Mathematica Moravica
Vol. 12 -2 (2008), 35–43
Sequence with K1 , K2 , Kn , Kn+1
Mutually Tangent Circles
Milorad R. Stevanović
Abstract. In this article is given the formula for radius of circle Kn ,
where in sequence {Kj }, four circles K1 , K2 , Kn , Kn+1 , for all n ≥ 3,
are mutually tangent. Radius rn is expressed in terms of radii r1 , r2 ,
r3 .
Four circles K1 , K2 , K3 , K4 , with centers and radii Oj , rj (j = 1, 2, 3, 4)
are mutually tangent what means that each of them is tangent to other
three. From Descartes-Soddy formula
1
1
1
1
=
+
+
+2
r4
r1 r 2 r 3
(1)
r
1
1
1
+
+
r1 r2 r2 r3 r3 r1
we get r4 = f (r1 , r2 , r3 ). If we, as in Fig. 1, inscribe circles K5 , K6 , . . . , Kn−1 , Kn
then we have rk = f (r1 , r2 , rk−1 ) for all k = 4, . . . , n.
The following problem appeares: Is it possible to express rn in closed form
as a function of first radii r1 , r2 , r3 ?
Theorem 1.
(2)
1
1
1
+
+ +
r 1 r2
r3
r
1
1
1
+ 2(n − 3)
+
+
,
r1 r2 r 2 r 3 r 3 r 1
1
= (n − 3)2
rn
(n ≥ 4).
2000 Mathematics Subject Classification. Primary: 51M99, 51M04.
Key words and phrases. Sequences of circles, arbelos, Pappus chain.
35
c
2008
Mathematica Moravica
36
Sequence with K1 , K2 , Kn , Kn+1 Mutually Tangent Circles
Fig. 1.
Proof. Formula (2) can be proved by induction. For n = 4 we have DescartesSoddy formula. Also, we have
r
1
1
1
1
1
1
1
=
+
+
+2
+
+
=
rk+1
r1 r2 rk
r 1 r 2 r2 rk
rk r1
1
1
1
1
1
2
=
+
+ (k − 3)
+
+ +
r1 r2
r1 r2
r3
s
r
1
1
1
1
1
1
+ 2(k − 3)
+
+
+2
+
+
M=
r1 r2 r 2 r 3 r 3 r 1
r1 r2
r1 r2
r
1
1
1
1
1
1
2
= [(k − 3) + 1]
+
+
+ 2(k − 3)
+
+
+
r1 r2
r3
r 1 r 2 r2 r3 r 3 r 1
r
1
1
1
1
1
+ 2 (k − 3)
+
+
+
+
=
r1 r 2
r1 r 2 r 2 r 3 r3 r1
r
1
1
1
1
1
1
2
= (k − 2)
+
+
+ 2(k − 2)
+
+
,
r 1 r2
r3
r 1 r 2 r 2 r 3 r3 r1
37
Milorad R. Stevanović
where
2
M = (k − 3)
1
1
+
r1 r 2
r
1
1
1
1
+
+ 2(k − 3)
+
+
,
r3
r1 r2 r2 r3 r3 r1
which proves formula (2).
Formula (2) can be used in finding radii rn in various configurations. Some
Arbelos configurations of inscribed circles will be considered.
Case 1. (Fig. 2):
r 1 + r2 = r 0 ,
1
1
1
1
+
=
+ ,
r 3 r0
r1 r2
r
1
1
1
1
1
+
+
=
+ ,
r 1 r 2 r 2 r 3 r3 r1
r1 r2
1
1
1
1
2
= (n − 2)
+
− .
rn
r1 r 2
r0
Fig. 2.
Case 2. (Fig. 3):
r 1 + r2 = r 0 ,
1
1
1
1
+
=
+ ,
r 3 r0
r1 r2
r
1
rk+1
1
1
1
1
1
−
−
=
− ,
r 1 r 3 r 3 r 0 r0 r1
r1 r0
r
1
1
1
1
1
1
=
−
+
+2
−
−
,
r 1 r0 r k
r1 rk
r k r0 r 0 r 1
(k ≥ 3).
Formula similar to formula (2) with r12 → − r10 in this case is
r
1
1
1
1
1
1
1
2
0
(2 )
= (n − 3)
−
+
+ 2(n − 3)
−
−
.
rn
r1 r0
r3
r1 r3 r 3 r 0 r 0 r 1
38
Sequence with K1 , K2 , Kn , Kn+1 Mutually Tangent Circles
which leads to the formula for radius of n-th left circle
1
1
1
1
2
= (n − 2)
−
+ ,
rn
r1 r 0
r2
and formula for radius of n-th right circle is
1
= (n − 2)2
rn
1
1
−
r2 r 0
+
1
.
r1
Fig. 3.
Case 3. (Fig. 4):
r 1 + r2 = r 0 ,
r
1
1
1
=
+ ,
r3
r1 r2
1
1
1
−
−
=
r 1 r 3 r3 r 0 r 0 r 1
s
1
r1
1
1
−
,
r 1 r0
From the same recurrence relation as in Case 2 and from formula (20 ) we
get formula for radius of n-th left circle
1
=
rn
r 2
r
1
1
1
1
−
+
+ .
