If we define in as the nth intercept of the series (Sn) (1, projection of O3S1 on AC), then (4) and (5) are the formulæ for the nth radius and intercept in the series (Sn). An interchange of r1 and rз will give the corresponding formulæ for the series (S'), while an interchange of r2 and rз with r2 considered negative will give the corresponding formulæ for the set (S,"). We will now establish the following theorem. THEOREM: If two circles (01) and (O2) are tangent internally, and a circle (S) is drawn tangent to these two, such that SO; (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', O1 and O2 determine a rectangle. Proof: Let SO2 be perpendicular to 0102, and let the coördinates of S and S' be x, y and x', y' respectively and let the radii be p and p'. Then THEOREM: If in the series (Sn), the points Sn, Sn+1, 01 and O2 determine a rectangle, then r1 = nr3. Proof: Equate the values of p given in equations (4) and (6). Let the foot of the perpendicular from S to 0102 be Pn. Let angle PnSnSn+1 = Bn. n Then if r1 = kr3 (k an integer or a rational fraction) tan Bn = (2k+ 1)(2n+1) 2(n2+nk - k2) (7) and the slopes of the lines of successive centers of the series (Sn) are all rational. Moreover if in (7) k is an integer, B = π/2, and From the identity (Pn + Pn+1)2 = in+12 + (Yn − Yn+1)2 we get by using equa tions (3), (4) and (5) the equation [(2n2+2n+1)г32 + 2r1rз + 2r12]2 = [(2n2 + 2n)r32 - 2r1rз 2r1212 + [(2n + 1)r3(r2 + r1)]2 giving a triply infinite set of rational right triangles. 3. Formulæ 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 r1 and of (02) be r2, and let the centers of the circles be Sn. Then from equations (1), (2) and (3), since y1 = pi we get by induction 4(n2 - n) (r2- r1)2 + (r2 + r1)2 32(n-1)rir2(r2 — r1)2(r2 + r1) [4(n2 — n) (r2 — r1)2 + (r2+r1)2][(4n2 — 12n + 8) (r2 − r1)2 + (r2 + r1)2] where for in n ≥ 2. (8) (9) THEOREM: If in this series the points Sn, Sn+1, O1 and O2 determine a rectangle Also the equation (Pn + Pn+1)2 = (in+1)2 + (yn yn+1)2 gives the triply infinite set of rational right triangles2 [4n2(r2 − r1)2 + (r2 + r1)2]2 = [4n2(r2 — r1)2 — (r2 + r1)2]2 + [4n(r22 — r12)]2. 1 The distance from the point of contact of S1 with the diameter AO2 to the end of this diameter, on the side opposite from the point O2,, is taken as i.-Editor 2 This result is but a special case of a rational right triangle with sides u2 + v2, u2 — v2, and 2uv.-Editor. 4. Formulæ 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 (O2) 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 (01) 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 Using equations (1), (2), (3) and (10) we have by induction. and if n = 1, r1: r2 = side of decagon inscribed in a circle of unit radius. Here also as in sections (2) and (3) we may obtain an equation which gives a doubly infinite set of rational right triangles. 5. Formulæ 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 O2, (S1) being also tangent to 01. The centers Sn lie on a parabola with vertex 01 and focus O2. Using 01 as origin the equation of the parabola is = Let r1 Ar2. Let Dn denote the sum of the odd-numbered terms in the expansion of (1 + √)": and N, the sum of the even-numbered terms, that is Then using equations (11), (12) and (14) and induction 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 r1 and r2 respectively, and let the radius of a circle tangent to these two and having its center on the line 0102 be rз and let its center be 03. Let O be the center of any circle tangent to (01) and (02) and let r be its radius. The conic associated with (0) and (O2) is an ellipse with the points 0 and 02 as foci, and passing through O1: the conic associated with (0) and (01) is a hyperbola passing through О, and having 0 and 01 as foci. Likewise we will have an ellipse passing through O and having O1 and O2 as foci. Draw from A the line AT tangent to the circles (01) and (02). Then with the three circles (01,) (02) and (0) the straight line AT there will be associated four parabolas1 two of which pass through A, one through 0, and the fourth through 02. 1 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. Call the conic through O2, H2, the one through O1, E1 and the one through O 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 O1 and O2 intersect on a line through 03 perpendicular to 0102. C2 F P 0. FIG. 4. 2 THEOREM: The tangents to H2 and E at the points 02 and 01 intersect on the line AT. THEOREM: If two circles are tangent internally, and any circle is tangent to 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. pass through a common point. Proof: Let the normal to E1 at 01 and to H2 at O2 intersect at N, and let the intersection of the tangents at these points be T. Let the point of contact of (0) and (01) be C1 and of (0) and (02) be C2. Then the angle AO1C1 is bisected by O1T. Therefore a line from T to C1 will be tangent to (01) and (0). Moreover since T and N are ex-centers of the triangle 00102, T, N, and O 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, 0, 01, C2, O2, C1. THEOREM: E1 and P1 have the same normal at 01. 1 1 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 E1 because 0 and O2 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 O1 and O2 are inscribed in a semicircle with TN as diameter. Call this circle C. Steiner has pointed out the fact that D, B, C1 and C2 are points of a circle Cn with center N.1 It is then evident that the tangents to C, at its points of intersection with C pass through T. We then have two sets of coaxial circles Cn and Ct, the centers. 1 See reference to Steiner in Introduction. |