A Novel Method for Drawing a Circle Tangent to Three Circles Lying on a Plane by Straightedge,Compass,and Inversion Circles
aa r X i v : . [ m a t h . HO ] M a y A Novel Method for Drawing a CircleTangent to Three Circles Lying on a Planeby Straightedge,Compass,and InversionCircles
Ahmad Sabihi ∗ Professor and researcher at some of Iranian Universities
Abstract
In this paper, we present a novel method to draw a circle tangent to three given cir-cles lying on a plane. Using the analytic geometry and inversion (reflection) theorems, the center andradius of the inversion circle are obtained. Inside any one of the three given circles, a circle of thesimilar radius and concentric with its own corresponding original circle is drawn.The tangent circle tothese three similar circles is obtained. Then the inverted circles of the three similar circles and thetangent circle regarding an obtainable point and a computable power of inversion (reflection) constantare obtained. These circles (three inverted circles and an inverted tangent circle)will be tangent to-gether.Just,we obtain another reflection point and power of inversion so that those three reflected circles(inversions of three similar circles) can be reflections of three original circles, respectively. In such acase,the reflected circle tangent to three reflected circles regarding same new inversion system will betangent to the three original ones. This circle is our desirable circle. A drawing algorithm is also givenfor drawing desirable circle by straightedge and compass. A survey of conformal mapping theory andinversion in higher dimensions is also accomplished. Although, Laguerre transformation might be usedfor solution of this problem, but we do not make use of this method. Our novelty is just for drawing acircle tangent to three given circles applying a tangent circle to three identical circles concentric withthree given ones and then inverting them as original ones by compass and straightedge not any thing else.
Keywords : Inversion Circles;Conformal Mapping;Transformation Geometry;Three-Circle TangencyProblem;Inversion in Higher Dimension; Laguerre transformation
The most important and interest problem,which was posed as a challenge by Apollonius [1] inthe third century BC is three-circle tangency problem.This problem is stated as ”given three circleslying on a certain plane in general,construct all circles tangent to them ”. Apollonius generated theconstruction for nine cases,briefly, given three objects, each of them being a circle, a line, or a point,construct a circle tangent to all three given objects. The simplest case is the elementary constructs ofa circle through three given points. The most challenging case is that one can draw a circle tangentto three given circles.That case is of the most emphasis here. Euclid and Archmedes had presented ∗ Permanent Address: First floor, Bldg.name: Anita, Golbarg blind alley, Ershad al-ley (No.33), Avecina St., Isfahan, Iran, Tel:+98(31)34486478, Cell Phone:+989131115784, E-mail: [email protected] construction for three points and three lines,but Apollonius gave the construction methods in hisown books I and II on three-circle problem as well. Altshiller-Court [2]in 1925 gave a construction ofsame problem by reducing it to the construction of PCC (Point Circle-Circle). Fran¸cois Vi`ete, IsaacNewton, and Joseph-Diaz Gergonne [3] individually solved the problem by different approaches.Vi`etemade use of the geometric fundamentals, Newton the conic approach, and Gergonne developed a newunderstanding of inversion geometry properties.Vi`ete’s work is fine for any three given circles,but not inan environment, where the given conditions can change after completion of the construction. Newton’ssolution is also a bit problematic when it comes to dynamic geometry software. A change in the sizeor location of one of the given circles is likely to capsize the construction. His construction method isnot unique so that decision must be made along the construction way. The method is based on theintersection of hyperbolas. An intersection of two hyperbolas is the case not three. Two hyperbolascan intersect at four points, and only two of which solve problem [3]. Gergonne in 1816 [3] publishedan inversion-based solution to the tangency problem. The Gergonne’s solution can be applied to manyof the nine other cases.His method is the most exhaustive solution,which is specially flexible regardingpositions of the given circles.Inversion was invented by J.Steiner around 1830 and transformation by inversion in circle was in-vented by L.I.Magnus in 1831 [5]. For further information, refer to [6-9]. The author,surveys on inversivegeometry,transformation theory, and conformal mapping in the Section 2. A novel geometrical methodfor drawing a circle tangent to the three given circles is given in the Section 3. A new algorithm fordrawing same by Compass,and Straightedge without using any other axillary tools comes in the Section4.
