121
A FIVE-CIRCLE PROBLEM
Hiroshi Okumura
The recent book [1] introduced the Japanese mathematics of the 18th and 19th
centuries to the West. This mathematics deals with such unhackneyed and challenging
problems as that book's Example 1.4 [1, pages 7 and 84]:
P R O B L E M : In the circle C of radius r let AB be a chord whose midpoint is M.
The circle C0 of radius r0 (r0 < r/2) touches AB at M and also touches C internally.
Let P be any point on AB distinct from A, B and M; a circle C2 of radius r0 (equal to
the radius of Co) touches AB at P on the other side of AB. Distinct circles C\ and C$
of radii r± and r 3 touch AB, and each touches C internally and C2 externally. Show that
r = r1 + r2 + r3.
Figure 1
T h e purpose of this note is to generalize the problem and display periodic patterns
of circles based on it. We will see that neither the restriction r 0 < r/2 nor P ^ A, 5 , M
are needed for the desired relation among the radii in the problem. We need the following
easy lemma.
L E M M A : (i) Common tangents of externally touching circles whose radii are a
and b have length 2\/ab.
(ii) Let C be a circle of radius r with a chord AB, and CQ a circle of radius r0 touching AB at the midpoint M of AB and touching C internally. Then AM = 2 y (r — rQ)r0.
W i t h the aid of the lemma, we get a generalization of the problem as follows.
T H E O R E M 1: Let i b e a secant of a circle C of radius r; let Ci, C 2 , &3 be circles
on one side of t and tangent to it, with Ci and C 3 internally tangent to C, while C 2 is
externally tangent to C\ and C3. Let C0 be the circle internally tangent to C on the other
side of t, and tangent to t at the midpoint of the segment cut from it by C. If C% has
radius rt- and r 0 = r 2 , then
r = rt + r2 + r 3 .
(1)
122
Our proof will be an adaptation to our more general setting of the proof found in
[1, p. 84], which the authors reproduced from an 1864 Japanese book.
Figure 2
Figure 3
Proof: Let 0 , 0 ; be the centers of the circles C and Ci and F,F{ the feet of the
perpendiculars from 0 , 0 ; to t. From the right triangle 00\G\ formed by the lines 0 0 i ,
0 i F i and the line.parallel to t through 0 , we get
FXF = G10 =
v
/(r-r1)2-(r-2r2-r1)2 =
2y/(r-r2-rx)r2.
Similarly we obtain JF 3 J F = 2w(r — r2 — r3)r2. Also we have F1F2 = 2- v /r 1 r 2 and F2F3 =
2<\/r2r3 by the lemma.
There are two cases to be considered according as F2 lies inside C or not (see Figures
2 and 3). We claim that in the former case, JF lies between F\ and F3. For otherwise, we may
assume t h a t FUF2, F3, F lie in this order. Then we have FXF3 = F1F2+F2FZ =
FtF-F3F.
Substituting, we get
ZVrV^ + 2^/r^
= 2<J(r-r2
- r ^
- 2y/(r - r 2 -
r3)r2
or
V^T - y/r - r
2
- rx = -y/rl
- y/r - r2 - r 3 .
Squaring this, we obtain —y/riy/r — r2 — rx = y/r^y/r — r2 — r 3 , a contradiction. Therefore
F lies between i ^ and F3 as claimed. In the other case we may assume that Fx lies between
F2 and JP3, which implies that JF3 must lie between JP and Fx\ For since r 2 + r 3 < r w e get
r 2 r 3 < ^ ( r ~ r2), or J P 2 F 3 = 2 y / r 2 r 3 < 2 y ( r — r 2 )r 2 . Therefore F 2 F 3 is less t h a n or equal
to one-half of the chord by the lemma; consequently F2F3 < F2F.
Therefore we have FXF3 = FXF2 + F2F3 = i ^ F + JFJF 3 in the former case, and
F1F3 = - F i i ^ + F 2 F 3 = FXF - F F 3 in the latter. Thus we get
123
Therefore we have
{±y/r\
- ^r-r2~
r x ) 2 = ( ± y V ~ r2 - r3 - y^s) 2 -
Expanding the equation and rearranging we get
(Vi - r3)(r - n - r
2
- r 3 ) = 0.
Therefore, when r^ ^ r3 we have (1). Suppose that rx = r3. This happens only in the
former case. Then we obtain FF3 = F2F3 or 2 y (r — r 2 — r 3 )r 2 = 2 % /r 2 r 3 . Hence we get
r = r 2 + 2r 3 and the proof is now complete.
•
If C2 touches C externally in the theorem, then C1 and C3 coincide. Therefore we
get:
C O R O L L A R Y : Let t be a secant of a circle C of radius r, and C 0 a circle touching
C internally and touching t at the midpoint of the segment cut from it by C. On the other
side, circles C\ and C2 touch t and touch C internally and externally respectively at the
same point. If the radius of C% is T{ and r0 = r2, we have
r = 2rx + r 2 .
The following theorem shows that the external tangents of C§,C\ and (7o,C 3 are
equal in Theorem 1. A related result can be seen in [2].
T H E O R E M 2: Let C and Co be circles with C0 inside and tangent to C, and let
the chord k of C be tangent at its midpoint to C0. Then
(i) If the circle C\ is tangent to k on the other side from C0 and touches C internally,
then an external common tangent of C0 and G\ is half as long as k\ and
(ii) if C\ touches the extension of k on the same side as C0 and touches C externally,
then an internal common tangent of C0 and C\ is half as long as k.
