Spatial Localization Problem and the of Apollonius. Joseph Cox1, Michael B. Partensky2

1 Stream Consulting, Rialto Tower, Melbourne, Victoria 3000 Australia email: [email protected] 2Brandeis University, Rabb School of Continuing Studies and Dept. of Chemistry, Waltham, MA, USA email:[email protected]

The Circle of Apollonius is named after the ancient geometrician . This beautiful geometric construct can be helpful when solving some general problems of mathematical physics, optics and electricity. Here we discuss its applications to “source localization” problems, e.g., pinpointing a radioactive source using a set of Geiger counters. The of Apollonius help one to analyze these problems in a transparent and intuitive manner. This discussion can be useful for high school physics and math curriculums.

1. Introduction 2. Apollonius of Perga helps to save

We will discuss a class of problems where the Sam position of an object is determined based on the analysis of some “physical signals” related to its Description of the problem location. First, we pose an entertaining problem Bartholomew the Frog with Precision Hopping that helps trigger the students’ interest in the Ability could hop anywhere in the world with a subject. Analyzing this problem. we introduce the thought and a leap [1]. Publicly, he was a retired , and show that this track and field star. Privately, he used his talent to geomteric insight allows solving the problem in help save the world. You see, Bartholomew had an elegant and transparent way. At the same become a secret agent, a spy - a spook. In fact, time, we demonstrate that the solution of the only two people in the whole world knew who inverse problem of localizing an object based on Bartholomew really was. One was Sam the readings from the detectors, can be nonunique. Elephant and the other was Short Eddy, a This ambiguity is further discussed for a typical fourteen-year-old kid who did not have a whole “source localization” problem, such as lot of normal friends but was superb in math and pinpointing a radioctive source with a set of science. One day an evil villain Hrindar the detectors. It is shown for the planar problem that platypus kidnapped Sam. Bartholomew, as soon the “false source” is the inverse point of the real as he realized Sam was missing, hopped straight one relative to the circle passing through a set of “to Sam the Elephant.” When he got there, he was three detectors. This observation provides an shocked to see Sam chained to a ship anchored in insight leading to an unambiguous pinpointing of the ocean. As soon as Sam saw Bartholomew the source.

1 he knew he was going to be okay. He quickly |SA|:|SB|:|SC|=1:2:3 (the square roots of 1/36, 1/9 and quietly whispered, “Bartholomew, I don’t and 1/4). Eddy always tried to break a complex exactly know where we are, but it is somewhere problem into smaller parts. Therefore, he decided near Landport, Maine.” It was dark out and to on the two lighthouses, A and B, first. Bartholomew could hardly see anything but the Apparently, S is one of all possible points P two blurred outline of the city on his left, and the times more distant from B than from A: lights from three lighthouses. Two of them, say A ||/||1/2PA PB = . This observation immediately and B, were on land, while the third one, C, was reminded Eddy of something that had been positioned on the large island. Using the discussed in his AP class. At that time photometer from his spy tool kit, Bartholomew he was very surprised to learn that in addition to found that their brightnesses were in proportion being the of points P equally distant from a 36:9: . He hopped to Eddy and told him what was 4 center, a circle can also be defined as a locus of up. Eddy immediately googled the map of the points whose distances to two fixed points A and area surrounding Landport that showed three B are in a constant ratio. Eddy opened his lecture lighthouses (see Fig. 1). ABC turned out to be a notes and... There it was! The notes read: “Circle right , with its legs |AB|=1.5 miles and of Apollonius ... is the locus of points P whose |AC|=2 miles. The accompanying description distances to two fixed points A and B are in a asserted that the lanterns on the lighthouses were constant ratioγ : the same. In a few minutes the friends knew the ||/||PA PB = γ (1) location of the boat, and in another half an hour, For convenience, draw the x-axis through the still under cover of the night, a group of points A and B. It is a good exercise in algebra commandos released Sam and captured the and geometry (see the Appendix) to prove that the villain. The question is, how did the friends of this circle is manage to find the position of the boat? ||AB R0 = γ (2) |1|γ 2 − Discussion and solution and its center is at Being the best math and science student in his γ 2 x − x class, Eddy immediately figured out that the ratio x = B A (3) O 2 of the apparent brightnesses could be transformed γ −1 into the ratio of the distances. According to the The examples of the Apollonius circles with the inverse square law, the apparent brightness fixed points A and B corresponding to different (intensity, luminance) of a point light source (a values of γ are shown in Fig. 2. Each of the reasonable approximation when the of Apollonius circles defined by Eq. 1 is the the source are small compared to the distance r inversion circle [3] for the points A and B from it) is proportional to P / r2 , where P is the (in other words, it divides AB harmonically): 2 power of the source. Given that all lanterns have (x A − xO ) ⋅ (xB − xO ) = RO (4) equal power P, the ratio of the distances between This result immediately follows from Eqs. 2 and the boat (“S” for Sam) and the lighthouses is 3. (Apollonius of Perga [261-190 b.c.e.] was

