How many circles does it take to subtend" an ellipse externally?

Transcription

1 How many circles does it take to subtend" an ellipse externally? Cassandra Lee 1 1 The Chinese University of Hong Kong Abstract It takes five circles, arranged suitably and with given radii and centers, to find out the ellipse that they all touch. We first discuss the case for a circle before moving on to the case for an ellipse. Uniqueness is first proven, and then the solution is found numerically. 1 Introduction There are many questions which fools can ask that wise men cannot answer. George Pólya This quote is here because this paper addresses a simple geometrical problem we are able to solve without computer assistance, until we change a single object in it. In one of the surviving factories of former industrial hub Hong Kong, a special machine measures the radius and center of a circular object C by moving cylinders C i around until they touch C, as shown in Figure 1. C is a particular solution to the three-circle Apollonius Problem:[1] given three circles, draw a circle touching each of them. Here we require that the solution circle can only touch the given circles externally, the three given circles must not all have a common tangent, and the disk one of them bounds must not contain the disk another of them bounds (i.e. the circles must not be inside" one another). If these circles are mutually tangent, touching only externally, then C is just the inner Soddy circle.[2] The construction of C, which has center P = (x, y) and radius r > 0, is as follows: we call the three given circles and the required circle C 1, C 2 and C 3. Under the Cartesian coordinate system, we let their centers and radii be P i = (x i, y i ) and r i respectively for i = 1, 2, 3. Then for each i, the distance between each known center P i and the unknown center P is the sum of the corresponding circle s radius r i and the unknown radius r, i.e. PP i = r + r i. In coordinate form, we obtain three equations in x, y and r for i = 1, 2, 3: (x x i ) 2 + (y y i ) 2 = r + r i = (x x i ) 2 + (y y i ) 2 = (r + r i ) 2 (i) Subtracting each equation (i) from equation (j), j = 1, 2, 3, j i yields three linear equations in x, y and r, which can then solved computationally in matrix form to get the center and radius of C. Consistency of this system of equations follows from the conditions imposed on the three given circles. 4

2 Figure 1: The three circles whose centers and radii are known, and the circle C with unknown center and radius. 2 Finding the subtended ellipse By replacing the C i s by ellipses in the original Apollonius Problem, one can still find C.[5] However, what if we replace C by an ellipse instead, keeping C i s circles? In the latter case, which is the focus of this paper, we look for the sufficient number of circles that can touch the ellipse on the outside and determine the equation of the ellipse. In this paper we refer to this construction as subtending the ellipse. This notion of subtending" is not to be confused with a circular or elliptic arc subtending an angle, as both meanings will be used in the following discussion, but it will be clear from the context which meaning is intended. A known result in Euclidean geometry is that five points, of which no three are collinear, suffice to determine a conic uniquely,[3] and so the intuition behind solving the titular question is: ( ) five suitably chosen points can give us the ellipse we want. 2.1 The rules the circles follow We need to set and justify the criteria on the circles that together subtend an ellipse. (a) There must be exactly five circles. We cannot deduce ( ) immediately from the property that five points determine a conic uniquely because circles have uncountably many points, and so there are uncountably many conics through any number of circles. Suppose we have four or fewer circles, and we wish to apply the property we ve just mentioned. By the pigeonhole principle, at least two points must lie on one of the circles. But for both points to 5

