I was recently revisiting my earlier post on A nice result in Geometry and got 'distracted' into the Classical Apollonius problem. The fact that this problem has attracted the interests some of the most famous mathematicians throughout history tells us how beautiful this problem really is.
As always, the wiki page was a great source of information for me and it gives a very nice account on Gergonne's solution to the problem which, needless to say, is ingenious. Meanwhile, I was also consulting Cut-The-Knot's version of Gergonne's solution.
Both the solutions seemed similar but there was one subtle difference. To make it more clear (and readable), I (re)created the diagrams on Wikipedia using Geogebra. Before we dwell further, let's list out the key labels used in the construction.
In what follows , $i \in \{1,2,3\}$ and $X \in \{A,B\}$
$C_1,C_2,C_3$ - The given circles. In general, there are eight circles that are mutually tangent to all these circles
$C_O$ - Circle orthogonal to all the three given circles
$G$ - Radical center of the given circles. Also, center of the Orthogonal circle $C_O$
$C_A,C_B$ - A pair (out of the possible eight) of solution circles
$X_i$ - Points where the solution circle $C_X$ touches Circle $C_i$
$l_i$ - Line passing through $A_i$ and $B_i$. As $A_i$ and $B_i$ are inverses to each other w.r.t to $C_O$, this line also contains the Radical center $G$; also the polar of point $T_i$ defined below
$p$ - (One out of the possible four) Homothetic Axis. This line also happens to be the radical axis of the solution circles
$P_i$ - Pole of the Homothetic axis $p$ w.r.t to circle $C_i$
$J_i, K_i$ - Points of intersection of the Orthogonal circle $C_O$ and circle $C_i$
$T_i$ - Point of intersection of the line joining $J_i$ and $K_i$, and the Homothetic axis $p$; Also, the pole of line $l_i$ defined above
$O_1,O_2$ - Points of intersection between the Orthogonal circle $C_O$ and Homothetic axis $p$
$C_I$ - Circle with center at $O_1$ and passing through $O_2$
Gergonne's Solution - Wikipedia
(i) Construct $p$, the Homothetic axes (or axes of similitude) (ii) Construct the Radical center $G$ of the three given circles (iii) Construct the poles $P_i$ of the Homothetic axes w.r.t the given circles.
Now the line joining $G$ and $P_i$ cuts the circle $C_i$ in $A_i$ and $B_i$ giving six points in total. These six points determine the solution circles.
Gergonne's Solution - Cut-The-Knot
(i) Construct $p$, the Homothetic axes (or axes of similitude) (ii) Construct the circle orthogonal to the given circles $C_O$ (iii) Construct the Radical axis of the Orthogonal circle and each of the given circles (that is, the line joining $J_i$ and $K_i$); this intersects the Homothetic axes in $T_i$.
Now construct tangent from $T_i$ to each of the given circles.These points then determine the solution circles.
The third step in these solutions were the source of my confusion. Especially, while Wikipeida's version needs an inversion (constructing the poles $P_i$), Cut-the-Knot's version needs no inversion at all. This baffled me and it took me some time (two days to be exact) to figure out how this was possible.
The explanation given in Cut-The-Knot did not seem to clarify this point. It quotes Four Touching Circles IV, which does not add anything in relation to the third step in the construction.
Gergonne's solution relies crucially on the inversive relation between poles and polars of conics. In particular, it exploits the fact that if the pole $P$ of a line $p$ w.r.t a conic lies on a line $q$, then the pole $Q$ of line $q$ w.r.t the same conic lies on line $p$. Wikipedia version (and most descriptions of Gergonne's solution) makes use of this relation.
Now, Cut-The-Knot's solution uses the mirror image of the same relation. That is, if the polar $p$ of a point $P$ contains a point $Q$, then the polar $q$ of point $Q$ w.r.t the same conic, contains the point $P$. And hence, it cleverly avoids any explicit inversion construction. Here's how it works.
From Wikipedia, we know that point $T_i$ is the pole of the (polar) line $l_i$ w.r.t circle $C_i$. Now the polar of point $T_i$ contains the point $G$. Hence, the polar of $G$ must contain $T_i$.
Here's comes the nice part. Given any pair of orthogonal circles, the polar of center of either circle w.r.t to the other circle is their radical axis. This is very obvious from the construction of polars and orthogonal circles.