(n − 3)
r1 r0
r1
r2
Similar is the formula for radius of n-th right circle
1
=
rn
r
r 2
1
1
1
1
(n − 3)
−
+
+ .
r2 r0
r2
r1
39
Milorad R. Stevanović
Fig. 4.
Case 4. (Fig. 5):
1
1
1
=
+ ,
r3
r1 r2
s
s
1
1
2
1
1
1
1
2
1
+
+
=
+
+
.
4r1 r2 2 r1 r3 r2 r3
2
r1 r2
r2 r1
r 1 + r2 = r 0 ,
Formula for radius rn is given by formula (2) if we take r1 → 2r1 , r2 → 2r2 .
s
2
1
(n − 3)2 + 2 1
1
1
2
1
=
+
+ (n − 3)
+
+
.
rn
2
r1 r2
r1 r 2
r2 r1
Fig. 5.
Case 5. (Fig. 6):
r 1 + r2 = r 0
1
1
=
r3
4
1
1
−
r 1 r0
+
1
,
r2
40
Sequence with K1 , K2 , Kn , Kn+1 Mutually Tangent Circles
r
1
1
1
1
−
−
=
r1 r3 r 3 r 0 r 0 r 1
2
1
1
−
r1 r0
.
From the same recurrence relation as in Case 2 and from formula (20 ) we
get
1
=
rn
5
n−
2
2 1
1
−
r1 r0
+
1
.
r2
Fig. 6.
Case 6. (Fig. 7):
1
1
1
√ =√ +√ ,
r3
r1
r2
r
1
1
1
1
1
1
+
+
=
+
+√
.
r 1 r 2 r 2 r 3 r3 r1
r 1 r2
r1 r2
Applying of formula (2) yields to
1
= (n − 2)2
rn
1
1
+
r1 r2
+ 2(n − 2) √
1
.
r1 r2
Milorad R. Stevanović
41
Fig. 7.
Theorem 2 (Pappus). If the center of n-th circle with radius rn inscribed
in arbelos (see Fig. 8) is on the distance dn of the base of arbelos then
dn = 2n · rn .
Fig. 8.
Proof. n-th circle inscribed in arbelos in our notation has radius rn+2 and
distance dn+2 from the base of arbelos.Semiperimeter of triangle O1 OOn is
equal to r0 . From Archimedes-Heron formula for area of triangle, because
42
Sequence with K1 , K2 , Kn , Kn+1 Mutually Tangent Circles
of r1 + r2 = r0 , we have
p
2 r0 r1 rn (r2 − rn ) = r2 · dn ⇒
s
1
r0 r1 1
−
=
dn =2rn
r2
rn r2
r
r0 r1
r2
=2rn
(n − 2)2
= 2(n − 2)rn .
r2
r0 r1
In the proof is used the formula for radius of n-th left circle in arbelos.
Theorem 3. If the circles Kn , Kn+1 , Kn+2 and Kn+3 are four consecutive
circles from our sequence then for theirs radii we have
1
1
1
1
−3
+3
−
= 0.
rn
rn+1
rn+2 rn+3
Proof. Radii of these circles are given by unique formula
r
1
1
1
1
1
1
1
2
= (k − 3)
+
+ 2(k − 3)
+
+
,
+
rk
r1 r 2
r3
r 1 r 2 r2 r3 r 3 r 1
(k = n, n + 1, n + 2, n + 3).
The desired formula follows after elimination of r1 , r2 and r3 from upper
four formulas.
References
[1] M.G. Gaba, Amer. Math. Monthly, Vol. 47, 1940,pp.19-24.
[2] V. Thebault, Amer. Math. Monthly, Vol. 47, 1940, p.640.
[3] L. Bankoff, How Did Pappus Do It, The Mathematical Gardner, Pridle, Weber &
Schmidt, 1981, pp.112-118.
[4] L. Bankoff, The Marvelous Arbelos,The Lighter Side of Mathematics, Mathematical
Association of America, 1994, pp.247-253.
[5] C.W. Dodge, Thomas Schoch, Peter Y. Woo, Paul Yiu, Those Ubiquitous
Archimedean Circles, Mathematics Magazine of Mathematical Association of America, Vol. 72, No 3, June 1999, pp.202-213.
[6] D. Klingens, http://www.pandd.demon.nl/arbelos.htm
[7] T. Schoch, A Dozen More Arbelos Twins,
http://www.biola.edu/academics/undergrad/math/woopy/arbel2.htm
[8] E.W. Weisstein,
http://mathworld.wolfram.com/arbelos.htm
Milorad R. Stevanović
43
[9] P. Woo, The Arbelos,
http://www.biola.edu/academics/undergrad/math/woopy/arbelos.htm
Milorad R. Stevanović
Technical Faculty
Svetog Save 65
32000 Čačak
Serbia
E-mail address: [email protected]