Radii of Circles in Apollonius' Problem

14 Pages·2017·0.12 MB·English

ForumGeometricorum
b
Volume17(2017)359–372.
bb b
FORUMGEOM
ISSN1534-1178
Radii of Circles in Apollonius’ Problem
MiloradR.Stevanovic´ ,PredragB.Petrovic´,andMarinaM.Stevanovic´
Abstract. ThepaperpresentstherelationforradiioftheeightcirclesinApollo-
nius’problemforcircleswhicharetangenttothreegivencircles. Analogously,
we derived the relations for radii of the 16 spheres which are tangent to four
givenspheres,withcoordinatesoftheircentersandwiththeirradii.
1. Introduction
Itiswellknownthatforthreegivencirclesgenericallythereareeightdifferent
circlesthataretangenttothem. Theproblemofrulerandcompassconstructability
oftheseeightcirclesiswell-known. TherearefamousApollonius’andGergonne’s
solutionstothisproblem[3]. Specialcasesofthethreegivencirclesareconsidered
and a number of other problems is known [2]. The first case is when we consider
three sides of the original triangle as three circles with infinite radii. The incircle
and three excircles of the original triangle are four solutions to Apollonius’ prob-
lem. The second case is when we have three excircles as a starting point. Three
sides of the original triangle are three solutions to Apollonius’ problem with infi-
nite radii [1]. The nine-point circle is tangent externally to the three excircles, by
Feuerbach theorem, and a relatively new object - the Apollonius circle is tangent
internallytothreeexircles(forsomeresultsaboutthiscirclesee[4]-[7]). Tothese
five circles we can add three Jenkins circles which are tangent to three excircles,
byaddingtwoofthemexternallyandthethirdoneinternally.
2. Mainresult
LetusassumethatthethreegivencirclesareK1(O1,r1),K2(O2,r2),K3(O3,r3),
Figure 1, with distances between centers O2O3 = a, O3O1 = b, O1O2 = c and
with the area (O1O2O3) = ∆, which is different from 0. The following theorem
holdstrue:
Theorem1. LetusassumethattheradiioftheeightcircleswithcentersS given
i
inFigure1((a),(b),(c)and (d))arep ,(j = 1,2,...,8). Then
j
1 1 1 1 1 1 1 1
− + − + − + − = 0. (1)
p1 p2 p3 p4 p5 p6 p7 p8
Proof. Letusintroducetheanglesϕ1 = ∠O2SO3,ϕ2 = ∠O3SO1,ϕ3 = ∠O1SO2,
whereS = S1 forFigure1(a),soastoobtain
2 2 2
cos ϕ1+cos ϕ2+cos ϕ3−2cosϕ1cosϕ2cosϕ3 = 1. (2)
PublicationDate:June25,2017.CommunicatingEditor:PaulYiu.
360 M.R.Stevanovic´,P.B.Petrovic´,andM.M.Stevanovic´
O3
O3
b a
S2S1 b a S4
S3
O1 c O2
O1 c O2
1A 1B
O3
S8
b a
O3
O1 c O2
S6 b a S7
S5
O1 c O2
1C 1D
Figure1. Theeightdifferentcirclesthataretangenttothethreegivencircles.
Ifwesubstitute
2 ϕ1 2 ϕ2 2 ϕ3
t1 = sin , t2 = sin , t3 = sin ,
2 2 2
fromrelation(2)wehave
2 2 2
t1+t2+t3−2(t1t2+t2t3+t3t1)+4t1t2t3 = 0. (3)
If we generally denote the center of the circle and radius by S and p, then for
the first unknown circle L1 (see Figure 1 (a)) is SO1 = p+r1, SO2 = p+r2,
RadiiofcirclesinApollonius’problem 361
SO3 = p+r3 and
2 2 2 2 2
(p+r2) +(p+r3) −a a −(r2−r3)
cosϕ1 = =⇒ t1 = ,
2(p+r2)(p+r3) 4(p+r2)(p+r3)
andanalogously
2 2 2 2
b −(r3−r1) c −(r1−r2)
t2 = , t3 = .