3 Figure 2: Five circles subtending an ellipse. The straight lines are the tangents common to the circles and the ellipse. To obtain this ellipse, we first pick five points (light blue) on the circles of fixed radii and draw the conic through those points. Then, we find the tangents at those points to the conic and the circles, and finally we move the five points around the circle until the two tangents at each light blue point overlap each other. While the solution in this figure is heuristic, the discussion that follows shows us how to find the light blue points. touch the ellipse, the circle that contains them both must lie inside the the ellipse, which is tangent to all circles. Yet we require that the circles all lie outside the ellipse. Hence it is impossible to have fewer than five circles. If there are six or more circles, a point from any five of the circles could determine a conic. Therefore one of the circles is left out. We cannot guarantee that the conic and the remaining circle(s) must intersect at exactly one point externally. Therefore, for an ellipse to touch all circles, each circle must contribute a contact point, of which five is necessary. (b) At most two circles share a common tangent. It is straightforward to show that two circles share two common tangents. If at least three circles share a common tangent, and these three or more circles touch the required ellipse externally, the common tangent has to lie away from the region where the ellipse is expected to be, or it degenerates into a straight line. This rule may be regarded as an extension to the condition of having no three collinear points in the property that five points determine a conic. (c) For i = 0, 1, 2, 3, 4, let T i be the collection of all the tangent lines to a circle C i, in which the circles are numbered anticlockwise (or clockwise). Then there exists a set of tangent lines {l i T i } i=0,1,2,3,4 such that the bounded pentagonal subset of the plane formed by the points of intersection of l i mod 5 and l (i+1) mod 5 is convex and only intersects each of the circles at a single point. In other words, a convex pentagon is formed by the points of intersection of each adjacent" pair of tangent lines taken from distinct circles anticlockwise (or clockwise if so chosen), as in Figure 2. 6

4 The reason for convexity: an ellipse is convex, so any five points on its boundary forms a convex polygon. The single-point-intersection condition means that the tangent lines bound" the pentagon on the outside, not on the inside in that case, at least one of the circles touches the ellipse internally. That phenomenon is not covered in this paper. (d) The conditions above are necessary but not sufficient. We can also find on parabolas or hyperbolas five points bounding a convex pentagon. As degenerate conics have been ruled out, the sufficient condition needed is that the 5 points on the C i s satisfy the inequality B 2 4AC < 0 on the conic discriminant,[4] where A, B, C are functions of those points and are defined in section Proof of uniqueness of the ellipse subtended The proof of existence of the subtended ellipse, to be denoted by ζ, will be postponed. For now, we assume that such a ζ exists. Recall that each circle C i has radius r i > 0 and center (h i, k i ) R 2 (i = 0, 1, 2, 3, 4). We write the equation of C i as (x h i ) 2 + (y k i ) 2 = r 2 i. (1) Parametrize each point of C i by (x i (t i ), y i (t i )) = (h i + r i cos t i, k i + r i sin t i ). Restrict C i to their corresponding arcs C i (end points included) contained in the polygon V formed by connecting the centers of C i anticlockwise (or clockwise). For the arguments that follow to be valid, we intend to have the arc C i contain the point at which C i and ζ share a common tangent and C i subtend an angle at least 0 and smaller than π. However, two other situations may exist: C i may not contain the point of tangency between C i and ζ (see Figure 3): Figure 3: A subtended ellipse on which points of tangency G (green) and H (blue) lie outside polygon V. We may consider each circle C i to be a union of two semicircles: one contains the point of contact between ζ and it, and another does not. 7

5 The process to be described is illustrated in Figure 4. Since part of ζ lies on V, we shall extend one of the sides of V joined to C i so that the extended side cuts C i into two semicircles we extend the side of V in order that the interior of one of the semicircular sectors in C i contains no subset of V. Then we shorten the semicircular arc that contains a subset in V by a sufficiently small δ > 0 along its length, and let this arc minus δ be C i. Figure 4: Construction of the new C i (pink) for Figure 3 at point G. Similar procedures apply to H. V may not be convex at (h i, k i ), the center of C i, so that the angle subtended by C i is at least π (see Figure 5): Figure 5: A subtended ellipse on which V is not convex at B. This implies that there are two larger circles adjacent to C i so that the polygon circumscribed by ζ, which touches all circles, is convex. Therefore, the range of possible values of t i such that (x(t i ), y(t i )) lies on ζ is an interval with length strictly less than π. Here we make use of the two large circles adjacent to C i to shorten C i and thus reduce the angle it subtends to a value smaller than π, so that the slope at every point on C i corresponds to a unique point on it. We cut off part of C i as follows, illustrated in Figure 6: join the two centers of the two adjacent 8

