Problem of Apollonius

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Figure 1: The eight solutions of Apollonius' problem. The three given circles are shown in black. The eight solution circles are shown as four pairs (blue, magenta, yellow and grey). In each pair, one solution circle encloses the given circles that are excluded by the other solution, and vice versa. For example, the larger blue solution encloses the two larger given circles, but excludes the smallest; the smaller blue solution does the reverse.
Figure 1: The eight solutions of Apollonius' problem. The three given circles are shown in black. The eight solution circles are shown as four pairs (blue, magenta, yellow and grey). In each pair, one solution circle encloses the given circles that are excluded by the other solution, and vice versa. For example, the larger blue solution encloses the two larger given circles, but excludes the smallest; the smaller blue solution does the reverse.

One of the most famous problems in Euclidean plane geometry, the problem of Apollonius is to construct circles that are tangent to ("touch") three given circles. The three given circles can be of any size and at any distance from each other; they may even be points (circles of zero radius) or lines (circles of infinite radius). Apollonius of Perga (ca. 262 BC – ca. 190 BC) posed and solved the problem and its limiting cases in his work Επαφαι ("Tangencies"); although this work is now lost, the 4th-century Hellenistic mathematician Pappus of Alexandria's report of his results has survived, and in the 16th century the Frenchman François Viète was able to reconstruct Apollonius' solution.

In general, Apollonius' problem has eight solutions—this is one of the oldest results in enumerative geometry—each of which include or exclude the given circles in diferent ways (Figure 1). These solution circles—sometimes called Apollonius circles—may be found by a variety of geometric and algebraic methods, including transformations such as circle inversion. The first modern solution was developed in the 16th century by Adriaan van Roomen (writing in Latin as Adrianus Romanus), but many eminent mathematicians have developed alternative methods, including Issac Newton, Carl Friedrich Gauss, Lazare Carnot, Augustin Louis Cauchy, Jean-Victor Poncelet and Joseph Diaz Gergonne. When the given circles are tangent to each other, René Descartes found a relation between the radii of the solution circles and those of the given circles now known as Descartes' theorem. This theorem was rediscovered three centuries later by Nobel laureate Frederick Soddy, who published it as a poem.

Apollonius' problem can be generalized in several ways. The three given circles may be chosen to lie on a sphere or other quadric surface, instead of the plane. Another generalization is to construct a circle that crosses the three given circles at three specified angles instead of being tangent to them. The three-dimensional analog of Apollonius' problem is to construct a sphere tangent to four given spheres; generalizations to even higher dimensions are also possible.

Although Apollonius' problem has historically been a test for novel methods in geometry, it has also found practical applications, mainly in trilateration and packing problems. One of the earliest fractals described in print was the Apollonian gasket proposed by Gottfried Leibniz, which is based on solving Apollonius' problem iteratively.

Contents

[edit] Statement and motivation

Figure 2: The pink and black circles (corresponding to the magenta circles in Figure 1) are a conjugate pair of solutions to Apollonius' problem. The given circles are drawn in red, green and blue. Inversion in the orange circle CG (centered on G) leaves the given circles invariant, but transforms the pink and black solution circles into one another. Since their points of tangency map into one another, the three lines L1–3 intersect at the radical center G.
Figure 2: The pink and black circles (corresponding to the magenta circles in Figure 1) are a conjugate pair of solutions to Apollonius' problem. The given circles are drawn in red, green and blue. Inversion in the orange circle CG (centered on G) leaves the given circles invariant, but transforms the pink and black solution circles into one another. Since their points of tangency map into one another, the three lines L1–3 intersect at the radical center G.

Apollonius' problem is to construct one or more circles that are tangent to three given objects in a plane, where an "object" may be a circle, line or point. These objects may be placed in any way and may cross one another; however, they are usually taken as distinct from one another. If one of the given circles is already tangent to the other two, it is counted as a solution, since a circle is assumed to be tangent to itself. Two geometrical objects are said to intersect if they share at least one point. If the angle between the objects at an intersection point is zero, the objects are said to be tangent there; the intersection point is called a tangent point or a point of tangency. The word "tangent" derives from the Latin present participle, tangens, meaning "touching". In practice, two circles are tangent if they intersect at only one point; this is also true for a line and a circle. A point is tangent to a circle or a line if and only if it intersects them. Two distinct points or two distinct lines cannot be tangent.

Apollonius' problem can also be formulated as locating a point such that the differences of its distances to three given points equal three given values. This second formulation is useful in trilateration and in the solution below of Isaac Newton.

Many geometrical constructions such as bisecting an angle or a line segment can be done "by eye", or improved iteratively from an approximate solution. Apollonius' problem is more difficult in this respect; it is difficult to see "by eye" where to place the center of the solution circle or how large to make its radius. Nevertheless, several exact solutions to Apollonius' problem have been developed by geometers over the centuries, with methods ranging from pure algebra to purely geometric compass and straightedge constructions.

[edit] Pairs of solutions by inversion

Figure 3: P ' is the inverse of P with respect to the circle.
Figure 3: P ' is the inverse of P with respect to the circle.

Solutions to Apollonius' problem generally occur in pairs; for each solution circle, there is a conjugate solution circle (Figure 2). One solution circle excludes the given circles that are enclosed by its conjugate solution, and vice versa. For example, in Figure 2, the pink solution circle encloses the red and blue given circles, but excludes the green; conversely, the conjugate black solution circle encloses the green given circle, but excludes the red and blue.