Let C be a circle passing through the point A perpendicular to C . Let ´ A be a second point ofintersection of OA with C ,then OC is tangent to C . This means that OA.O ´ A = OC = R where R is the radius of C . C and O are called the reflection circle and center respectively. Also,the location ofpoint ´ A does not depend on circle C ,because ´ A lies on the line OA at distance equal to R OA . The point´ A is known as the inverse , inversion ,or reflection of A with respect to C . Therefore, ´ A is the reverse orreflect of A and vice versa. Except the center of C ,which is called as the center of inversion,all pointsas A in the plane have inverse images as ´ A . The term inversion also applies to define transformationof the plane.This transformation maps the points inside C to the points in its exterior and vice versa.The center of inversion is often [4-5] left over as a point with no inverse image,but sometimes is saidto be mapped to the point at infinity. The origin or zero in the circle inversion mapping requires aspecial definition to adjoin a point at infinity as ∞ . R is called the radius of inversion, R as the power ofinversion. As stated in the introduction,inversive geometry refers to a study of transformation geometryby Euclidean transformations together with inversion in an n-sphere so: x i r x i P nj =1 x j (1-2)where r denotes the radius of the inversion. In two dimensions,with r = 1,this is indeed a circle inversionwith respect to the unit circle so that x r x | x | . If two inversions in concentric circles are combined,thena similarity,homothetic transformation will be resulted by the ratio of the circle radii: x r x | x | = y k y | y | = ( kr ) x (2-2) he algebraic form of the inversion in a unit circle is w = z because¯ z | z | . z | z | = R = 1 (3-2)and the reciprocal of Z is ¯ zz . In the complex plane,reciprocation leads to the complex projective lineand is often called the Riemann sphere. The subspaces and sub groups of this space and group of map-pings,were applied to produce early models of hyperbolic geometry given by Arthur Cayley,Felix Klein,Henri Poincar´e. Therefore,inverse geometry includes the ideas originated by Lobachevsky and Bolyai intheir plane geometry. More generally, a map f : U −→ V is called conformal or angle-preserving at v ,ifit preserves oriented angles between curves through v with respect to their orientations. Conformalmaps preserve both angles and shapes of infinitesimally small figures,but not necessarily their size. Someexamples of conformal maps could be made from complex analysis. Let A be an open subset of the com-plex plane, C ,then a function g : A −→ C is defined as a conformal map if and only if it is holomorphicand its derivation is everywhere non-zero on A .But if g is anti-holomorphic (conjugate to a holomorphicfunction),it still preserves angles except it reverses their orientation. Conformal mapping applicationin electromagnetic potential,heat conduction,and flow of fluids are taken into considerations. The ideasare that the problems at hand with a certain geometry could be mapped into a problem with simplergeometry or a geometry which have already been solved. A such mappings are found by transformation[10]. For further study,refer to [11-12]. Laguerre transformation [13] might be used for solving ourproblem using a plane spanned by three points in a three dimensional space and transforming it into ahorizontal plane, but we do not make use of such transformations. Also, methods of Cyclography mightbe applied to solve it. Our novelty is just for drawing a circle tangent to three given circles applyinga tangent circle to three identical circles concentric with three given ones and then inverting them asoriginal ones by compass and straightedge not any thing else. Let C , C ,and C be three given circles of optional radii R < R < R . Let’s draw three similarcircles of radii R concentric with each one of circles so that inequality R < R < R < R holds. Adjointhe centers of circles together by straight-lines. Then draw perpendicular bisector