UNIVERSITY OF PENNSYLVANIA SOME EXTENSIONS OF THE WORK OF PAPPUS AND STEINER ON TANGENT CIRCLES BY J. H. WEAVER A THESIS PRESENTED TO THE FACULTY OF THE GRADUATE SCHOOL IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY PRESS OF THE NEW ERA PRINTING COMPANY LANCASTER, PA. 1920 [Reprinted from THE AMERICAN MATHEMATICAL MONTHLY, Vol. XXVII., January 1920.] SOME EXTENSIONS OF THE WORK OF PAPPUS AND STEINER ON TANGENT CIRCLES. By J. H. WEAVER, Ohio State University. Introduction. The figure of three mutually tangent semicircles with their centers in the same straight line was known among the Greeks as the "Shoemaker's Knife" (iap/3oXos). A few of the properties of the figure are found in the works of Archimedes.' Others occur in the Collection of Pappus.2 After the Greek period we find no work done on the problem until Steiner generalized the results of Pappus and added several others dealing with the perspective properties of the figure.3 Later Sir Thomas Muir added a theorem giving formulae for various sets of radii involved.4 Habicht has discussed some of the properties of elliptic functions connected with the figure5 while M. G. Fontene has generalized certain formule arising from sets of tangent circles.6 In the present paper formule for the radii of certain sets of circles are developed and used to build up several types of infinite series which may be summed geometrically. Then some general properties of tangents and normals to conies associated with three mutually tangent circles are set forth.7 These properties lead to a quadrangular-quadrilateral configuration and S// \ incidentally furnish some methods for constructing conies. And finally some theorems connected with centers of perspectivity of the A- 01 0, —, a, various sets of circles are proved. 1FIG. 1 3 1. General Considerations. Let there be two circles tangent internally, with centers 0, and 02 and let a circle with center Sn (in = 1, 2,... ) (Fig. 1) be tangent to these. Then Sn lies on an ellipse with foci 01 and 02. If we take the midpoint of 0102 as origin and 0102 as the x-axis, the equation of the ellipse will be 4x2 s 2 -= -1, (1) (rl +- r2)2 rlr2 where ri and r2 (r2 > ri) are the radii of the circles (01) and (02) respectively. 1 Works of Archimedes, ed. Heath, Cambridge, 1897, Lemmas, 4-6. 2 Collectio, ed. Hultsch., Berlin, 1876-8, Vol. I, pp. 209 and ff. 3 Steiner, Gesammelte Werke, Berlin, 1881, Vol. 1, pp. 47-76. 4Proceedings of Edinburgh Math. Soc., Vol. 3, p. 119. In the same volume, pages 2-11, J. S. Mackay has collected some of the simpler theorems connected with the problem. 5 Konrad Habicht, Die Steinerschen Kreisreihen, Berne, 1904, 35 pp. In this work are found extensive references bearing on the subject. 6 " Sur les cercles de Pappus," Nouvelles Annales de Math6matiques (4), tome 1918, pp. 383-90. 7 The center of a circle tangent to two given circles lies on a conic having the centers of the two given circles as foci. This is, of course, equivalent to the definition that the sum or difference of the focal radii is constant. I have called such conics "associated" conics. 8 In what follows circles will be designated by their centers in brackets. 2 WORK OF PAPPUS AND STEINER ON TANGEN TCIRCLES. [Jan., Let pa be the radius of (Sn) and let the coordinates of the point Sn be x, and yn. From a fundamental property of the ellipse we have ri + p =n a + ex., (2) where 2a = ri + r2, and - '2?'1 e = '2 + i'+ Pappus has shown that if another circle (Sn+i) with radius pn+, center at point Xn+l, yn+l and coming after (Sn) in the positive direction around the circles, is tangent to (Sn), the following relation holds Yn + 2Pn Yn+l Pn Pn+l or n n-1 + 2 - +2(n- 1). (3)1 Pn Pn-1 Pi 2. Formulae arising from the figure of three mutually tangent circles with their centers in the same straight line. Let there be three mutually tangent circles (01, 02) and (03) having their centers in the same straight line and radii rl, r2, and r3 respectively (Fig. 1). Then let a series of circles (Sn) be drawn tangent to (01) and (02), the first circle in the series being also tangent to (03) and each of the others tangent to the one preceding it in the series. There are two other such sets of tangent circles. The set tangent to (02) and (03) we will designate as (Sn'), and the set tangent to (01) and (03) as (Sn"). Let the radii of the various sets be Pn, pn' and p," respectively, and the coordinates of the centers be Xn, yn, xn', Yn' and xn", yI" respectively. We will now consider the set (Sn). By means of equations (1), (2) and (3) and the use of induction we have in this particular case, since the y-coordinate of 03 = 0 r1r2r3 p = n2r32 + r1r3 + r12 ( This result is arrived at by Muir and Fontene by different methods.2 Also from equations (2) and (4) (2n - 1)r32rr2(rl +- r2) Xn-I - Xn = i = 24, say. (5) xm-l -- Zx = [(n - 1)2r32 + r1r3 + r12][n2r32 + r1r3 + r12] = say. (5) From the geometric properties of the figure c0 Zpn (6) n=l is a convergent series, and if r3 approaches the limit 0, then (6) approaches the value 7rr2/2 but is 0 at the limit. 1 Pappus, Collectio, p. 224. 2 See Introduction. 1920.] WORK OF PAPPUS AND STEINER ON TANGENT CIRCLES. 3 Also 00 Zin= 2r1+ r3. n=l If we define in as the nth intercept of the series (Sn) (ii, projection of 03S1 on AC), then (4) and (5) are the formulae for the nth radius and intercept in the series (Sn). An interchange of ri and r3 will give the corresponding formulae for the series (Sn'), while an interchange of r2 and r3 with r2 considered negative will give the corresponding formulae for the set (Sn,). If r2 =" 2rl we have the special case 00 E Pn 22/2. n=l We will now establish the following theorem. THEOREM: If two circles (01) and (02) are tangent internally, and a circle (S) is drawn tangent to these two, such that SOi (i = 1, 2) is perpendicular to 0102 then it is possible to draw a circle (S') tangent to (01), (02) and (S) such that the four centers S, S', 01 and 02 determine a rectangle. Proof: Let S02 be perpendicular to 0102, and let the coordinates of S and S' be x, y and x', y' respectively and let the radii be p and p'. Then r2 - r 2rlr2 r2(r2 - rl) x= 2 ' r1+r2' P (6) and by virtue of equations (1), (2) and (3) y' =y and x' = -x, which proves the theorem. THEOREM: If in the series (Sn), the points Sn, Sn+1, 1O and 02 determine a rectangle, then ri = nr3. Proof: Equate. the values of p given in equations (4) and (6). Let the foot of the perpendicular from Sn to 0102 be Pn. Let angle PnSnSn+i = Z Bn. Then if r1 = kr3 (k an integer or a rational fraction) (2k -+ 1)(2n + 1) tan Bn = 2(n2 +- n - k - ) (7) and the slopes of the lines of successive centers of the series (S,) are all rational. Moreover if in (7) k is an integer, Bk = 7r/2, and 00 E= (Bn — Bn) =Bkli m Bn = 7/2. n=-k+1 n=)o From the identity (pn + Pn+l)2 = i~n+2 + (yn yn+1)2 we get by using equi 4 WORK OF PAPPUS AND STEINER ON TANGENT CIRCLES. [Jan.,, tions (3), (4) and (5) the equation r/~2 _ C) _Ll~r32 _IC)?i.12]2 [(2n2 + 2n ~ 1)r32 + 2rlr3 + 21] = [(2n2 + 2n)r32 - 2r1r3 - 2r 12]2 ~ [(2n + 1)r3(r2 + r)12 giving a triply infinite set of rational right triangles. 3. Formulae arising from three mutually tangent circles, one of which is tangent to the line of centers of the other two. Let there be two circles (01) and (02) (Fig. 2) tangent internally at A and let a series of circles be drawn tangent to these, the first one in the series being tangent to the line 0102 and each of the others tangent to the one preceding it in the series. Let the radius of (01) be rl and of (02) be r2, and let the centers of the circles be S,. Then from equations (1), (2) and (3), since Yi = pi we get by induction 4rlr2(r2 - ri) Pn 8 4(n2 - n)(r2 - r1)2 + (r2 + r1)2 32(n - l)r1r2(r2 - r1)2(r2 -- r1) in 212+12] 2(9) [4(n2 - n)(r2 - r1)2 + (r2 + r1)2][(4n2 - 12n, + 8) (r2 - r1)2 + (r2 + ri)2] where for i, n: 2. Here co Z zn = 27'2 il n=2 S and if ri approaches r2 00 rA 01 0,2 z=2 FiG. 2. approaches rrr2/2. THEOREM: If in this series the points S,, Sn+l, 01 and 02 determine a rectangle then 2n t I r2 ri. 2n + 1 Let 2k + 1 r2 ri. 2k - 1 Then 2nk tan Bn = 2 Moreover, Z (Bn-1 Bn) = Bk -lim Bn 7r/2. n=-le1 Also the equation (p, + Pn3i)2 (jnf1)2 + (Yn - Yn+12 gives the triply infinite Qset of rational right triangles2 [4R2(r2 - r1)2 - (r2 -I ri)2]2 = [4n2(r2 - ri)2 - (r2 -- r1)2]2 r - [4n(r22 - ri2)]2. 1 The distance from the point of contact of Si with the diameter A02 to the end of this diamet ~r, on the side opposite from the point 0,,, is taken as i,.