2 known to contemporaries as “The Great soon Big Sam was released. Once again, the Geometer”. Among his other achievements is the knowledge of physics and math turned out to be famous book Conics where he introduced such very handy. commonly used terms as , and [2])”. 3. The question of ambiguity in some

source localization problems Equipped with this information, Eddy was able to draw the Apollonius circle L1 for the points A Our friends have noticed that the solution of their and B, satisfying the condition γ =1/2 (Fig. 3). problem was not unique. The issue was luckily Given |AB|=1.5 and Eq. 2, he found that the radius resolved, however, because the “fictitious” of this circle R1 = 1 mile. Using Eq. 3, he also location happened to be inland. In general, such an ambiguity can cause a problem. Had both the found that xO − x A = −0.5mile which implies intersection points appeared in the ocean, the that the center O of the circle L1 is half a mile to evil villain would have had a 50:50 chance to the south from A. In the same manner Eddy built escape. Thus, it is critical to learn how to deal the Apollonius circle L2 for the points A and C, with this ambiguity in order to pinpoint the real corresponding to the ratio γ =|PA|/|PC|=1/3. Its target and to discard false solutions. radius is R2 = 0.75 mile and the center Q is 0.25 We address this issue using a slightly different mile to the West from A. Eddy put both circles setting, quite typical for the localization on the map. Bartholomew was watching him, and problems. In the previous discussion a measuring holding his breath. “I got it!”- he suddenly tool, the photo detector, was positioned right on shouted. “Sam must be located at the point that the object (the boat) while the physical signals belongs simultaneously to both circles, i.e. right used to pinpoint the boat were produced by the in their intersection. Only in this point his lanterns. More commonly, the signal source is the distance to A will be 2 times smaller than the searched object itself, and the detectors are distance to B and at the same time 3 times smaller located in known positions outside the object. than the distance to C ”. “Exactly!”- responded Practical examples are a radioactive source whose Eddy, and he drew two dots, gray and orange. position must be determined using Geiger Now his friend was confused: “If there are two counters, or a light source detected by the light possible points, how are we supposed to know sensors. Assuming that the source and detectors which one is the boat?” “That’s easy”- Eddy are positioned in the same , there are three smiled joyfully- “The gray dot is far inland which unknown parameters in the problem: two leaves us with only one possible location!”. And coordinates, and the intensity of the source P. Eddy drew a large bold “S” right next to the One can suggest that using three detectors should orange dot. Now it was peanuts to discover that be sufficient for finding all the unknowns. The the boat with poor Big Sam was anchored corresponding solution, however, will not be approximately 0.35 miles East and 0.45 miles unique: in addition to the real source, it will North from A. Bartholomew immediately return a false source, similar to the gray dot found delivered this information to the commandos, and

3 This immediately follows from the observation by Eddy and Bartholomew. Let us now that B is the inverse point of A (and vice versa) discuss the of this ambiguity and relative to an Apollonius circle with the fixed possible remedies. Consider a source S of points A and B (see Eq. 4). In other words, power P located at the point ( xS , yS ), and three obtaining B by inverting A in an arbitrary circle isotropic detectors Dk (k =1, 2, 3) placed at the L, automatically turns L into the Apollonius Circle for A and B. points ( xk , yk ) (see Fig. 4). The intensities Ik sensed by the detectors are related to the source Fig. 4 shows a circle L passing through the parameters through the inverse square law, three detectors. Inverting the source S in L leading to the system of three algebraic produces the point S ' . Its distance from the equations: center of the circle O follows from Eq. 4:

2 P / d s,k = Ik , k = 1,2,3. (5) 2 xRxS'= O /(6)S

2 2 Here d s,k = (xS − xk ) + ( yS − yk ) is the The corresponding parameter γ is obtained by distance between the k-th detector and the applying Eq. 1 to the point P shown in Fig. 3: source. x − R γ = S (7) Finding the source parameters based on the R − xS' observed (e.g. by solving Eqs. 5), is often As explained above, L is the Circle of called the “inverse problem”. To address the question of ambiguity, we chose a more direct Apollonius with the fixed points S and S ' . and intuitive approach, allowing a simple This observation is directly related to the geometric relation between two (due to the non- question of ambiguity. From the definition of uniqueness) solutions of Eqs. 5. the Apollonius Circle, any chosen point on L is

Treating the source as given, we use Eqs. 5 to exactly γ times closer to S ' than it is to the generate the observables Ik (usually, a much real source S. In conjunction with the inverse easier task than resolving the source based on square law it implies that a “false” source of the the observations). Using the Circles of power P'= P /γ 2 placed in S ' would produce Apollonius, we show that another (image or exactly the same intensity of radiation at all the false) source exists that exactly reproduces I k points on the circle L as does the real source S. generated by the real source. Clearly, the Therefore, it is generally impossible to existence of such an image signifies the non- distinguish between the real and the false uniqueness of the inverse problem. Finally, a sources based on the readings from three simple geometric relation between the real and (isotropic) detectors. Apparently, this is also image sources will prompt a remedy for treating true for any number of detectors placed on the the ambiguity and pinpointing the source. To same circle. This is exactly the reason for the proceed, we first notice that for any point A ambiguity (nonuniqueness) of the inverse and any circle L, a second point B exists such problem. Notorious for such ambiguities, the that L is an Apollonius Circle with the fixed inverse problem is often characterized as being points A and B. “ill-posed”.