The two conjugate solution circles are related by a transformation known as inversion, which plays an important role in some solutions to Apollonius' problem. Inversion in a circle with center O and radius R consists of the following operation (Figure 3): every point P is mapped into a new point P' such that O, P, and P' are collinear, and the product of the distances of P and P' to the center O equal the radius R squared

\overline{\mathbf{OP}} \cdot \overline{\mathbf{OP^{\prime}}} = R^{2}.

Thus, if P lies outside the circle, then P' lies within, and vice versa. When P is the same as O, the inversion is said to send P to infinity. (In complex analysis, "infinity" is defined precisely in terms of the Riemann sphere.) Inversion has the useful property that lines and circles are always transformed into lines and circles, and points are always transformed into points. Circles are generally transformed into other circles under inversion; however, if a circle passes through the center of the inversion circle, it is transformed into a straight line, and vice versa. Importantly, if a circle crosses the circle of inversion at right angles (intersects perpendicularly), it is left unchanged by the inversion; it is transformed into itself.

In general, any three distinct circles have a unique circle that intersects all of them perpendicularly; the center of that circle is the radical center of the three circles. For illustration, the orange circle CG in Figure 2 crosses the red, blue and green given circles at right angles. Inversion in CG leaves the given circles unchanged, but transforms the conjugate pink and black solution circles into one another. For this reason, solutions to Apollonius' problem generally occur in pairs.

Under this inversion, the corresponding points of tangency of the two solution circles are transformed into one another. In Figure 2, the points A1 and B1 are transformed into one another, and similarly for the two other pairs (A2, B2) and (A3, B3). Hence, the lines L1–3 connecting these conjugate tangent points must pass through the radical center G, which is the center of CG.

[edit] Special cases

[edit] Limiting cases: points and lines

Main article: Limiting cases of Apollonius' problem

Apollonius problem is to construct one or more circles tangent to three given objects in a plane, which may be circles, points or lines. Thus, there are ten types of Apollonius' problem, which may be labeled with three letters, either C, L or P, to denote whether the given elements are a circle, line or point, respectively (Table 1). As an example, the type of Apollonius problem with a given circle, line and point is denoted as CLP. The problems of constructing circles tangent to three points (PPP) and to three lines (LLL) were solved first by Euclid in his Elements.

Points and lines may be viewed as special cases of circles; a point can be considered as a circle of infinitely small radius, and a line may be thought of an infinitely large circle whose center is also at infinity. From this perspective, the general Apollonius problem is that of constructing circles tangent to three given circles. The nine other cases involving points and lines may be viewed as limiting cases of the general problem.[1] These limiting cases often have fewer solutions than the general problem; for example, the replacement of a given circle by a given point halves the number of solutions.

Table 1: Ten Types of Apollonius' Problem
Index Code Given Elements Number of solutions
(in general)		 Example
(solution=pink; givens=red/green/blue)
1 PPP three points 1
2 LPP one line and two points 2
3 LLP two lines and a point 2
4 CPP one circle and two points 2
5 LLL three lines 4
6 CLP one circle, one line, and a point 4
7 CCP two circles and a point 4
8 CLL one circle and two lines 8
9 CCL two circles and a line 8
10 CCC three circles (the classic problem) 8
Figure 4: An Apollonius problem with no solutions.
Figure 4: An Apollonius problem with no solutions.

[edit] Number of solutions

The general number of solutions for each of the ten types of Apollonius' problem is given in Table 1 above. However, special arrangements of the given elements may change the number of solutions. For illustration, Apollonius' problem has no solution if one circle separates the two (Figure 4); to touch both the red and blue circles, the solution circle would have to cross the green circle; but that it cannot do, if it is to touch the green circle tangentially. Conversely, if three given circles are all tangent at the same point, then any circle tangent at the same point is a solution; thus, such Apollonius problems have an infinite number of solutions. If any of the given circles are identical, there is likewise an infinity of solutions. If only two given circles are identical, there are only two distinct given circles; the centers of the solution circles form a hyperbola, as used in one solution to Apollonius' problem.

An exhaustive enumeration of the number of solutions for all possible configurations of three given circles, points or lines was first undertaken by Muirhead in 1896,[2] although earlier work had been done by Stoll[3] and Study.[4] However, Muirhead's work was incomplete; it was extended in 1974[5] and a definitive enumeration was published in 1983.[6] Although solutions to Apollonius' problem generally occur in pairs related by inversion, an odd number of solutions is possible in some cases, e.g., the single solution for PPP, or when one or three of the given circles are themselves solutions. (An example of the latter is given in the section on Descartes' theorem.) However, there are no Apollonius problems with seven solutions.[3] Alternative solutions based on Lie geometry have been developed and used for higher dimensions.[7][8]

[edit] Mutually tangent given circles: Soddy's circles and Descartes' theorem

Figure 5: If the given three circles (shown in black) are mutually tangent, Apollonius' problem has only two other solutions (shown in red). The inner solution circle "gets three kisses from without"; the outer circle is "thrice kissed internally". Each circle is labeled with its curvature.
Figure 5: If the given three circles (shown in black) are mutually tangent, Apollonius' problem has only two other solutions (shown in red). The inner solution circle "gets three kisses from without"; the outer circle is "thrice kissed internally". Each circle is labeled with its curvature.

