7 Apollonius problem

How to construct a cycle $\pi_{0}$ tangent to three given cycles $\pi_{1},\,\pi_{2},\,\pi_{3}$ ? Apollonius of Perga (ca. 262 BC - ca. 190 BC) posed and solved this famous problem.

Theorem 7.1 (Free translation from Gergonne, 1816)   . The two special cases of Apollonius problem are cycles from the same pencil (cf Proposition 7.3) and cycles through the same point (cf Proposition 7.4). Otherwise, apexes of the three given cycles and their common orthogonal $\pi_{4}$ form a basis that splits the problem into four pairs of solutions. Using the cross product to define $X_{4}$ (cf Proposition 3.16), one of the solutions is given by :

$$k_{1} = \left({\it w_{2}}\,{\it w_{3}}-{\it W_{23}}\right)\left(-{\it w_{... ...w_{1}}\,{\it W_{23}}+{\it w_{2}}\,{\it W_{13}}+{\it w_{3}}\,{\it W_{12}}\right)$$

$$k_{2} = \left({\it w_{1}}\,{\it w_{3}}-W_{13}\right)\left(-{\it w_{1}}\,{... ...+{\it w_{1}}\,{\it W_{23}}-{\it w_{2}}\,{\it W_{13}}+w_{3}\,{\it W_{12}}\right)$$

$$k_{3} = \left({\it w_{1}}\,{\it w_{2}}-W_{12}\right)\left({\it -w_{1}}\,{... ...+w_{1}\,{\it W_{23}}+{\it w_{2}}\,{\it W_{13}}-{\it w_{3}}\,{\it W_{12}}\right)$$

$$k_{4} = \sqrt{2\left({\it w_{2}}\,{\it w_{3}}-{\it W_{23}}\right)\left({\it w_{1}}\,{\it w_{3}}-W_{13}\right)\left({\it w_{1}}\,{\it w_{2}}-W_{12}\right)}$$ (7.1)

and the others are obtained by changing $k_{4}$ into $-k_{4}$ (inversion through $\pi_{4}$) or changing the signs of $w_{1},w_{2},w_{3}$. A solution is real/imaginary or "unimaginable" (object that would have a non real center) according to the sign of $k_{4}^{2}$. Globally, the number of "imaginable" solutions changes when the tangency condition $\prod tang_{jk}$ vanishes.

Proof. Let $X_{4}$ be an independent point ( $X_{4}=X_{\perp}$ is not assumed for the moment). When $\pi_{i}$ is a circle and $X_{i}$ in standard form, then $w_{i}=2\, r_{i}$ giving a geometric reason to split $W_{jj}$ into $w_{j}^{2}$. Write $\pi_{0}$ as :

$$X_{0}=k_{1}X_{1}+k_{2}X_{2}+k_{3}X_{3}+k_{4}X_{4}$$ (7.2)

and substitute in (). This leads to a set of three equations whose Maple-length is $256$ each. A first elimination gives a rational expression for $k_{1}$ ($L=1010)$ and a $L=256,\, L=3765$ system. But the second equation splits in two $L=1879$ factors (the other being obtained by changing the sign of either $w_{2}$ or $w_{3}$). Another elimination involving only one factor leads to a rational expression for $k_{2}$ ($L=1247)$ and the remaining equation in $k_{3}$ splits in two $L=3170$ factors of second degree. Changing signs of $w_{1},\, w_{2},\, w_{3}$ swaps between these four $L=3170$ sized equations.

This proves the constructibility by straightedge and compass even in the special cases. The discriminants of the former four equations have the following lcm :

$$\operatorname{lcm}_{1..4}\Delta_{i}=8\, k_{4}^{2}\, Gram_{1..4}\left(\pi_{j}\right)\,\prod_{3}tang_{jk}\,\left(something\right)^{2}$$

The choice $X_{4}=X_{\perp}$ cancels all odd powers of $k_{4}$, allowing the transformation $k_{4}\mapsto-k_{4}$ in all formulas. Moreover $Gram_{1..4}$ splits into $w_{4}^{4}$ while "$something$" becomes the product OOKV of the four factors like $\left({\it w_{1}}\,{\it w_{2}}\,{\it w_{3}}+{\it w_{1}}\,{\it W_{23}}+{\it w_{2}}\,{\it W_{13}}+{\it w_{3}}\,{\it W_{12}}\right)$. Substituting back into $k_{2}$ and then into $k_{1}$ leads to the symmetrical (7.1), proving that the "only one k vanishes" condition describes a local instability of the elimination process rather than a property of the solutions. $\qedsymbol$

Remark 7.2   When one tangency condition is satisfied, then two pairs of solutions degenerate at the same time, like in .

Proposition 7.3   When $\pi_{1},\,\pi_{2},\,\pi_{3}$ are from the same pencil $P$, the either $P$ is a tangent pencil (and $\pi_{0}\in P$) or $\pi_{0}$ is a point-circle of $P^{\perp}$.

Proof. To obtain a basis of decomposition, introduce two extra circles $\pi_{4},\,\pi_{5}$ and decompose $X_{0}$ as $\sum_{1,2,4,5}k_{i}\, X_{i}$. Cycles $\pi_{4},\,\pi_{5}$ can be chosen as point-circles to reduce the size of the equations. When cycles $\pi_{1},\,\pi_{2}$ are tangent, we obtain three equations of first degree in $k_{1},\, k_{2}$ that are incompatible unless $k_{4}=k_{5}=0$. Otherwise, obtain $k_{2}$ from first equation and $k_{1}$ from the second. Both are second degree equations and share the same discriminant $\Delta_{45}$, an homogeneous polynomial of second degree in $k_{4},\, k_{5}$. Substituting into the last equation leads to $\Delta_{45}=0$ and there are no solutions apart the two indicated. $\qedsymbol$

Fig. 8: Special cases

[Two cycles are tangent]

Proposition 7.4   When $\pi_{1},\,\pi_{2},\,\pi_{3}$ are through the same point $\omega_{\perp}$ (and not from the same pencil) the eight solutions degenerate into four cycles and four times $\left\{ \omega_{\perp}\right\}$.

Proof. This situation is described in Fig. 9. One of the vertices of a triangle has been replaced by a circle of great radius. A near-sighted vision shows the in/ex circles of the triangle (left) while a long-sighted vision (right) shows four cycles tending to $\left\{ \infty\right\}$. $\qedsymbol$

Fig. 9: Cycles through the same point as a limiting case

[near-sighted vision]

[long-sighted vision]

Construction 7.5   Construct a cycle tangent to three given cycles.

Fig. 10: Apollonius problem for the excircles

Fig. 11: Lie coating of a pencil

[Isoptic pencil]

[Isotomic pencil]