4(p+r3)(p+r1) 4(p+r1)(p+r2)
Fromrelation(3)wenowhaverelation(4):
2 2 2 2 2 2 2 2
(a −(r2−r3) ) (p+r1) +(b −(r3−r1) ) (p+r2)
2 2 2 2
+(c −(r1−r2) ) (p+r3)
2 2 2 2
−2(a −(r2−r3) )(b −(r3−r1) )(p+r1)(p+r2)
2 2 2 2
−2(a −(r2−r3) )(c −(r1−r2) )(p+r1)(p+r3)
2 2 2 2
−2(b −(r3−r1) )(c −(r1−r2) )(p+r2)(p+r3)
2 2 2 2 2 2
+(a −(r2−r3) )(b −(r3−r1) )(c −(r1−r2) )
= 0. (4)
orinanotherform
F1(p, r1, r2, r3) = 0. (5)
Equation(4)isoftheseconddegreeandisoftheform
2
A1p +B1p+C1 = 0, (6)
where
2 2 2
A1 = 4(a (r1−r2)(r1−r3)+b (r2−r3)(r2−r1)+c (r3−r1)(r3−r2))
2
−16∆
= f(r1,r2,r3), (7)
2 2 2 2 2 2 2 2 2
B1 = 2{r1(a −(r2−r3) ) +r2(b −(r3−r1) ) +r3(c −(r1−r2) )
2 2 2 2
−(a −(r2−r3) )(b −(r3−r1) )(r1+r2)
2 2 2 2
−(b −(r3−r1) )(c −(r1−r2) )(r2+r3)
2 2 2 2
−(c −(r1−r2) )(a −(r2−r3) )(r3+r1)}
= g(r1,r2,r3), (8)
2 4 2 4 2 4 2 2 2
C1 = r1a +r2b +r3c +a b c
2 2 2 2 2 2 2 2 2 2 2 2
−a b (r +r )−b c (r +r )−c a (r +r )
1 2 2 3 3 1
2 2 2 2 2 2 2 2 2 2
+a (r −r )(r −r )+b (r −r )(r −r )
1 2 1 3 2 3 2 1
2 2 2 2 2
+c (r −r )(r −r )
3 1 3 2
= h(r1,r2,r3). (9)
362 M.R.Stevanovic´,P.B.Petrovic´,andM.M.Stevanovic´
For the second unknown circle L2 (see Figure 1 (a)) we have SO1 = p−r1,
SO2 = p−r2,SO3 = p−r3andacorrespondingequationintheformofequation
(4),and
F1(p, −r1, −r2, −r3) = 0, (10)
2
A2p +B2p+C2 = 0, (11)
with
A2 = f1(−r1, −r2, −r3) = A1,
B2 = g1(−r1, −r2, −r3) = B1,
C2 = h1(−r1, −r2, −r3) = C1,
whichimpliesthat
2 2
A1p1+B1p1+C1 = 0, A1p2−B1p2+C1 = 0,
and
1 1 B1
− = − . (12)
p1 p2 C1
ForthethirdcircleL3 (seeFigure1(b))wehaveSO1 = p−r1,SO2 = p+r2,
2
SO3 = p+r3 and we get F1(p, −r1, r2, r3) = 0 with A3p +B3p+C3 = 0,
A3 = f1(−r1, r2, r3),B3 = g1(−r1, r2, r3),C3 = h1(−r1r2, r3) = C1.
ForthefourthcircleL4(seeFigure1(b))wehaveSO1 = p+r1,SO2 = p−r2,
2
SO3 = p−r3 and we get F1(p, r1, −r2, −r3) = 0 with A4p +B4p+C4 =
0, A4 = f1(r1, −r2, −r3) = A3, B4 = g1(r1, −r2, −r3) = −B3, C4 =
h1(r1,−r2,−r3) = C1 andagainweget
1 1 B3
− = − . (13)
p3 p4 C1
Analogously,wehave
1 1 B5
− = − , (14)
p5 p6 C1
1 1 B7
− = − , (15)
p7 p8 C1
where
B5 = g1(r1, −r2, r3), B7 = g1(r1, r2,−r3).
Formula(1)followsfrom(12),(13),(14),(15)because
g1(r1, r2, r3)+g1(−r1, r2, r3)+g1(r1, −r2, r3)+g1(r1, r2, −r3) = 0.