If the three given circles are mutually tangent, Apollonius' problem has five solutions. Three solutions are the given circles themselves, since each is tangent to itself and to the other two given circles. The remaining two solutions (shown in red in Figure 5) correspond to the inscribed and circumscribed circles, and are called Soddy's circles. Either Soddy circle, when taken together with the three given circles, produces a set of four circles that are mutually tangent at six points. The radii of these four circles are related by an equation known as Descartes' theorem. In a 1643 letter to Princess Elizabeth of Bohemia,[9] René Descartes showed that

\left( k_{1}+k_{2}+k_{3}+k_{s} \right)^{2} = 2\, \left( k_{1}^{2} + k_{2}^{2} + k_{3}^{2} + k_{s}^{2} \right)

where ks = 1/rs and rs are the curvature and radius of the solution circle, respectively, and similarly for the curvatures k1-3 and radii r1-3 of the three given circles. For every set of four mutually tangent circles, there is a second set of four mutually tangent circles that are tangent at the same six points.[10][11]

Descartes' theorem was rediscovered independently in 1826 by Jakob Steiner,[12] in 1842 by an amateur mathematician, Philip Beecroft,[10][11] and again in 1936 by Nobel laureate Frederick Soddy.[13] Soddy published his findings in the scientific journal Nature as a poem, The Kiss Precise, of which the first two stanzas are reproduced below. The first stanza describes Soddy's circles, whereas the second stanza gives Descartes' theorem. In Soddy's poem, two circles are said to "kiss" if they are tangent, whereas the term "bend" refers to the curvature k of the circle.

For pairs of lips to kiss maybe
Involves no trigonometry.
'Tis not so when four circles kiss
Each one the other three.
To bring this off the four must be
As three in one or one in three.
If one in three, beyond a doubt
Each gets three kisses from without.
If three in one, then is that one
Thrice kissed internally.
Four circles to the kissing come.
The smaller are the benter.
The bend is just the inverse of
The distance from the center.
Though their intrigue left Euclid dumb
There's now no need for rule of thumb.
Since zero bend's a dead straight line
And concave bends have minus sign,
The sum of the squares of all four bends
Is half the square of their sum.

Sundry extensions of Descartes' theorem have been derived by Daniel Pedoe.[14]

[edit] Solution methods

Apollonius' problem has had a rich history and numerous solution methods have been developed over the centuries.[15][16][17] The original geometrical method used by Apollonius of Perga has been lost, but reconstructions have been offered by François Viète and others, based on the clues in the description by Pappus.[18] After its re-introduction in the 16th century, Adriaan van Roomen solved for the centers of the solution circles as the intersection of two hyperbolae, a method that was refined by Isaac Newton in his Principia; Newton's solution was re-discovered independently by John Casey in 1881. Apollonius' problem has also been solved algebraically; this approach was pioneered by René Descartes and Princess Elisabeth of Bohemia, and subsequently refined by Leonhard Euler, Nicolas Fuss, Carl Friedrich Gauss, Lazare Carnot, Augustin Louis Cauchy and Jean-Victor Poncelet. Direct geometrical solutions have been published by Joseph Diaz Gergonne and by Poncelet, with alternative formulations being offered by Maurice Fouché and others. Methods using circle inversion were pioneered by Julius Petersen and are sometimes considered to be the most intuitive approach for lay-people; in some versions, the solution circle is transformed into a line, or confined between two lines or two concentric circles. These various approaches are outlined below.

Figure 6: The difference in center-to-center distances d1 and d2 between the solution circle (black) and two given circles (red and blue) does not depend on the radius rs of the solution circle; it depends only on the difference of the given radii, r2−r1.
Figure 6: The difference in center-to-center distances d1 and d2 between the solution circle (black) and two given circles (red and blue) does not depend on the radius rs of the solution circle; it depends only on the difference of the given radii, r2−r1.

[edit] Intersecting hyperbolas

The first solution of Apollonius' problem in modern times was that of Adriaan van Roomen and is based on the intersection of two hyperbolas.[19][20] Let the given circles be denoted as C1, C2 and C3. Van Roomen solved the general problem by solving a simpler problem, that of finding the circles that are tangent to two given circles, such as C1 and C2. He noted that the center of a circle tangent to both given circles must lie on a hyperbola whose foci are the centers of the given circles. To understand this, let the radii of the solution circle and the two given circles be denoted as rs, r1 and r2, respectively (Figure 6). The distance d1 between the centers of the solution circle and C1 is either rs + r1 or rs − r1, depending on whether these circles are chosen to be externally or internally tangent, respectively. Similarly, the distance d2 between the centers of the solution circle and C2 is either rs + r2 or rs − r2, again depending on their chosen tangency. Thus, the difference d1 − d2 between these distances is always a constant that is independent of rs. This property, of having a fixed difference between the distances to the foci, characterizes hyperbolas, so the possible centers of the solution circle lie on a hyperbola. A second hyperbola can be drawn for the pair of given circles C2 and C3, where the internal or external tangency of the solution and C2 should be chosen consistently with that of the first hyperbola. An intersection of these two hyperbolas (if any) gives the center of a solution circle that has the chosen internal and external tangencies to the three given circles. The full set of solutions to Apollonius' problem can be found by considering all possible combinations of internal and external tangency of the solution circle to the three given circles.

