_2r-a_2x \rightarrow y=\frac{d_2-c_2 r -a_2 x}{b_2}$

Plugging into the second equation gives us

$a_3 x+b_3\left(\frac{d_2-c_2 r -a_2 x}{b_2}\right) +c_3 r =d_3$

$a_3 b_2 x+b_3 d_2 -b_3 c_2 r-a_2 b_3 x +b_2 c_3 r=b_2 d_3$

$x(a_3 b_2-a_2 b_3)=b_2 d_3-b_3 d_2 +(b_2 c_3-b_3 c_2) r$

• $x=\frac{b_2 d_3-b_3 d_2 +(b_2 c_3-b_3 c_2) r}{a_3 b_2-a_2 b_3}$.

With $x=\frac{b_2 d_3-b_3 d_2 +(b_2 c_3-b_3 c_2) r}{a_3 b_2-a_2 b_3}$ and $y=\frac{a_2 d_3-a_3 d_2 +(a_2 c_3 -a_3 c_2)r}{a_2 b_3-a_3 b_2}$, we plug in values for the a's, b's and c's (which we get from the original information about the centers and radii of the three circles) to calculate x and y. Using our very first equations for the circles, we can then solve for r.

I need an example!

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The Apollonian gasket is an example of one of the earliest studied fractals and was first constructed by Gottfried Leibniz. It can be constructed by solving the problem of Apollonius iteratively. It was a precursor to Sierpinski's Triangle, and in a special case, it forms Ford Circles.
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Constructing the gasket begins with three mutually tangent circles. By solving this case of the problem of Apollonius we know that there are two other circles that are tangent to the three given circles. We now have five circles from which to start again.

Repeat the process with two of the original circles and one of the newly generated circles. Again, by solving Apollonius' problem we can find two circles that are tangent to this new set of three circles. Although, we already know one of the two solutions for this set of three circles; it is the other of the three circles that we started with.

Repeating this process over and over again with each set of three mutually tangent circles will create the Apollonian gasket.

References

Math Pages, Apollonius' Tangency Problem

MathWorld, Apollonius' Problem