6 circles to get a line L. Draw a line L parallel to L and passing through the center of C i. For sufficiently small ε > 0, on the closed disk D ε of radius ε around the center of C i, find the point on D ε furthest from L and outside the strip between L and L, and call L ε the line parallel to L and through that point. Then cut off the portion(s) of C i that lie between L and L ε, and take the remaining arc to be C i as it subtends an angle less than π. Figure 6: Construction of the new C i (pink) for Figure 5 at point B. The yet-undetermined point of tangency exists because it is also the only point of intersection between ζ and a C i. The slope of the tangent line along C i is a strictly monotone function f i (t i ) with the domain being the closed interval I = [0, π δ], δ > 0. If the slope" at a point is taken as its conventional meaning the infinitesimal change in vertical position over horizontal position at that point then f i may have a vertical asymptote where the slope approaches + when t i moves along one portion of the arc and along the other. Moreover, any two distinct points on each C i have different slopes, because every interior angle in V is in I, so the tangent trigonometric function on I gives a strictly monotone function with a vertical asymptote at t i = π 2. One-point compactifying the real line (and/or a suitable coordinate change) eliminates further mention of asymptotes and we can safely assume that f i is continuous and defined on a compact interval of t i. If points P i C i are chosen to generate ζ, then f i and the slope of the tangent line of ζ at P i agree with each other. Let this slope be m i R (the possibility of m i = may be eliminated with a suitable coordinate change). Then it is easy to show that if ζ is an ellipse, there exists exactly two points at which the slope of the tangent is m i ; but even if the slopes at both points are the same, only one of them is allowed to lie on C i for some i because the other such point lies on the opposite side of ζ and hence is disjoint from C i C i. In light of the preceding discussion, by the Intermediate Value Theorem, there exists exactly one t i in the domain of f i such that f i (t i ) = m i. As the same argument applies to all five points on the five C i s, only one conic may be drawn through those five points. Q.E.D. 2.3 Constructing the ellipse Now that we know that the subtended ellipse ζ is unique if it exists, we shall see if it does. Let the equation of ζ be Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0, (2) 9

7 with (A, B, C) (0, 0, 0). Then we substitute points (x i, y i ) on C i to (2) to obtain a system of linear equations equivalent to the determinant equation x 2 xy y 2 x y 1 x1 2 x 1 y 1 y 2 1 x 1 y 1 1 x2 2 x 2 y 2 y 2 x 2 y 2 1 x3 2 x 3 y 3 y 2 3 x 3 y 3 1 = 0 (3) x4 2 x 4 y 4 y 2 4 x 4 y 4 1 x5 2 x 5 y 5 y 2 5 x 5 y 5 1 so A, B, C, D, E, F are functions of x i and y i. We shall parametrize our circles C i as in the uniqueness proof with bounded t i. Then (3) becomes a function in x, y and t i for i = 0, 1, 2, 3, 4. For each i, we shall find the points at which the tangent lines at ζ and C i match. Now, implicit differentiation with respect to x on (2) gives the slope at (x, y) ζ: dy dx = 2Ax + By + D Bx + 2Cy + E while implicit differentiation on (1) gives the slope at (x, y) C i : It seems intuitive to then equate (4) and (5) to get dy dx = x h i y k i (5) (4) (2Ax + By + D)(y k i ) = (Bx + 2Cy + E)(x h i ) Bx 2 2(A C)xy By 2 + (2Ak i Bh i + E)x +(Bk i + 2Ch i D)y + (Dk i Eh i ) = 0 (6) but there is a catch, because we are not taking all solutions to this equation. Instead, we are only taking the points that lie on the C i s. In fact we get (7), a nonlinear system of five degree 8 polynomial equations in five unknowns t i, i = 0, 1, 2, 3, 4: Bx 2 i 2(A C)x i y i By 2 i + (2Ak i Bh i + E)x i +(Bk i + 2Ch i D)y i + (Dk i Eh i ) = 0 (7) We solve this system with a fixed point iteration which captures the idea behind the heuristic method used to obtain Figure 2. Alternative solution techniques involving linear and advanced algebra may be found in [6], [7], [8] and [9]. 2.4 An algorithm to get the ellipse The t i s are bounded, so we can get C i after Section 2.2 s adjustments (pink in Figure 7), make an initial guess (t i ) (0) C i satisfying the requirements in Section 2.1, compute (x i) (n) := x i ((t i ) (n) ), (y i ) (n) := y i ((t i ) (n) ) and find the conic ζ (n). For n = 0: ζ (n) : A (n) x 2 + B (n) xy + C (n) y 2 + D (n) x + E (n) y + F (n) = 0. (8) We get its slope at (t i ) (n) : (marked in red in Figure 7) dy dx ζ,(n) = 2A (n)(x i ) (n) + B (n) (y i ) (n) + D (n) B (n) (x i ) (n) + 2C (n) (y i ) (n) + E (n) (9) 10