Isaac Newton refined van Roomen's solution, so that the solution-circle centers were located at the intersections of a line with a circle.[21][22] Newton formulates Apollonius' problem as a problem in trilateration: to locate a point Z from three given points A, B and C, such that the differences in distances from Z to the three given points have known values. These four points correspond to the center of the solution circle (Z) and the centers of the three given circles (A, B and C). Instead of solving for the two hyperbolas, Newton constructs their directrix lines instead. For any hyperbola, the ratio of distances from a point Z to a focus A and to the directrix is a fixed constant called the eccentricity. The two directrices intersect at a point T, and from their two known distance ratios, Newton constructs a line passing through T on which Z must lie. However, the ratio of distances TZ/TA is also known; hence, by Apollonius' definition of the circle, Z also lies on a known circle. Thus, the solutions to Apollonius' problem are the intersections of a line with a circle. Newton's solution was re-discovered independently by John Casey in 1881.[23]

[edit] Viète's reconstruction

Figure 7: The tangency of a set of circles is preserved if their radii are changed by the same amount. Two internally tangent circles (such as 3B and C2) must shrink or swell together to preserve their tangency, whereas two externally tangent circles (such as 3B and C1) must do the opposite: if one circle shrinks in radius by Δr, the other must swell by the same amount.
Figure 7: The tangency of a set of circles is preserved if their radii are changed by the same amount. Two internally tangent circles (such as 3B and C2) must shrink or swell together to preserve their tangency, whereas two externally tangent circles (such as 3B and C1) must do the opposite: if one circle shrinks in radius by Δr, the other must swell by the same amount.

A prized property in classical Euclidean geometry was solution of a problem using compass and straightedge constructions.[24] Many relatively basic constructions are not possible using only these tools, such as angle trisection and doubling the cube. In general, a geometrical construction can be done with compass and straightedge if and only if the ratios of its distances involve only square roots and not cube roots, higher roots, or transcendental numbers such as π. Since van Roomen's solution uses the intersection of two hyperbolas, which are quadratic curves, his solution may involve a fourth root and thus could violate the straightedge-and-compass requirement.

Van Roomen's friend François Viète, who had urged van Roomen to work on the problem in the first place, noted[25] that many problems that are impossible to solve using a ruler and straightedge are possible by intersecting conic sections, such as the hyperbolas that van Roomen used in his solution. For instance, doubling the cube cannot be done using a straightedge and compass, but Menaechmus showed that it is possible to do using the intersections of two parabolas. Viète produced a compass and straightedge construction for all ten special cases of Apollonius' problem,[25] by working upwards from the simplest limiting cases. According to the 4th-century report of Pappus of Alexandria, Apollonius' own book on this problem—entitled Επαφαι ("Tangencies", Latin: De tactionibus, De contactibus)—follows a similar approach. Hence, Viète's solution is considered to be a plausible reconstruction of Apollonius' solution, although another reconstruction has been published, independently by three different authors.[26] Prior to Viète's solution, the mathematician Regiomontanus doubted whether Apollonius' problem could be solved by straightedge and compass.[27]

Viète begins by solving the PPP case (three points) following the method of Euclid in his Elements. From this, Viète derives a lemma corresponding to the power of a point theorem, which he uses to solve the LPP case. He then solves the LLL case using the angle bisectors, similar to Euclid. He then derives a lemma for constructing the line perpendicular to an angle bisector that passes through a point, which he uses to solve the LLP problem. This accounts for the first four cases of Apollonius' problem, those that do not involve circles.

To solve the remaining problems, Viète exploits the fact that the given circles and the solution circle may be re-sized in tandem while preserving their tangencies (Figure 7). If the solution-circle radius is changed by an amount Δr, the radius of its internally tangent given circles must be likewise changed by Δr, whereas the radius of its externally tangent given circles must be changed by −Δr. Thus, as the solution circle swells, the internally tangent given circles must swell in tandem, whereas the externally tangent given circles must shrink, to maintain their tangencies.

Viète uses this approach to shrink at least one of the given circles to a point, thus reducing the problem to a simpler limiting case. He first solves the CLL case by shrinking the circle into a point, rendering it a LLP case. He then solves the CLP case using three lemmas. Viète then solves the CCL case by shrinking one circle to a point, rendering it a CLP case. Next he solves the CPP and CCP cases, the latter case by two lemmas. Finally, Viète solves the general CCC case by resizing one circle to a point, rendering it a CCP case.

Figure 8: The black circle is one solution of Apollonius' problem, being internally tangent to the green given circle C2 and externally tangent to the red and blue circles, C1 and C3. The signs for this solution are "− + −"; the solution encloses C2 and excludes C1 and C3.
Figure 8: The black circle is one solution of Apollonius' problem, being internally tangent to the green given circle C2 and externally tangent to the red and blue circles, C1 and C3. The signs for this solution are "− + −"; the solution encloses C2 and excludes C1 and C3.

[edit] Algebraic solutions

Apollonius' problem can be framed as a system of three coupled quadratic equations in three variables xs, ys and rs[28]

\left( x_{s} - x_{1} \right)^{2} + \left( y_{s} - y_{1} \right)^{2} = \left( r_{s} - s_{1} r_{1} \right)^{2}
\left( x_{s} - x_{2} \right)^{2} + \left( y_{s} - y_{2} \right)^{2} = \left( r_{s} - s_{2} r_{2} \right)^{2}
\left( x_{s} - x_{3} \right)^{2} + \left( y_{s} - y_{3} \right)^{2} = \left( r_{s} - s_{3} r_{3} \right)^{2}

Here, r1–3 represent the radii of the three given circles, whereas (x1–3, y1–3) represent their center positions in Cartesian coordinates. The three signs s1–3 on the right-hand side may equal ±1, giving eight possible sets of equations (2 × 2 × 2 = 8), each one corresponding to one of the eight types of solution circles. If a particular sign (say, s1) is positive, the solution circle will be internally tangent to the corresponding given circle (in this case, C1); the solution circle encloses the given circle. Conversely, if a particular sign is negative, the solution is externally tangent to the corresponding given circle; in other words, the solution circle excludes the given circle (Figure 8).