-Editor 2 This result is but a special case of a rational right triangle with sides u2 + v2, u2 - v', and 2uv.-Editor. 1920.] WORK OF PAPPUS AND STEINER ON TANGENT CIRCLES. 5 4. Formulae arising from a series of tangent circles, tangent to two given circles, the first circle in the series being tangent to a line tangent to the smaller circle and perpendicular to the line of centers. Let there be two circles (01) and (02) tangent internally at A and let (01) < (02) and let the tangent to (01) perpendicular to 0102 be drawn and let a series of tangent circles be drawn tangent to (0O) and (02), the first circle in the series being also tangent to the perpendicular just drawn (Fig. 3). Then from Pappus, Book IV., lemma XIX, we have "2 S1^ \ _r2 12 Let A o0 02 B then FIG. 3. 2pi = myi. (10) Using equations (1), (2), (3) and (10) we have by induction rim2 Pn = (n- 1)2m4 + 2(n- 1)m3 + 1' rim3 (2 - m2)[(2n - 3)m + 2] n = [(n - 2)2m4 + 2(n - 2)m3 + 11[(n - 1)2m4 + 2(n - 1)m3 + 1]' If ri approaches r2, the sum sn=i pn approaches ir2 r/2 but is zero at the limit. Also ie= i = 2rl, if ii = pi. Let L Bn = r/2. Then since sin Bn = Pn + Pn+1 we have the relation 1 —m and if n = 1, ri: r = side of decagon inscribed in a circle of unit radius. Here also as in sections (2) and (3) we may obtain an equation [(2n2 - 2n + 1)m4 + (4n - 2)m3 + 212 = [2n2 - 2n)m4 + (4n - 2)m3 + 4m2 - 2]2 + [m(2 - m2)((2n - 1)m + 2)]2, which gives a doubly infinite set of rational right triangles. 5. Formulae arising from a series of tangent circles tangent to a given circle and a given straight line. Let (Cf. figure 3) the series of circles (Sn) be tangent to the perpendicular at B and to 02, (Si) being also tangent to 01. The centers Sn lie on a parabola with vertex 01 and focus 02. Using 0 as origin the equation of the parabola is y2 = 4(r2- rl)x. (11) 6 WORK OF PAPPUS AND STEINER ON TANGENT CIRCLES. [Jan., Moreover Pn = rl - Xn (12) and (Yn - Yn-)2 = (Pn + Pn-1)2 - (Xn - xn-1)2. (13) By a substitution from (12) equation (13) reduces to (Yn - Yn-1)2 = 4PnPn-. (14) Let ri = Xr2. Let Dn denote the sum of the odd-numbered terms in the expansion of (1 + i/X)n: and Nn the sum of the even-numbered terms, that is Dn + Nn = (i + /X)n, Dn- Nn= (l -X)" Then using equations (11), (12) and (14) and induction ( Nn2\ = 1 —n2 r___, 2N, Yn = D +1(r2 - ri). (15) From (15) 2A (1 - X)?'lr2 - Yn ( r2- 4rj)n 2 4(1 - X)rlr2 + yn ( 2 + - i)' Therefore 2 j(1 -X)rll2 - Yn r2 -r n=i 2 (l - X)rTr2 + y,, 2 r Also we have c0 Z (Yn - Yn-1) - 2 4ri(r2- ri). n=l 6. Some properties of conics associated with three mutually tangent circles. Let there be two circles (01) and (02) tangent internally at A (Fig. 4) and let the radii of these circles be ri and r2 respectively, and let the radius of a circle tangent to these two and having its center on the line 0102 be r3 and let its center be 03. Let 0 be the center of any circle tangent to (01) and (02) and let r be its radius. The conic associated with (0) and (02) is an ellipse with the points 0 and 02 as foci, and passing through 01: the conic associated with (0) and (01) is a hyperbola passing through 02 and having 0 and 01 as foci. Likewise we will have an ellipse passing through 0 and having 01 and 02 as foci. Draw from A the line A T tangent to the circles (01) and (02). Then with the three circles (01,) (02) and (0) the straight line A T there will be associated four parabolas1 two of which pass through A, one through 01 and the fourth through 02. See article "Some Properties of a Straight Line and Circle and their Associated Parabolas," Annals of Math., second series, Vol. 19, pp. 174-5. Also " Some properties of circles and related conics," Annals of Math, second series, vol. 20, pp. 279-280. 1920.] WORK OF PAPPUS AND STEINER ON TANGENT CIRCLES. 