124
Proof: Let C\ be an arbitrary circle touching t on the other side and C internally,
and F,Fi,Oi,r,r{
as in the proof of Theorem 1 (see Figure 4). Then we have (FFi)2 =
4(r — r0 — ri)r0 as we saw in the proof of Theorem 1, and (OoOi) 2 = (FFi)2 + (r 0 + n ) 2
by the Pythagorean theorem. Therefore the square of the external tangents of C0 and C\
is equal to
(OQOX)2
- (r 0 - r 2 ) 2 = 4(r - rQ -
ri)r0
+ (r 0 + rxf
- (r 0 - rxf
= 4(r -
r0)r0.
Hence the tangents are equal to 2 y (r — r0)r0, one-half of k. (ii) is proved similarly.
•
Returning to Theorem 1, if (73 degenerates to a point circle in t h a t theorem, C2
touches t at an intersection of C and t. Since r = r 0 + r\ in this situation, the centers of
C, Co and Ci are collinear. This suggests a converse: Let C be a circle and k its chord,
Co and C\ circles touching C internally and also touching k at the midpoint on opposite
sides. Then the radius of the circle touching k at an end of k on the other side from Co
and touching C\ externally is equal to the radius of C0. But we can prove a more general
result (see Figure 5):
T H E O R E M 3: Let Co and C\ be externally (resp. internally) touching circles,
and C the circle touching the two at points different from the point of tangency of Co and
C\ such t h a t the three centers are collinear. For any chord k of C perpendicular to the
line through the three centers, there is a circle of radius equal to the radius of Co touching
k at an end of k and touching C\ externally (resp. internally).
Figure 5
Figure 6
Proof: Let us first assume that Co and C\ touch externally; let r 0 and rx be their
radii. O and 0\ are the centers of C and C\ respectively (see Figure 6). Let us draw
a chord k of C perpendicular to the line through the three centers and a new circle of
radius r0 and center A touching C\ externally. If the circle touches k at B and the circle
is drawn such that 0\0 and AB have the same orientation, then 0\0 and AB are equal
and parallel, since 0\0 = r0. Hence 00\AB
is a parallelogram and we get OB = r 0 + 7V
Therefore B lies on C. Thus the theorem follows from the uniqueness of the figure. The
internal case is proved similarly.
•
125
Figure 7
Figure 8
We now extend the three-circle pattern of Theorem 3 to the entire plane (as in
Figures 7 and 8). Let • • •, I } _ 2 5 ^ - i 5 A ) 5 A b £ > 2 ? ' " ' be distinct circles such that all the
centers lie on a line, and Di and D{+i touch externally (resp. internally) and the radii of
D2i and 2?2t+i a r e equal to ro and r i , the radii of DQ and D\ respectively. For each pair
of Di and Z?t-+1 let us draw another circle D2>"+i touching to the two at points different
from the point of tangency of Di and Z)t-+1 such that the three centers are collinear. By
Theorem 3, there is a translation T such that DQ touches Dt externally (resp. internally)
and intersects D 0 ,i at a point where the tangent of DQ is perpendicular to the line through
the centers of DQ and D\. If we draw the images of the whole figure by the translations
T , T 2 , • • •, and I 1 " 1 , T~ 2 , • • •, we get Figure 7 (resp. Figure 8).
Let us observe that T equals the product of two translations Tx and Ty, where Tx
has the direction perpendicular to the line joining the centers of D0 and D l 9 and Ty has
the direction parallel to the line (see Figure 9). If the lengths of Tx and Ty are dx and dyi
then we have dx = Jr2 — (dy — r)2 = Jdy(2r — dy) by the Pythagorean theorem, where
r is the radius of Do,i- With this fact it is easily seen that our pattern has the following
properties.
1. ''',D^2,DotuD\^1,DT4^3i-',D^^2n+1,-'
are tangent successively and the
centers are collinear,
2. • • •, jDjf
, A ) , £^i? DZ.21'''
3. D*Ei o a n d Df2
are
tangent successively and the centers are collinear,
touch at a point where D„li0
and J91?2 also touch.
126
Figure 9
Figure 10
In Figure 7, if the line joining the centers of D^i o ai*d D\ is perpendicular to the
line joining the centers of D0 and D1, then D^10 passes through the point where D1 and D2
touch. This will happen if and only if dy = r1 + r = r0 + 2r x . Moreover we can adjust the
ratio of r 0 and rx so that D3^ and DQX touch externally in this situation. Since JD3?4 can be
obtained by translating Do,i through the distance 2r—2r1 = 4r 0 +2ri along the line through
the centers of D0 and Di, and d2x = dy(2r — dy) = (r 0 + 2ri)r 0 , the square of the distance
between the centers of D3?4 and D^ is equal to (2dx)2 + (2dy — 4r 0 — 2r1)2 = 4(2r*o + r\).
We also get
4(2r* + rl) - (2r) 2 = A(2r20 + r\) ~ (2r0 + 2r x ) 2 = 4r 0 (r 0 - 2r x ).
Therefore D 3)4 and DQX touch externally if and only if r 0 : r\ = 2 : 1 (see Figure 10).
Acknowledgement: The author would express his thanks to the referees for their
helpful comments and for providing the neat drawings of Figures 7, 8 and 10.
References:
[1] H. Fukagawa and D. Pedoe, Japanese Temple Geometry Problems, Charles Babbage
Research Center, Winnipeg, Canada, (1989).
[2] Solution of the Problem 1155, Mathematics Magazine 57 (1984), p. 47.
Maebashi City College of Technology
460 Kamisadori Maebashi Gunma 371, Japan