Whatever the choice of signs, this system of three equations may be solved by the method of resultants. Multiplying out the three equations and canceling the common terms yields formulae for the coordinates xs and ys

xs = M + Nrs
ys = P + Qrs

where M, N, P and Q are known functions of the given circles and the choice of signs. Substitution of these formulae into one of the initial three equations gives a quadratic equation in rs, which can be solved by the quadratic formula. Substitution of the numerical value of rs into the linear formulae yields the corresponding values of xs and ys.

Algebraic solutions to Apollonius' problem were pioneered by René Descartes and Princess Elisabeth of Bohemia, although their solutions were rather complex.[15] The equations were subsequently refined by Leonhard Euler,[29] Nicolas Fuss,[15] Carl Friedrich Gauss,[30] Lazare Carnot,[31] Augustin Louis Cauchy[32] and Jean-Victor Poncelet.[33]

The signs s1–3 on the right-hand sides of the equations may be chosen in eight possible ways, and each choice of signs gives up to two solutions, since the equation for rs is quadratic. This might suggest (incorrectly) that there are up to sixteen solutions of Apollonius' problem. However, due to a symmetry of the equations, if rs is a solution, so is –rs; these represent the same circle, but with opposite signs si. Therefore, Apollonius' problem has at most eight independent solutions, consistent with Bézout's theorem. These eight solutions are depicted in Figure 1. One way to simplify the equations to make this more obvious is to consider only the four sets of signs s1–3 that are positive, i.e., s1 + s2 + s3 > 0. In that case, the two solutions of the quadratic equation for rs correspond to the two conjugate solutions related by inversion (Figure 2).

The two roots of any quadratic equation may be of three possible types: two different real numbers, two identical real numbers (i.e., a degenerate double root), or a pair of complex conjugate roots. The first case corresponds to the usual situation; each pair of roots corresponds to a pair of mutually inversive (conjugate) solutions, as described above. In the second case, both roots are identical, corresponding to a solution circle that transforms into itself under inversion. In this case, one of the given circles is itself a solution to the Apollonius problem, and the number of distinct solutions is reduced by one. In the third case, there is no solution for Apollonius' problem, since the solution circle cannot have an imaginary radius; therefore, the number of solutions is reduced by two. Interestingly, Apollonius' problem cannot have 7 solutions, although it may have any other number of solutions between zero and eight.

[edit] Lie sphere geometry

The same algebraic equations can be derived in the context of Lie sphere geometry.[7] That geometry represents circles, lines and points in a unified way, as a five-dimensional vector X = (v, cx, cy, w, sr), where c = (cx, cy) is the center of the circle, and r is its (non-negative) radius. If r is not zero, the sign s may be positive or negative; for visualization, s is imagined as an "orientation" of the circle, with counterclockwise circles having a positive s and clockwise circles having a negative s. The parameter w is zero for a straight line, and one otherwise.

In this five-dimensional world, there is an unusual product similar to the dot product:

\left( X_{1} | X_{2} \right) \equiv v_{1} w_{2} + v_{2} w_{1} + \mathbf{c} \cdot \mathbf{c} - s_{1} s_{2} r_{1} r_{2}.

The Lie quadric is defined as those vectors whose product with themselves (their norm) is zero, (X|X) = 0. Let X1 and X2 be two vectors belonging to this quadric; the norm of their difference equals

\left( X_{1} - X_{2} | X_{1} - X_{2} \right) = 2 \left( v_{1} - v_{2} \right) \left( w_{1} - w_{2} \right) + \left( \mathbf{c}_{1} - \mathbf{c}_{2} \right) \cdot \left( \mathbf{c}_{1} - \mathbf{c}_{2} \right) - \left( s_{1} r_{1} - s_{2} r_{2} \right)^{2}.

The product distributes over addition and subtraction:

\left( X_{1} - X_{2} | X_{1} - X_{2} \right) = \left( X_{1} | X_{1} \right) - 2 \left( X_{1} | X_{2} \right) + \left( X_{2} | X_{2} \right).

Since (X1|X1) = (X2|X2) = 0 (both belong to the Lie quadric) and since w1 = w2 = 1 for circles, the product of any two such vectors on the quadric equals

- 2 \left( X_{1} | X_{2} \right) = \left| \mathbf{c}_{1} - \mathbf{c}_{2} \right|^{2} - \left( s_{1} r_{1} - s_{2} r_{2} \right)^{2}.

This formula shows that if two quadric vectors X1 and X2 are orthogonal (perpendicular) to one another—mathematically, if (X1|X2) = 0—then their corresponding circles are tangent. For if the two signs s1 and s2 are the same (i.e. the circles have the same "orientation"), the circles are internally tangent; the distance between their centers equals the difference in the radii

\left| \mathbf{c}_{1} - \mathbf{c}_{2} \right|^{2} = \left( r_{1} - r_{2} \right)^{2}.

Conversely, if the two signs s1 and s2 are different (i.e. the circles have opposite "orientations"), the circles are externally tangent; the distance between their centers equals the sum of the radii

\left| \mathbf{c}_{1} - \mathbf{c}_{2} \right|^{2} = \left( r_{1} + r_{2} \right)^{2}.

Therefore, Apollonius' problem can be re-stated in Lie geometry as a problem of finding perpendicular vectors on the Lie quadric; specifically, the goal is to identify solution vectors Xsol that belong to the Lie quadric and are also orthogonal (perpendicular) to the vectors X1–3 corresponding to the given circles.