7 Call the conic through 02, H2, the one through 01, E1 and the one through 0 and 03, E3, and the parabolas through 01 and O2, P1 and P2 respectively. With this notation the following may be readily proved analytically: THEOREM: The normals to E1 and H2 at the points 01 and 02 intersect on a line through 03 perpendicular to 0102. THEOREM: The tangents to H2 and ~'^vZ / j~~E1 at the points 02 and 01 intersect on the line A T. / \\\\THEOREM: If two circles are tangent internally, and any circle is tangent to \2a \? these two, there are associated with these circles three non-degenerate conics, six points of which are the three points of / \ \ contact of the three circles and the three \ \ \//centers of the three circles, and the six tangents to the conics at these six points A 1'f -2 DP UO _B pass through a common point. Proof: Let the normal to E1 at 01 ^~G/V^~~ ~and to H12 at 02 intersect at N, and let the FIG. 4. intersection of the tangents at these points be T. Let the point of contact of (0) and (01) be Ci and of (0) and (02) be C2. Then the angle AOIC1 is bisected by OiT. Therefore a line from T to Ci will be tangent to (01) and (0). Moreover since T and N are ex-centers of the triangle 00102, T, N, and 0 are collinear. It is also evident that the tangent to (02) at C2 will pass through T. We have therefore the six lines TA, TO, TO1, TC2 TO2, and TC1, and these lines are the six tangents to E3, E1 and H2 at the points A, O, 01, C2, 02, C0, THEOREM: E1 and P1 have the same normal at 01. Proof: Since 0 is the focus of P1 and the axis is parallel to 0102, then the bisector of the angle 00102 will be normal to P1. But this is also normal to El because 0 and 02 are the foci of E1. THEOREM: The three axes of the three non-degenerate conics associated with three tangent circles, and the three normals at the centers of the circles, meet in points that are collinear. Proof: Let the normal and axis of E3 intersect in F, the normal and axis of H2 intersect in G and the normal and axis of E1 in K. Then since we have the triangle 01020 and the two bisectors of two interior angles and the bisector of the opposite exterior angle, the points F, K and G are collinear. The right angles formed by the tangents and normals at 01 and 02 are inscribed in a semicircle with TN as diameter. Call this circle Ct. Steiner has pointed out the fact that D, B, C1 and C2 are points of a circle C, with center N.1 It is then evident that the tangents to Cn at its points of intersection with Ct pass through T. We then have two sets of coaxial circles Cn and Ct, the centers 1 See reference to Steiner in Introduction. 8 WORK OF PAPPUS AND STEINER ON TANGENT CIRCLES. [Jan., of one being on a line through 03 perpendicular to 0102 and the diameters of the other being segments of tangents to E3 cut off by the tangents at A and 03.1 It should also be noted in this connection that the point T is the pole of the line drawn from the point of tangency of any two of the circles to the center of the third circle with respect to the conic passing through that center. Also if there is drawn at D a line perpendicular to 0102 and TC1 is produced intersecting this line in S, then N, S and 01 are collinear. THEOREM: If three circles are mutually tangent and tangents and normals be drawn to the three associated conies at the points of contact and the centers of the three circles, and if the normals of two of the conics be chosen, these will intersect by twos on a tangent to the third conic. Proof: Consider the lines 01N, 02N, 001, 002. These intersect in the points 0 and N which are on the tangent to E3. THEOREM: The axes and normals to two of the conic", together with the tangents to these two conics drawn at the centers of the circles determine two perspective triangles whose center of perspectivity is the intersection of the axis and normal to the third conic. Proof: Consider the lines 01N, 02N, 991, 002, 01T, 02T. These intersect by twos on the line TN. They may therefore be considered as the sides of two perspective triangles. Let the corresponding sides be 1 T, 02 T, 02N, 01N, 001, 002. These determine the perspective triangles A12A13A23 and B12B13B23 and the center of perspectivity is the point F on 0102 (see Fig. 4 where F is marked). But this point is also on the normal OB12. Corollary: There will be three such T sets of perspective triangles and the centers of perspective will be collinear (second theorem before the last). THEOREM: N and T are double points of an involution, of which 0 and the point (Y), where TN intersects 0102, are a conjugate pair, and therefore 0 and Y are inverse points with respect to the circle Ct. A O The proof of this theorem follows A immediately from a consideration of FIG. 5. the quadrangle 01A1202B12. The following theorems are also evident. THEOREM: If the three axes and the three normals to the three associated 1 See Conics of Apollonius, Book III, prop. 45 (Heath's ed., p. 114). 