So in essence, we are actually constructing a polar (rather than a pole in Wikipedia's version) of point. However, because of the clever choice, it just happens to be the point of intersection of circles already constructed avoiding the need for any inversion.
That was indeed neat. Hope you agree.
This whole process of revisiting one of the classical problems in Geometry was really nice experience for me. I greatly enjoyed creating the Geogebra drawing of the solution. In fact, I came up with something on my own.
Whenever, the Homothetic axis $p$ meets the Orthogonal circle $C_O$, something very interesting happens. As the solution circles are inverses of each other under $C_O$, the points of intersection between $p$ and $C_O$ also happens to be the points of intersection between $C_A$ and $C_B$.
Now draw a circle $C_I$ with center at one of the points of intersection (Point $O_1$) and passing through the other (Point $O_2$). Use this as a circle of inversion.
As the Orthogonal circle passes through the center of inversion, it becomes a straight line $C_O'$.
The three given circles, being orthogonal to the $C_O$, transforms into circles with their center on the line above.
The solution circles too passes through the inversion center and they too transform into straight lines ($C_A'$ and $C_B'$).
The (other) point of intersection $O_2$ of the solution circles passes through the inversion circle $C_I$ and hence this point is where the solution lines $C_A'$ and $C_B'$ meet. That is, point $O_2$ becomes the Homothetic center of the (inverted) given circles.
In summary, if $p$ meets $C_O$, we can construct a circle $C_I$ with center at $O_1$ and passing through $O_2$. Invert any of the given circle and draw tangent lines from $O_2$ to the inverted circle. Inverting back these tangent lines gives the required solution circles.
While constructing inversion is harder (needs more steps) while doing manually, this method is relatively simpler if we have an 'macro' for inversion (as in Computer graphics systems like Geogebra).
But the main drawback about this method is that it only works when the Homothetic axes intersects the Orthogonal circle which, sadly, is not guaranteed when the Homothetic axis contains two external Homothetic centers. Even in this case, we can actually develop a solution using the concept of mid-circles. This is not complicated but is more laborious. Gergonne's solution can be used in these cases.
That's all for today. Hope you enjoyed it.
Until then
Yours aye
Me
As always, the wiki page was a great source of information for me and it gives a very nice account on Gergonne's solution to the problem which, needless to say, is ingenious. Meanwhile, I was also consulting Cut-The-Knot's version of Gergonne's solution.
Both the solutions seemed similar but there was one subtle difference. To make it more clear (and readable), I (re)created the diagrams on Wikipedia using Geogebra. Before we dwell further, let's list out the key labels used in the construction.
In what follows , $i \in \{1,2,3\}$ and $X \in \{A,B\}$
$C_1,C_2,C_3$ - The given circles. In general, there are eight circles that are mutually tangent to all these circles
$C_O$ - Circle orthogonal to all the three given circles
$G$ - Radical center of the given circles. Also, center of the Orthogonal circle $C_O$
$C_A,C_B$ - A pair (out of the possible eight) of solution circles
$X_i$ - Points where the solution circle $C_X$ touches Circle $C_i$
$l_i$ - Line passing through $A_i$ and $B_i$. As $A_i$ and $B_i$ are inverses to each other w.r.t to $C_O$, this line also contains the Radical center $G$; also the polar of point $T_i$ defined below
$p$ - (One out of the possible four) Homothetic Axis. This line also happens to be the radical axis of the solution circles
$P_i$ - Pole of the Homothetic axis $p$ w.r.t to circle $C_i$
$J_i, K_i$ - Points of intersection of the Orthogonal circle $C_O$ and circle $C_i$
$T_i$ - Point of intersection of the line joining $J_i$ and $K_i$, and the Homothetic axis $p$; Also, the pole of line $l_i$ defined above
$O_1,O_2$ - Points of intersection between the Orthogonal circle $C_O$ and Homothetic axis $p$
$C_I$ - Circle with center at $O_1$ and passing through $O_2$
Gergonne's Solution - Wikipedia
(i) Construct $p$, the Homothetic axes (or axes of similitude) (ii) Construct the Radical center $G$ of the three given circles (iii) Construct the poles $P_i$ of the Homothetic axes w.r.t the given circles.
 |
| Points $P_i$ are the objects of interest in Wikipedia version of Gergonne's Solution |
Now the line joining $G$ and $P_i$ cuts the circle $C_i$ in $A_i$ and $B_i$ giving six points in total. These six points determine the solution circles.