4 and special analytical methods (e.g., nonlinear Eqs. 5 typically (except for the case where S is regression). Nevertheless, the geometric ideas placed right on L) return two solutions, one for described above can still be useful in these the real and the other for the false source. This applications. ambiguity can be resolved by adding a fourth detector positioned off the circle L. Repeating 5. Appendix the previous analysis for the second triad of With the x-axis passing through A and B (see detectors (e.g., 1, 3 and 4 positioned on the Fig. 3), the coordinates of these points are circle L1, see Fig. 4), we can find a new pair of correspondingly (xA,0) and (xB,0). Let (x,y) solutions: the original source S and its image be the coordinates of a point P satisfying Eq. 2. S '' . Comparing this with the previous result Squaring Eq.1 and expressing |PA| and |PB| allows pinpointing the source S, which is the through the coordinates we find: common solution obtained for the two triades ()x −+=−+xy222γ [()] xxy 22 (A1) of detectors, and filtering out the false AB 2 solutions. Expanding the squares, dividing by 1−γ (the case γ = 1 is discussed separately) and 4. Conclusions performing some simple manipulations, we Using a simple geometric approach based on can derive the following equation: the Circles of Apollonius*, we have shown that 22 2 ()x −+=xyROO (A2) (a) A planar isotropic (with three unknowns) with source localization problem posed for a set of 2 three detectors is typically non-unique. γγ()xA −−xxxBAB RxOO==, (A3) γγ22−−11 (b) The “real” (S) and the “false” ( S ' ) Eq. A2 describes the circle of radius R , with solutions are the mutually inverse points O relative to the circle L through the detectors its center at xO . Eqs. A3 are equivalent to (the Apollonius circle for S and S ' ). Eqs. 2 and 3, which proves the validity of those equations. (c) Placing additional detectors on the same The solution for γ = 1 obtained directly from circle (e.g., in the vertexes of a polygon) does not help pinpoint the real source uniquely. Eq. A1 is the straight to AB and from the points A and B. (d) With a fourth detector placed off the circle ______L, the real source can be found uniquely as a * Some similar geometric ideas also inspired common solution obtained for two different by Apollonius of Perga, are discussed in ref. sets of three detectors chosen out of four. Two [4] in application to GPS. other solutions (see S ' and S '' in Fig. 4) must be rejected. Finally note that our analysis completely ignored the statistical fluctuations (noise) in the source and detectors, which is another important cause of ambiguity. Dealing with the noise usually requires additional detectors

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References

1. J. Cox. "Grobar and the Mind Control Potion" ( Suckerfish Books, 2005). 2. E.W. Weisstein. "Apollonius Circle." From MathWorld –A Wolfram Web Resource. http://mathworld.wolfram.com/ApolloniusCircle.html 3. E.W. Weisstein. “Inversion”, MathWorld –A Wolfram Web Resource http://mathworld.wolfram.com/Inversion.htm

4. J. Hoshen. “The GPS equations and the ”

IEEE Transactions on Aerospace and Electronic Systems. 32, 1116 (1996).

Acknowledgements

We are grateful to Jordan Lee Wagner for helpful and insightful discussion. MBP is thankfull to Arkady Pittel and Sergey Liberman for introducing him to some aspects of the source localization problem, and to Lee Kamentsky, Kevin Green, James Carr and Philip Backman for valuable comments and suggestions.

6 Figures

Fig.1 The map of the Landport area showing three Fig. 2: The Circles of Apollonius (some are truncated) lighthouses marked A, B and C. Other notations are for the points A(-1,0) and B(1,0) corresponding to ratio explained in the text. γ=k (right) and γ=1/k (left), with k taking integer values from 1 (red straight line) through 6 (bright green).

Pinpointing the source S. L is the circle through the Fig. 3: Construction of the Apollonius circle L1 for the Fig. 4 points A and B. Distance |AB| = 1.5, R=1, |OA|= 0.5 first three detectors; it is the Apollonius Circle for the (miles). For any point P on the circle, |PA|/|PB| =1/2. It original source S and the false source S ' (two sulutions of is clear from the text that the lantern A looks from P four the inverse problem). The detector D4 is positioned off L. times brighter than B. The x-coordinates of the points A, The circle L1 passes through the detectors 1, 3 and 4. The B and O are shown relative to the arbitrary origin x =0. corresponding solutions are S and S " . Note that only the ratio of brightnesses is fixed on the circle while their absolute values vary.

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