\left( X_{\mathrm{sol}} | X_{\mathrm{sol}} \right) = \left( X_{\mathrm{sol}} | X_{1} \right) = \left( X_{\mathrm{sol}} | X_{2} \right) = \left( X_{\mathrm{sol}} | X_{3} \right) = 0

The advantage of this re-statement is that one can exploit theorems from linear algebra on the maximum number of linearly independent, simultaneously perpendicular vectors. This gives another way to calculate the maximum number of solutions and extend the theorem into higher dimensional spaces.[7][8]

[edit] Inversive methods

The basic strategy of inversive methods is to transform a given Apollonius problem into another Apollonius problem that is simpler to solve; the solutions to the original problem are found from the solutions of the transformed problem by undoing the transformation. Candidate transformations must one Apollonius problem into another; therefore, it must transform the given points, circles and lines to other points, circles and lines, and no other shapes. Circle inversion has this property and allows the center and radius of the inversion circle to be chosen judiciously. Other candidates include the Euclidean plane isometries; however, they do not simplify the problem, since they merely shift, rotate, and mirror the original problem.

The application of circle inversion to Apollonius' problem was pioneered by Julius Petersen.[34] Circle inversions correspond to a subset of Möbius transformations on the Riemann sphere. The planar Apollonius problem can be transferred to the sphere by a inverse stereographic projection; hence, solutions of the planar Apollonius problem also pertain to its counterpart on the sphere. Other inversive solutions to the planar problem are possible besides the common ones described below.[35]

[edit] Inversion to an annulus

Figure 9: The solutions (pink circle) of the first family lie between the concentric given circles (red and blue). There are four solutions, depending on the sign of θ and whether the solution circle is internally or externally tangent to the non-concentric given circle (green).
Figure 9: The solutions (pink circle) of the first family lie between the concentric given circles (red and blue). There are four solutions, depending on the sign of θ and whether the solution circle is internally or externally tangent to the non-concentric given circle (green).

If two of the three given circles are disjoint, a center of inversion can be chosen so that those two given circles become concentric.[10] Under this inversion, the solution circles must fall within the annulus between the two concentric circles. Therefore, they belong to two one-parameter families. In the first family (Figure 9), the solutions do not enclose the inner concentric circle, but rather revolve like ball bearings in the annulus. In the second family (Figure 10), the solution circles enclose the inner concentric circle. There are generally four solutions for each family, yielding eight possible solutions, consistent with the algebraic solution.

Figure 10: The solutions (pink) of the second family touch the inner and outer concentric circles, but enclose the inner circle. There are again four solutions.
Figure 10: The solutions (pink) of the second family touch the inner and outer concentric circles, but enclose the inner circle. There are again four solutions.

When two of the given circles are concentric, Apollonius' problem can be solved easily using a method of Gauss.[30] The radii of the three given circles are known, as is the distance dnon from the common concentric center to the non-concentric circle (Figure 9). The solution circle can be determined from its radius rs, the angle θ, and the distances ds and dT from its center to the common concentric center and the center of the non-concentric circle, respectively. The radius and distance ds are known (Figure 9), and the distance dT = rs ± rnon, depending on whether the solution circle is internally or externally tangent to the non-concentric circle. Therefore, by the law of cosines,

\cos \theta = \frac{d_{\mathrm{s}}^{2} + d_{\mathrm{non}}^{2} - d_{\mathrm{T}}^{2}}{2 d_{\mathrm{s}} d_{\mathrm{non}}} \equiv C_{\pm}.

Here, a new constant C has been defined for brevity, with the subscript indicating whether the solution is externally or internally tangent. A simple trigonometric rearrangement yields the four solutions

\theta = \pm 2 \ \mathrm{atan}\left( \sqrt{\frac{1 - C}{1 + C}} \right).

This formula represents four solutions, corresponding to the two choices of the sign of θ, and the two choices for C. The remaining four solutions can be obtained by the same method, using the substitutions for rs and ds indicated in Figure 10. Thus, all eight solutions of the general Apollonius problem can be found by this method.

Any initial two disjoint given circles can be rendered concentric as follows. The radical axis of the two given circles is constructed; choosing two arbitrary points P and Q on this radical axis, two circles can be constructed that are centered on P and Q and that intersect the two given circles orthogonally. These two constructed circles intersect each other in two points. Inversion in one such intersection point F renders the constructed circles into straight lines emanating from F and the two given circles into concentric circles, with the third given circle becoming another circle (in general). This follows because the system of circles is equivalent to a set of Apollonian circles, forming a bipolar coordinate system.

[edit] Resizing and inversion

The usefulness of inversion can be increased significantly by resizing. As noted in Viète's reconstruction, the three given circles and the solution circle can be resized in tandem while preserving their tangencies. Thus, the initial Apollonius problem is transformed into another problem that may be easier to solve. For example, the four circles can be resized so that one given circle is shrunk to a point; alternatively, two given circles can often be resized so that they are tangent to one another. Thirdly, given circles that intersect can be resized so that they become non-intersecting, after which the method for inverting to an annulus can be applied. In all such cases, the solution of the original Apollonius problem is obtained from the solution of the transformed problem by undoing the resizing and inversion.

Shrinking one given circle to a point

In the first approach, the given circles are shrunk or swelled (appropriately to their tangency) until one given circle is shrunk to a point P.[36] In that case, Apollonius' problem degenerates to the CCP limiting cases, the problem of finding a solution circle tangent to the two remaining given circles that passes through the point P. Inversion in a circle centered on P transforms the two given circles into new circles, and the solution circle into a line. Therefore, the transformed solution is a line that is tangent to the two transformed given circles. There are four such solution lines, which may be constructed from the external and internal homothetic centers of the two circles. Re-inversion in P and undoing the resizing transforms such a solution line into the desired solution circle of the original Apollonius problem. All eight general solutions can be obtained by shrinking and swelling the circles according to the differing internal and external tangencies of each solution; however, different given circles may be shrunk to a point for different solutions.