8 We also get the slope at (t i ) (n) on C i : (marked in green in Figure 7) dy dx C i,(n) = (x i) (n) h i (y i ) (n) k i (10) Let ε > 0 be the margin for error. For each i, if dy dx ζ,(n) dy dx C i < ε we pause the iteration. (We can,(n) only stop when this inequality is true for all i.) Otherwise, ζ (n) and C i intersect at more than one point, and we continue below. We find (s i ) (n) such that (x i ((s i ) (n) ), y i ((s i ) (n) )) C i is another point of intersection. There may be more than one such point, so we may have to compute more than one value of (s i ) (n) distinct from (t i ) (n). However, if there are two or more choices for (s i ) (0), the ellipse is no longer touching C i but has gone too deep into the circle, and so we will have to find another (t i ) (n) C i until we only have two intersection points in total. (A circle and an ellipse can intersect each other at 4 points at most.) We also compute the quantities (9) and (10) at (s i ) (n) ( (t i ) (n) in general), calling them (9) and (10) respectively. As C i is a circle, (s i ) (n) gives rise to two possible mid-points along C i " which we call M i,n,k (k = 1, 2) between (s i ) (n) and (t i ) (n) and these two points are antipodal on C i. We pick the (s i ) (n) such that M i,n,k is inside ζ (n), i.e. the winding number of ζ (n) around M i,n,k is 1. Basically we are doing a mid-point iteration until the mid-point is close enough to the solution curve ζ. We set (x i ) (n+1) := x i Mi,n,k and (y i ) (n+1) := y i Mi,n,k along C i, and repeat the processes starting from equation (8) for other i s and increasing n. It is expected that { dy dx ζ,(n) dy dx C i } is a decreasing sequence,(n) tending to 0 as n tends to infinity. If there is more than one choice of (s i ) (n), we compute two sets of (x i ) (n+1) and (y i ) (n+1) and pick the one that decreases dy dx ζ,(n) dy dx C i the most.,(n) Suppose B(n) 2 4A (n)c (n) < 0 throughout all iterations (and this condition is to be checked for every iteration). To show that this algorithm converges to the solution A, B, C, D, E (F obtainable by backsubstitution), we translate the circles and the ellipse it subtends until they all lie within the same quadrant, away from the origin and coordinate axes, so that none of the solution x i, y i equals 0 for i = 0, 1, 2, 3, 4. Consider the left side of equation (7). For each i, replacing A to F with A (n) to F (n) respectively, the expression on the left side of (7) measures" how close an iteration is to the real solution, with a value of 0 for the true solution. By considering the difference between (7) for the iterated A (n) to F (n) and (7) for the true values, we get the expression (we write x (n), y (n) for (x i ) (n), (y i ) (n) respectively below) +[(A (n) A)x (n) y (n) + A x (n) y i x i y (n) [(B (n) B)x(n) 2 + B(x (n) + x i )(x (n) x i )] (C (n) C)x (n) y (n) C x (n) x i ] y i y (n) +[(B (n) B)y 2 (n) + B(y (n) + y i )(y (n) y i )] +2k i [(A (n) A)x (n) + A(x (n) x i )] h i [(B (n) B)x (n) + B(x (n) x i )] +[(E (n) E)x (n) + E(x (n) x i )] + k i [(B (n) B)y (n) + B(y (n) y i )] +2h i [(C (n) C)y (n) + C(y (n) y)] + [(D (n) D)y (n) + D(y (n) y i )] +[k i (D (n) D) h i (E (n) E)] (11) but this is equal to the left side of (7) with A (n), B (n), C (n), D (n), E (n) in place of A, B, C, D, E (because the true solution makes this expression vanish). Note also that x(n) 2, y2 (n), x (n)y (n), (x (n) + x i ), (y (n) + y i ) are nonzero as x (n), y (n), (x i ), (y i ) lie in the same quadrant. As a result, all of (Λ (n) Λ) (where Λ is a placeholder for each of these symbols: A, B, C, D, E, (x i ), (y i )) tends to 0 as n. (Oh, and x (n) x i y i y (n) 0 as n too, so the rows and columns in this determinant become equal" to each other eventually.) 11