of the created line-segments. The intersection point of perpendicular bisectors is indeed the center of tangent circle to thethree similar circles of the radii R since it is equidistance from the circles’ centers. Also, having equalradii,the distances of the intersection point from three circles are identical. This means that we able todraw a tangent circle to the three similar circles from the intersection point. Just, let the problem besolved. Using the reflection theorems and methods,the reflected objects (or reflected circles) of threesimilar circles of radii R with respect to a fixed point (where will be obtained later by drawing) areobtained. In this paper,we carry out following steps 3.1.1 through 3.1.7 based on Fig.1 Consider the three given circles C , C ,and C of radii R , R and , R , respectively. Draw the concentric circles of radii R denoted by C , C ,and C (these symbols are not shown inFig.1) inside the circles C , C ,and C , respectively. Draw the tangent circle C to the three similar circles of the radii R , if their centers are noton a line Draw the reflected circles of the three similar circles of radii R denoted by ´ C , ´ C ,and ´ C withrespect to the reflection center O .1.5 Draw the reflected circle of the tangent circle C i.e. ´ C . This reflected circle is also tangentto the reflected circles of the three similar circles of the radii R . Define a new center and a power of inversion so that the three reflected circles ´ C , ´ C ,and ´ C (created by three similar circles of radii R ) can be as the new reflected circles of C , C ,and C . Thenew reflection center is identical to the old one. Draw the reflection of the tangent circle ´ C with respect to the new given center (or same old one)and new reflection constant (power of inversion value). This circle denotes C ′′ and is a desirable circlein this paper, where is tangent to the three given circles. If the centers of the three given circles (and three similar circles) C , C ,and C are on a straightline,then we should draw a tangent line to the three similar circles instead of a tangent circle and obtainthe reflected circle of this line and continue steps 3.1.5 to 3.1.7. Let we have the first reflection center O referring Fig.1. We know the following rela-tions hold: OA.O ´ A = OB.O ´ B = OC.O ´ C = OD.O ´ D = OE.O ´ E = Of.O ´ f = k (1-3)where k denotes the power of inversion of the three similar circles. The points ´ A, ´ B, ´ C, ´ D, ´ E and ´ f denote reflected points for points A , B , C , D , E and f on the three reflection circles ´ C , ´ C ,and ´ C , re-spectively. Then, to make the reflected circles ´ C , ´ C ,and ´ C as the reflections of the circles C , C , and C , respectively, a new center of reflection can be defined so that lies on the previous one. Hence,thecenter O should be the reflection center of both reflection systems so that: OI.O ´ A = OH.O ´ B = OL.O ´ C = OM.O ´ D = OJ.O ´ E = OK.O ´ f = ´ k (2-3) Proof
To have O as the common center of both reflection systems: OI.O ´ A.OH.O ´ B = ´ k (3-3) OI.OH. k OA . k OB = ´ k ⇒ OH.OIOA.OB = OL.OMOC.OD = OJ.OKOE.Of = ´ k k = M (4-3)where M denotes the ratio of the second power of the inversions of both reflection systems. This is aconstant value. It also denotes the ratio of the powers of the point O with respect to concentric circlesof radii ( R , R ) , ( R , R ),and ( R , R ). This means that ratio of the tangent line segments drawn fromthe point O over both concentric circles ( R , R ) , ( R , R ),and ( R , R ) should be identical. The problemis that we wish to find geometrical location of all points whose powers are identical with respect to twogiven concentric circles. Referring Fig. 1 and Fig.2,we have the relations (5-3). OH.OIOA.OB = ( AT AT ) , OL.OMOC.OD = ( AT AT ) OJ.OKOE.Of = ( AT AT ) (5-3) K and ´ K Therefore,using (4-3) and (5-3)( AT AT ) = ( AT AT ) = ( AT AT ) = M (6-3)where AT to AT denote tangent lines drawn from the point A or same O (a typical reflectioncenter) to the three similar circles of the radii R ,respectively. Also, AT to AT