Resizing two given circles to tangency

In the second approach, the radii of the given circles are modified appropriately by an amount Δr so that two of them are tangential (touching).[37] Their point of tangency is chosen as the center of inversion in a circle that intersects each of the two touching circles in two places. Upon inversion, the touching circles become two parallel lines: Their only point of intersection is sent to infinity under inversion, so they cannot meet. The same inversion transforms the third circle into another circle. The solution of the inverted problem must either be (1) a straight line parallel to the two given parallel lines and tangent to the transformed third given circle; or (2) a circle of constant radius that is tangent to the two given parallel lines and the transformed given circle. Re-inversion and adjusting the radii of all circles by Δr produces a solution circle tangent to the original three circles.

Figure 11: The two tangent lines drawn from the two tangent points of a given circle intersect on the radical axis R of the two solution circles. The three points of intersection on R are the poles of L1–3 in their respective given circles C1–3.
Figure 11: The two tangent lines drawn from the two tangent points of a given circle intersect on the radical axis R of the two solution circles. The three points of intersection on R are the poles of L1–3 in their respective given circles C1–3.

[edit] Gergonne's solution

The solution to Apollonius' problem published by Joseph Diaz Gergonne in 1814[38] is widely considered to be the most elegant.[36] Gergonne's approach is to consider the solution circles in pairs. Let a pair of solution circles be denoted as CA and CB (the pink and black circles in Figure 2), and let their tangent points with the three given circles be denoted as A1, A2, A3, and B1, B2, B3, respectively. Gergonne's solution aims to locate these six points, and thus solve for the two solution circles.

Gergonne's insight was that if a line L1 could be constructed such that A1 and B1 were guaranteed to fall on it, those two points could be identified as the intersection points of L1 with the given circle C1 (Figure 2). The remaining four tangent points would be located similarly, by finding lines L2 and L3 that contained A2 and B2, and A3 and B3, respectively. To construct a line such as L1, two points must be identified that lie on it; but these points need not be the tangent points. Gergonne was able to identify two other points for each of the three lines. One of the two points has already been identified: the radical center G lies on all three lines (Figure 2).

To locate a second point on the lines L1–3, Gergonne noted a reciprocal relationship between the radical axis R of the solution circles, CA and CB, and the lines L1–3. To understand this reciprocal relationship, consider the two tangent lines to the circle C1 drawn at its tangent points A1 and B1 with the solution circles; the intersection of these tangent lines is the pole point of L1 in C1. Since the distances from that pole point to the tangent points A1 and B1 are equal, this pole point must also lie on the radical axis R of the solution circles, by definition (Figure 11). The relationship between pole points and their polar lines is reciprocal; if the pole of L1 in C1 lies on R, the pole of R in C1 must conversely lie on L1. Thus, if we can construct R, we can find its pole P1 in C1, giving the needed second point on L1 (Figure 12).

Figure 12: The poles P1–3 (shown in orange) of R in the three given circles C1–3 lie on the lines L1–3. Together with the radical center G, these poles define the lines L1–3.
Figure 12: The poles P1–3 (shown in orange) of R in the three given circles C1–3 lie on the lines L1–3. Together with the radical center G, these poles define the lines L1–3.

Gergone found the radical axis R of the unknown solution circles as noted. Any pair of circles has two centers of similarity; these two points are the two possible intersections of two tangent lines to the two circles. Therefore, the three given circles have six centers of similarity, two for each distinct pair of given circles. Remarkably, these six points lie on four lines, three points on each line; moreover, each line corresponds to the radical axis of a potential pair of solution circles. To show this, Gergonne considered lines through corresponding points of tangency on two of the given circles, e.g., the line defined by A1/A2 and the line defined by B1/B2. Let X3 be a center of similitude for the two circles C1 and C2; then, A1/A2 and B1/B2 are pairs of antihomologous points, and their lines intersect at X3. It follows, therefore, that the products of distances are equal

\overline{X_{3}A_{1}} \cdot \overline{X_{3}A_{2}} = \overline{X_{3}B_{1}} \cdot \overline{X_{3}B_{2}}

which implies that X3 lies on the radical axis of the two solution circles. The same argument can be applied to the other pairs of circles, so that three centers of similitude for the given three circles must lie on the radical axes of pairs of solution circles.

In summary, the desired line L1 is defined by two points: the radical center G of the three given circles and the pole in C1 of one of the four lines connecting the homothetic centers. Finding the same pole in C2 and C3 gives L2 and L3, respectively; thus, all six points can be located, from which one pair of solution circles can be found. Repeating this procedure for the remaining three homothetic-center lines yields six more solutions, giving eight solutions in all. However, if a line Lk does not intersect its circle Ck for some k, there is no pair of solutions for that homothetic-center line.

[edit] Generalizations

Apollonius' problem can be extended to construct all the circles that intersect three given circles at a precise angle θ, or at three specified crossing angles θ1, θ2 and θ3;[12] the ordinary Apollonius' problem corresponds to a special case in which the crossing angle is zero for all three given circles. Another generalization is the dual of the first extension, namely, to construct circles with three specified tangential distances from the three given circles.[7]

Apollonius' problem can be extended from the plane to the sphere and other quadratic surfaces. For the sphere, the problem is to construct all the circles (the boundaries of spherical caps) that are tangent to three given circles on the sphere.[38][39][40] This spherical problem can be rendered into a corresponding planar problem using stereographic projection. Once the solutions to the planar problem have been constructed, the corresponding solutions to the spherical problem can be determined by inverting the stereographic projection. Even more generally, one can consider the problem of four tangent curves that result from the intersections of an arbitrary quadratic surface and four planes, a problem first considered by Charles Dupin.[15]

Figure 13: A symmetrical Apollonian gasket, also called the Leibniz packing, after its inventor Gottfried Leibniz.
Figure 13: A symmetrical Apollonian gasket, also called the Leibniz packing, after its inventor Gottfried Leibniz.

