Circle Chains Inside a Circular Segment
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1 Forum eometricorum Volume 9 (009) FRUM EM ISSN ircle hains Inside a ircular Segment iovanni Lucca bstract. We consider a generic circles chain that can be drawn inside a circular segment and we show some geometric properties related to the chain itself. We also give recursive and non recursive formulas for calculating the centers coordinates and the radius of the circles.. Introduction onsider a circle with diameter center and a chord perpendicular to (see Figure ). Point is the intersection between the diameter and the chord. Inside the circular segment bounded by the chord and the arc it is possible to construct a doubly infinite chain of circles each tangent to the chord and to its two immediate neighbors. (X Y ) (X 0 Y 0) (X Y ) Figure. ircle chain inside a circular segment Let (a + b) be the diameter of the circle and b the length of the segment. We set up a cartesian coordinate system with origin at. eginning with a circle with center (X 0 Y 0 ) and radius r 0 tangent to the chord and the arc we construct a doubly infinite chain of tangent circles with centers (X i Y i ) and radius r i for integer values of i positive and negative. Publication Date: ugust ommunicating Editor: Paul Yiu. The author thanks Professor Paul Yiu for his help in improving this paper.
2 74. Lucca. Some geometric properties of the chain We first demonstrate some basic properties of the doubly infinite chain of circles. Proposition. The centers of the circles lie on the parabola with axis along focus at and vertex the midpoint of. Q Figure. enters of circles in chain on a parabola Proof. onsider a circle of the chain with center (x y) radius r tangent to the arc at Q. Since the segment Q contains (see Figure ) we have by taking into account that has coordinates (b a0) and From these we have Q = a + b = (x b + a) + y Q = r = x = Q Q. (x b + a) + y = a + b x which simplifies into y = 4a(x b). () This clearly represents the parabola symmetric with respect to the x-axis vertex (b 0) the midpoint of and focus (b a0) which is the center of the given circle.
3 ircle chains inside a circular segment 75 Proposition. The points of tangency between consecutive circles of the chain lie on the circle with center and radius. T i Q i Figure 3. Points of tangency on a circular arc Proof. onsider two neighboring circles with centers (X i Y i ) (X i+ Y i+ ) radii r i r i+ respectively tangent to each other at T i (see Figure 3). y using Proposition and noting that has coordinates ( a0) we have i = (X i + a) + Y i = r i = X i = ( Y i ) 4a + b. ( Y i 4a + b + a ) + Y pplying the Pythagorean theorem to the right triangle i T i we have T i = i r i = 4a(a + b) = =. It follows that T i lies on the circle with center and radius. Proposition 3. If a circle of the chain touches the chord at P and the arc at Q then the points P Q are collinear. Proof. Suppose the circle has center. It touches at P and the arc at Q (see Figure 4). Note that triangles Q and PQ are isosceles triangles with Q = P Q. It follows that Q = QP and the points P Q are collinear. i
4 76. Lucca Q P Figure 4. Line joining points of tangency Remark. Proposition 3 gives an easy construction of the circle given any one of the points of tangency. The center of the circle is the intersection of the line Q and the perpendicular to at P. 3. oordinates of centers and radii Figure 5 shows a right triangle i i K i with the centers i and i of two neighboring circles of the chain. Since these circles have radii r i = X i and r i = X i respectively we have (X i X i ) + (Y i Y i ) = (r i + r i ) = (X i + X i ) Making use of () we rewrite this as or 4a(a + b) Y i (Y i Y i ) = 4X i X i. (Y i Y i ) = 4 (b Y i )(b Y 4a i 4a ) (a + b)y i 4ab a 4a Yi Y i Y i + = 0. () If we index the circles in the chain in such a way that the ordinate Y i increases with the index i then from () we have Y i = ( Y Y i i a ) 4b + b a ( + b a Y i 4a ). (3a) This is a recursive formula that can be applied provided that the ordinate Y 0 of the first circle is known. Note that Y 0 must be chosen in the interval ( ab ab).
5 ircle chains inside a circular segment 77 i K i i Figure 5. onstruction for determination of recursive formula Formulas for the abscissa of the centers and radii are immediately derived from () i.e.: X i = r i = Y i + b. 4a (4) Now it is possible to transform the recursion formula (??) into a continued fraction. In fact after some simple algebraic steps (which we omit for brevity) we have Y i = a + b a. Y i a + + b a (5) Defining α = + b a and Z i = Y i a + b a (6) for i =... we have Z i =. α + Z i Thus for positive integer values of i Z i = α where we have used Z 0+ in place of Z 0 α Z 0+ = Y 0 a... α+z 0+ + b a.
6 78. Lucca This is to distinguish from the extension of the chain by working the recursion (3a) backward: ( ) Y Y i + i a 4b + b a Y i = ). (3b) ( + b a Y i 4a Thus for negative integer values of i with Z i = Y i a + + b a we have Z i = α α... α+z 0 where Z 0 = Y 0 a + + b a. It is possible to give nonrecursive formulas for calculating Y i and Y i. For brevity in the following we shall consider only Y i for positive integer indices because as far as Y i is concerned it is enough to change in all the formulae involved α into α Z i into Z i and Z 0+ into Z 0. Starting from (5) and by considering its particular structure one can write for i = 3... Z i = Q i (α) Q i (α) where Q i (α) are polynomials with integer coefficients. ere are the first ten of them. Q 0 (α) Q (α) α + Z 0+ Q (α) (α ) + αz 0+ Q 3 (α) (α 3 α) + (α )Z 0+ Q 4 (α) (α 4 3α + ) + (α 3 α)z 0+ Q 5 (α) (α 5 4α 3 + 3α) + (α 4 3α + )Z 0+ Q 6 (α) (α 6 5α 4 + 6α ) + (α 5 4α 3 + 3α)Z 0+ Q 7 (α) (α 7 6α 5 + 0α 3 4α) + (α 6 5α 4 + 6α )Z 0+ Q 8 (α) (α 8 7α 6 + 5α 4 0α + ) + (α 7 6α 5 + 0α 3 4α)Z 0+ Q 9 (α) (α 9 8α 7 + α 5 0α 3 + 5α) + (α 8 7α 6 + 5α 4 0α + )Z 0+ ccording to a fundamental property of continued fractions [] these polynomials satisfy the second order linear recurrence Q i (α) = αq i (α) Q i (α). (7) Equation (3b) can be obtained by solving equation () for Yi.
7 ircle chains inside a circular segment 79 We can further write Q i (α) = P i (α) + P i (α)z 0+ (8) for a sequence of simpler polynomials P i (α) each either odd or even. In fact from (7) and (8) we have P i+ (α) = αp i+ (α) P i (α). Explicitly i = 0 P i (α) = i k=0 ( ) i +k( i +k ) k α k i = i+ i+ i ( ) +k( +k ) α k i = From (6) we have for i =.... References k= k ( Y i = a α Q ) i (α) Q i (α) []. Davenport igher rithmetic 6-th edition ambridge University Press 99. iovanni Lucca: Via orvi Piacenza Italy address: vanni lucca@inwind.it
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