9 Initialization Solution Figure 7: The algorithm in action, ε = 0.10 From there we can find the foci, major and minor axes, eccentricity, etc. of the ellipse. If it happens to be a circle, the center and radius are easily obtained and accuracy may be checked using any three of the circles as outlined in Section 1. 3 Possible real-life applications and future work The work may be considered an optimization problem along an ellipse the resource [10] provides an example of an elliptical table subtended by multiple circles representing chairs. The inverse problem of finding circles tangent to a given ellipse, on the other hand, has been solved.[11] Another application may be data analytics, as the circles can represent margins of error in data points (for instance in Voronoi diagrams, and Support Vector Machine classifiers in image processing). Suitably generalizing to n-space and replacing the ellipse by some well-defined compact manifold can help determine the required closed body of data/information. 4 Acknowledgements The author would like to thank Dr. L.F. Cheung, from the Chinese University of Hong Kong, for posing the titular question in August He said it was an open question. The graphics were drawn with GeoGebra 5 ( References [1] Weisstein, E. W. Apollonius Problem", MathWorld A Wolfram Web Resource. Retrieved from http: //mathworld.wolfram.com/apolloniusproblem.html [2] Weisstein, E. W. Soddy Circles", MathWorld A Wolfram Web Resource. Retrieved from mathworld.wolfram.com/soddycircles.html [3] Dixon, A. C. (1908). The Conic through Five Given Points." The Mathematical Gazette, 4(70),

10 [4] Anton, H., Givens, I., & Davis, S. (2012). "Appendix K: The Discriminant", Calculus (10th ed.). Wiley. [5] Emiris, I. Z., Tzoumas, G. M. (2005) Algebraic Study of the Apollonius Circle of Three Ellipses. 21st European Workshop on Computational Geometry Eindhoven. Retrieved from tue.nl/ewcg2005/proceedings/38.pdf [6] Williams, M.P. (2010) Solving Polynomial Equations Using Linear Algebra." Johns Hopkins APL Technical Digest 28(4), pdf [7] Manocha, D. (1994) Solving systems of polynomial equations." Computer Graphics and Applications, IEEE, 14(2), [8] Sottile, F. (2002) "From Enumerative Geometry to Solving Systems of Polynomial Equations." Eisenbud, D., Stillman, M., Grayson, D. R., Sturmfels B. (Eds.) Computations in Algebraic Geometry with Macaulay 2, 8, [9] Sturmfels, B. (2002) Solving Systems of Polynomial Equations". CBMS Regional Conference Series in Mathematics, 97. Rhode Island: American Mathematical Society. cbms/097 [10] Brice, A. A mathematical digression". Retrieved from 18/a-mathematical-digression/ [11] How to draw ellipse and circle tangent to each other?", Mathematics Stack Exchange. Retrieved from how-to-draw-ellipse-and-circle-tangent-to-each-other 13