By solving Apollonius' problem repeatedly to find the inscribed circle, the interstices between mutually tangential circles can be filled arbitrarily finely, forming an Apollonian gasket, also known as a Leibniz packing or an Apollonian packing.[41] This gasket is a fractal, being self-similar and having a dimension d that is roughly 1.3,[42], which is higher than that of a regular (or rectifiable) curve (d=1) but less than that of a plane (d=2). The Apollonian gasket was first described by Gottfried Leibniz in the 17th century, and is a curved precursor of the 20th-century Sierpiński triangle.[43] The Apollonian gasket also has deep connections to other fields of mathematics; for example, it is the limit set of Kleinian groups.[44]

The configuration of a circle tangent to four circles in the plane has special properties, which have been elucidated by Larmor (1891)[45] and Lachlan (1893).[46] Such a configuration is also the basis for Casey's theorem,[23] itself a generalization of Ptolemy's theorem.[36]

The extension of Apollonius' problem to three dimensions, namely, the problem of finding a fifth sphere that is tangent to four given spheres, can be solved by analogous methods.[15] For example, the given and solution spheres can be resized so that one given sphere is shrunk to point while maintaining tangency.[37] Inversion in this point reduces Apollonius' problem to finding a plane that is tangent to three given spheres. There are in general eight such planes, which become the solutions to the original problem by reversing the inversion and the resizing. This problem was first considered by Pierre de Fermat,[47] and many alternative solution methods have been developed over the centuries.[48]

Apollonius' problem can even be extended to d dimensions, to construct the hyperspheres tangent to a given set of d + 1 hyperspheres. Following the publication of Frederick Soddy's re-derivation of the Descartes theorem in 1936, several people solved (independently) the mutually tangent case corresponding to Soddy's circles in d dimensions.[49]

[edit] Applications

Apollonius' problem is a pure problem in Euclidean plane geometry, and Apollonius himself was famously indifferent to the practical applications of his work.[50]

“ [This] subject is one of those that seems worthy of study for their own sake. ”

—Apollonius of Perga, from the preface of Book V of his Conics

Nonetheless, Apollonius' problem and its generalizations to higher dimensions have found some applications. Its principal application lies in trilateration, which is the problem of determining a receiver position from the differences in distances to at least three transmitters or conversely, from the differences in distances between a transmitter and at least three receivers. This problem is equivalent to Apollonius' problem, as expressed by Isaac Newton: to determine a point in a place (resp. in space) from its distances to three (resp. four) known points. Solutions to Apollonius' problem were used for trilateration in World War I to determine the positions of artillery pieces from the time required to hear the sound of the gun firing at three different positions.[15] Trilateration is a key component of modern navigational systems such as GPS[51] and the earlier LORAN and Decca Navigator System;[52] sometimes the solution is used without mentioning Apollonius' problem itself.[53] Trilateration is also used to determine the position of calling animals (such as birds and whales), although Apollonius' problem does not pertain if the speed of sound varies with direction (i.e., is not isotropic).[54]

Apollonius' problem has other applications as well. In Book 1, Proposition 21 in his Principia, Isaac Newton used his solution of Apollonius' problem to construct an orbit in celestial mechanics from the center of attraction and observations of tangent lines to the orbit corresponding to instantaneous velocity.[15] The special case of the problem of Apollonius when all three circles are tangent is used in the Hardy-Littlewood circle method of analytic number theory to construct H. Rademacher's contour for complex integration, given by the boundaries of an infinite set of Ford circles each of which touches several others.[55] Finally, Apollonius' problem has been applied to some types of packing problems, which arise in disparate fields such as the error-correcting codes used on DVDs and the design of pharmaceuticals that bind in a particular enzyme of a pathogenic bacterium.[56]

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[edit] Further reading

  • Boyd DW (1973). "The osculatory packing of a three-dimensional sphere". Canadian J. Math. 25: 303–322. 
  • Callandreau É (1949). Célèbres Problêmes mathématiques. Paris: Unknown, pp. 219–226. 
  • Coolidge JL (1916). A Treatise on the Circle and the Sphere. Oxford: Clarendon Press, 167–172.  Calls Apollonius' problem "the most famous of all" geometry problems.
  • Dörrie H (1965). "The Tangency Problem of Apollonius", 100 Great Problems of Elementary Mathematics: Their History and Solutions. New York: Dover, pp. 154-160 (§32). 
  • Gisch D, Ribando JM (2004). "Apollonius’ Problem: A Study of Solutions and Their Connections". American Journal of Undergraduate Research 3: 15–25. 
  • Oldknow A (1996). "The Euler-Gergonne-Soddy Triangle of a Triangle". The American Mathematical Monthly 103: 319–329. doi:10.2307/2975188. 
  • Simon M (1906). Über die Entwicklung der Elementargeometrie im XIX. Jahrhundert. Berlin: Teubner, pp. 97–105. 
  • Ver Eecke P (1933). Pappus d'Alexandrie: La Collection Mathématique. Brussels: Unknown, pp. lxvi–lxix, 483–485. 
  • Wells D (1991). The Penguin Dictionary of Curious and Interesting Geometry. New York: Penguin Books, pp. 3–5. ISBN 0-14-011813-6. 

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