1920.] WORK OF PAPPUS AND STEINER ON TANGENT CIRCLES. 9 conies be drawn the axis and normal to each conic being taken as corresponding sides, they form two perspective triangles with T as center of perspective. THEOREM: The four axes of perspective of the three circles and the six lines, three of which are normals and the other three are the tangents to the three conies at the three centers of the three circles, determine a quadrangular-quadrilateral configuration, whose diagonal triangle is the triangle determined by the three centers of the three circles. In connection with the above discussion it should be noted that it furnishes a method for constructing points on a conic. For let 01, 02 and 03 be any three points on a line, and let the perpendicular be drawn at 03 and let N be any point in the perpendicular. Let 01 and 02 be points such that we have the order 010203 or 020103. Then from N draw lines to 01 and 02, making the angles N0103 and N0203, and draw from 01 and 02 the lines 010 and 020 such that z N0103 = 001N, Z N0203 = z 002N. Then the point 0 is on an ellipse. If we have the order 010302, 0 will be on a hyperbola, and if 01 or 02 is at infinity we have a parabola. And in each instance 01 and 02 are foci of the conic. This also gives a method for establishing a (1, 1) correspondence between the points of a conic and the points of a straight line. 7. Some Projective Properties of the Figure in Section 2. Let A and C be the ends of the diameter 0102 of the circle (02) (Fig. 1), and let there be drawn from A lines to Sn and from C lines to Sn/, and let C and C' be the angles that these lines make with AC. Then 2nra3 tan C = - ' (16) ri + r2 2nrl tan C'= 4 —. (17) r2 - rs By means of equations (16) and (17) we may find the equations of the lines ASn and CSn', a solution of which reveals the fact that the line ASn and the line CSn' intersect on a line through S1 perpendicular to AC in points whose ordinates are 2pn.l By very simple analytical considerations we may prove the following THEOREM: The triangles Sn+-lSSnSn and Sn+l'Sn'Sn-' are perspective and their center of perspective is the external center of perspective of the circles 01 and 03. THEOREM: The locus of the point of contact of two tangent circles which are tangent to two given tangent circles (internally tangent) is a circle whose center is the center of perspective of the two given circles and whose radius is the harmonic mean between the radii of the two given circles. 1 In this connection see Steiner, p. 69 and ff. 10 WORK OF PAPPUS AND STEINER ON TANGENT CIRCLES. Proof: The center of perspective, P3, of the two given circles, 01 and 02, has the same power with respect to all circles tangent to these two in a given way. Therefore, the locus of the point of contact of any two such circles which are tangent to each other is a circle orthogonal to them all. THEOREM: The circle with P3 as center and P3A as radius cuts every C, orthogonally. Proof: Let there be drawn with T as center and TA as radius a circle. This will pass through C1 and C2. Therefore C1C2P3 will be the radical axis of the circle just drawn and Cn. The circle with P3 as center and P3A as radius is orthogonal to the circle with center T. It is therefore orthogonal to every C,. LIFE I, JAMES HENRY WEAVER, was born in Madison County, Ohio, June 19, 1883. I received my elementary education in the public schools of Ohio. I was granted the A.B. degree at Otterbein College in 1908 and the M.A. degree at Ohio State ITniversity in 1911. I taught in the rural schools of Ohio during the years 1900-1904; was a Teacher and Principal in the High School at Plain City, Ohio, 1908-1910; was substitute instructor in the Department of Mathematics at Ohio State University, 1910-1912; and Head of Mathematics Department of West Chester High School, West Chester, Pa., 1912-1917. I have published a number of short articles in the Annals of Mathematics, Bulletin of the American Mathematical Society, and School Science and Mathematics. The present paper was undertaken at the suggestion of Professor M. J. Babb, whose criticisms and suggestions have been most valuable.