Gergonne's Solution - Cut-The-Knot
(i) Construct $p$, the Homothetic axes (or axes of similitude) (ii) Construct the circle orthogonal to the given circles $C_O$ (iii) Construct the Radical axis of the Orthogonal circle and each of the given circles (that is, the line joining $J_i$ and $K_i$); this intersects the Homothetic axes in $T_i$.
 |
| Points $T_i$ are the objects of interest in Cut-The-Knot version of Gergonne's Solution |
Now construct tangent from $T_i$ to each of the given circles.These points then determine the solution circles.
The third step in these solutions were the source of my confusion. Especially, while Wikipeida's version needs an inversion (constructing the poles $P_i$), Cut-the-Knot's version needs no inversion at all. This baffled me and it took me some time (two days to be exact) to figure out how this was possible.
The explanation given in Cut-The-Knot did not seem to clarify this point. It quotes Four Touching Circles IV, which does not add anything in relation to the third step in the construction.
Gergonne's solution relies crucially on the inversive relation between poles and polars of conics. In particular, it exploits the fact that if the pole $P$ of a line $p$ w.r.t a conic lies on a line $q$, then the pole $Q$ of line $q$ w.r.t the same conic lies on line $p$. Wikipedia version (and most descriptions of Gergonne's solution) makes use of this relation.
Now, Cut-The-Knot's solution uses the mirror image of the same relation. That is, if the polar $p$ of a point $P$ contains a point $Q$, then the polar $q$ of point $Q$ w.r.t the same conic, contains the point $P$. And hence, it cleverly avoids any explicit inversion construction. Here's how it works.
From Wikipedia, we know that point $T_i$ is the pole of the (polar) line $l_i$ w.r.t circle $C_i$. Now the polar of point $T_i$ contains the point $G$. Hence, the polar of $G$ must contain $T_i$.
Here's comes the nice part. Given any pair of orthogonal circles, the polar of center of either circle w.r.t to the other circle is their radical axis. This is very obvious from the construction of polars and orthogonal circles.
So in essence, we are actually constructing a polar (rather than a pole in Wikipedia's version) of point. However, because of the clever choice, it just happens to be the point of intersection of circles already constructed avoiding the need for any inversion.
That was indeed neat. Hope you agree.
This whole process of revisiting one of the classical problems in Geometry was really nice experience for me. I greatly enjoyed creating the Geogebra drawing of the solution. In fact, I came up with something on my own.
Whenever, the Homothetic axis $p$ meets the Orthogonal circle $C_O$, something very interesting happens. As the solution circles are inverses of each other under $C_O$, the points of intersection between $p$ and $C_O$ also happens to be the points of intersection between $C_A$ and $C_B$.
Now draw a circle $C_I$ with center at one of the points of intersection (Point $O_1$) and passing through the other (Point $O_2$). Use this as a circle of inversion.
As the Orthogonal circle passes through the center of inversion, it becomes a straight line $C_O'$.
The three given circles, being orthogonal to the $C_O$, transforms into circles with their center on the line above.
The solution circles too passes through the inversion center and they too transform into straight lines ($C_A'$ and $C_B'$).
The (other) point of intersection $O_2$ of the solution circles passes through the inversion circle $C_I$ and hence this point is where the solution lines $C_A'$ and $C_B'$ meet. That is, point $O_2$ becomes the Homothetic center of the (inverted) given circles.
 |
| The given circles (not shown here) inverted under $C_I$ form a set of Homothetic circles with center $O_2$. The Orthogonal circles and both the solution circles transforms to straight lines. |
In summary, if $p$ meets $C_O$, we can construct a circle $C_I$ with center at $O_1$ and passing through $O_2$. Invert any of the given circle and draw tangent lines from $O_2$ to the inverted circle. Inverting back these tangent lines gives the required solution circles.
While constructing inversion is harder (needs more steps) while doing manually, this method is relatively simpler if we have an 'macro' for inversion (as in Computer graphics systems like Geogebra).
But the main drawback about this method is that it only works when the Homothetic axes intersects the Orthogonal circle which, sadly, is not guaranteed when the Homothetic axis contains two external Homothetic centers. Even in this case, we can actually develop a solution using the concept of mid-circles. This is not complicated but is more laborious. Gergonne's solution can be used in these cases.
That's all for today. Hope you enjoyed it.
Until then
Yours aye
Me