denote tangent linesfrom same point to the circles C , C ,and C , respectively. For example, for circles C and C (twoconcentric circles shown in Fig.2) of the center O we have: OL.OMOC.OD = ( AT AT ) = M = AO − R AO − R ⇒ AO − R = M .AO − M .R ⇒ AO = s R − M .R − M (7-3)The geometrical location of the point A is a circle of the radius AO and the center O . For the twoother concentric circles ( C , C ),and ( C , C ), the method is similar. Due to 0 < M < M = . We know O and O are the centers and R and R , the radii of twoother concentric circles, respectively, therefore regarding (7.3), we have AO − R = − R , AO − R = − R , AO − R = − R (8-3)Let the representing parametric equations of C , C ,and C as follows: C ( x, y ) : x + y + a x + b y + c = 0 , C ( x, y ) : x + y + a x + b y + c = 0 , C ( x, y ) : x + y + a x + b y + c = 0(9-3)Let A ( x , y ) be the mentioned typical reflection center. Using the equations (9-3),we obtain the circlecenters O ( − a , − b ), O ( − a , − b ), and O ( − a , − b ) then AO = ( x + a +( y + b , R = 14 ( a + b ) − c ⇒ AO − R = x + y + a x + b y + c +( c −
14 ( a + b ))(10-3)By the same method,we obtain AO − R = x + y + a x + b y + c + ( c −
14 ( a + b )) (11-3) O − R = x + y + a x + b y + c + ( c −
14 ( a + b )) (12-3) Theorem2, let K ( x, y ), K ( x, y ) and K ( x, y ) denote the new circle’s equations eliminating zeroindices as follows, then the center of inversion for three given circles is obtained by intersection ofnormal lines created by these circles. K ( x, y ) = AO − R + R = C ( x, y ) + ( c −
14 ( a + b )) + R = 0 (13-3) K ( x, y ) = AO − R + R = C ( x, y ) + ( c −
14 ( a + b )) + R = 0 (14-3) K ( x, y ) = AO − R + R = C ( x, y ) + ( c −
14 ( a + b )) + R = 0 (15-3) Proof
Consider K − K : C ( x, y ) − C ( x, y ) + c − c + 14 ( a − a + b − b ) = 0 (16-3) K − K : C ( x, y ) − C ( x, y ) + c − c + 14 ( a − a + b − b ) = 0 (17-3) K − K : C ( x, y ) − C ( x, y ) + c − c + 14 ( a − a + b − b ) = 0 (18-3)The equations (16-3) to (18-3) are indeed the equations of the normal lines to the lines which adjointhe centers of both circles ( C , C ), ( C , C ), and ( C , C ), respectively. Take the points M , M and M placing at the bisector of the line segments constructed, then referring to Fig.3, we could make therelations (19-3). AO − AO = 2 O O .M H , AO − AO = 2 O O .M H , AO − AO = 2 O O .M H (19-3)Therefore, the lines passing through the points H , H ,and H denote our desirable geometrical loca-tions of the points. These locations point out to the set of the points, whose difference of their powerof distances from the centers of each two circles are constant values. These are the lines perpendicularto the line segments adjoining their two centers. The distance between the perpendicular line and thecenter of the line segment for each two circles is: M H = R − R O O , M H = R − R O O , M H = R − R O O (20-3)The given lines presented by the equations (16-3) to (18-3) should intersect each other at a uniquepoint because ( K − K ) + ( K − K ) = ( K − K ) (21-3)Therefore, from the intersection of the two lines ( K − K ) and ( K − K ),the line ( K − K ) isobtained.Thus,the point ( x, y ) satisfies three line equations (16-3) through (18-3). This means that thecenter of inversion is found and the problem is solved. A The Novel Drawing Method by the Straightedgeand Compass in form of a New Algorithm
The algorithm is given regarding Figs.1,3
Draw three similar concentric circles C , C ,and C of radii R inside the three given circles C , C ,and C , respectively so that R < R < R < R . Adjoin the centers of the circles C , C ,and C to each other. Draw the perpendicular bisectors of the line segments adjoining each of two centers to meet ata point as O ′′ (not shown in Fig.3). Adjoin the point O ′′ and center of one of the circles,say, C (Fig.3) to meet at the point Z . Draw a circle of the center O ′′ and radius O ′′ Z to be tangent to the three similar circles C , C ,and C . Regarding Fig.3,the reflection center point A could be found as follows:Draw the radical axis of both circles ( C , C ),( C , C ), and ( C , C ).This can be done by drawingtwo optional circles to be intersected with circles,say, C ,and C . These two radical axes intersect withone another at the point ´ O . A perpendicular line should be drawn from this point to the line segment O O to meet it at a point. This point is called ´ H . Lie the compass’s needle on the ´ H and open itsother arm by ´ HM based on what is shown in the Fig.3. Then, draw an arc toward smaller circleside,say, C , to be intersected with O O at the point H . Draw a perpendicular line to O O fromthe point H . This line is indeed representing the line ( K − K ). The method should be repeated fordrawing the lines ( K − K ), and ( K − K ) by same method. These lines will intersect each other atthe point A ,which denotes the reflection center. Obtain the reflections of three similar circles of equal radii R (i.e. C , C , C )and C with re-spect to the reflection center A and power of inversion K . These reflections denote ´ C , ´ C , ´ C ,and ´ C according to Fig.1. The circle ´ C is tangent to the circles ´ C , ´ C ,and ´ C . According to given proof, three circles ´ C , ´ C ,and ´ C can also be as reflections of the three cir-cles C , C ,and C with regard to same reflection center and by different power of inversion ´ K . Based on the explanations given in 4.8 and referring to the relation (4-3), we find out that ´ k k = M .Therefore, the new power of inversion is obtained by the relation ´ k = M.k . Both k and M areknown,thus ´ k is also known, because, we can obtain size M by the relation M = and by means ofstraightedge and compass tools. This means that the reflections of the three given circles ´ C , ´ C ,and´ C of the center A and constant ´ k can be the three main circles C , C ,and C , respectively. Referring to Fig.1,the reflection of the circle ´ C is C ′′ ,which is a desirable solution. C ′′ is tan-gent to the three circles C , C ,and C due to C ′′ is also reflection of ´ C with regard to the point A and the reflection constant ´ k .Therefore,the problem of three-circle tangency is completely drawn bycompass and straightedge.Only tools applying in this method to draw desirable circle,are both Compass and straightedge without ny additional ones. Our method may be extended to three-dimension space by turning the plane ofeach one of circles about their related diameter. In such a case,we can make a sphere tangent to threegiven spheres. Also,this method would also be interested if it could be extended to higher dimensionspheres. If three original circles are of concentric circles or only one of three circles is placed insideanother one and the other one is outside of both,then we will not be able to draw a tangent circle tothem. Further investigations about drawing limitations are left for readers. References [1] T.Heath,A History of Greek Mathematics,Dover Publications,New York,Two Vols.(1981).[2] N.Altshiller-Court,College Geometry:A Second Course in Plane Geometry for Colleges and NormalSchools,Johnson Publishing,Atlanta,(1925).[3] P.Kunkel,The Tangency Problem of apollonius:Three Looks,BSHM Bulletin.,22(2007)34-46.[4] R.Courant and H.Robbins,What is Mathematics,Oxford University Press,(1996).[5] H.S.M. Coxeter,Introduction to Geometry,John Wiley and Sons,(1961).[6] H.S.M. Coxeter and S.L.Greitzer,Geometry Revisited,MAA,(1967).[7] Liang-Shin Hahn,Complex Numbers and Geometry,MAA,(1994).[8] D.Hilbert and S.Cohn-Vossen,Geometry and Imagination,Chesla Publishing Co.,New York.(1990).[9] H.Rademacher and O.Toeplitz,The Enjoyment of Mathematics,Dover Publications,(1990).[10] R.V.Churchill,Complex Variable and Applications,2nd Edition,McGraw-Hill,NY,(1960).[11] K.J.Binns and P.J.Lawrenson,Analysis and Computation of Electric and Magnetic Field Prob-lems,2nd Edition,Pergamon Press NY,(1973).[12] W.R.Smythe,Static and Dynamic Electricity,3rd Edition,Taylor and Frances,(1989).[13] H. Pottmann, P. Gorhs, and B. Blaschitz, Edge offset meshes in Laguerre geometry, GeometryPreprint 190, Technische Universit¨ a t Wien, (2008).t Wien, (2008).