(cid:3)
Remark1. Iftheindexj ofthecirclewithradiusp isaneven(odd)number,then
j
1/p (informula(1))hasthesign+(−).
j
RadiiofcirclesinApollonius’problem 363
Remark2. Inthefirstcaseofthethreegivencirclesmentionedintheintroduction,
wegettheformula
1 1 1 1
= + +
r r1 r2 r3
wherer,r1,r2,r3 aretheinradiusandexradiioftriangleABC.
Remark3. Inthesecondcaseweget
1 1 1 1 1
+ + + = ,
p1 p2 p3 q m
wherep1,p2,p3aretheradiiofJenkinscircles,qistheradiusofApollonius’circle
andm = R/2istheradiusofEulercircleornine-pointcircle.
Thisformulacanbeprovedindependentlysince
a b c
p1 = q, p2 = q, p3 = q.
b+c c+a a+b
Remark 4. In the same way, the same result can be proved for the three given
circles,providedthattwoofthemareinsideofthethirdone.
3. Positionsof8circles
TheradicalcircleofthethreegivencirclesK1(O1,r1),K2(O2,r2),K3(O3,r3),
is the circle orthogonal to all of them, and pairs of circles (L1,L2), (L3,L4),
(L5,L6), (L7,L8) – Figure 1, are inversive with respect to the radical circle. For
thisradicalcircleK0(S0,r0),Figure2,thefollowingholdstrue.
O3
b a
S1
S0
S2
O1 c O2
Figure2. TheradicalcircleofthethreegivencircleswiththecirclesL1andL2,
inversivetotheradicalcircle,basedonFigure1(a)
364 M.R.Stevanovic´,P.B.Petrovic´,andM.M.Stevanovic´
Proposition 2. (1) The center S0(x0 : y0 : z0) has barycentric coordinates with
respecttotriangleO1O2O3
2 2 2 2
x0 = a (b +c −a )+u0,
2 2 2 2
y0 = b (c +a −b )+v0,
2 2 2 2
z0 = c (a +b −c )+w0, (16)
where
2 2 2 2 2 2 2 2
u0 = (r2 +r3 −2r1)a +(r −r3)(b −c ), (17)
2 2 2 2 3 2 2 2
v0 = (r3 +r1 −2r2)b +(r −r1)(c −a ), (18)
2 2 2 2 1 2 2 2
w0 = (r1 +r2 −2r3)a +(r −r2)(a −b ). (19)
(2)Theradiusoftheradicalcircleisgivenbyformula
2 C1
r = . (20)
0 16∆2
(3)ThecoefficientsB1,B3,B5,B7 areexpressedintermsofthecoordinatesof
S0,i.e.,
B1 = −2(r1x0+r2y0+r3z0), B3 = −2(r1x0−r2y0−r3z0), (21)
B5 = −2(−r1x0+r2y0−r3z0), B7 = −2(−r1x0−r2y0+r3z0). (22)
(4) The line S0O0, where O0 represents the circumcenter of triangle O1O2O3,
2 2 2
is orthogonal to the line q : r x + r y + r z = 0. This line passes through
1 2 3
the midpoints of segments M11M12, M21M22 and M31M32, where M11 and M12
are the inner and outer centers of similarity of circles (K2) and (K3), which can
analogouslybeappliedtotheotherpoints.
FortheeightsolutionsL (S ,p )ofApollonius’problem,withS (x : y : z )
j j j j j j j
wehave
Proposition3. (1)ThecoordinatesofthecentersS areasfollows:
j
x1 = 2p1u1+x0, y1 = 2p1v1+y0, z1 = 2p1w1+z0,
x2 = −2p2u1+x0, y2 = −2p2v1+y0, z2 = −2p2w1+z0,
where
2 2 2
u1 = (r2+r3−2r1)a +(r2−r3)(b −c ) = u1(r1,r2,r3),
2 2 2
v1 = (r3+r1−2r2)b +(r3−r1)(c −a ) = v1(r1,r2,r3),
2 2 2
w1 = (r1+r2−2r3)c +(r1−r2)(a −b ) = w1(r1,r2,r3).
x3 = 2p3u3+x0, y3 = 2p3v3+y0, z3 = 2p3w3+z0,
x4 = −2p4u3+x0, y4 = −2p4v3+y0, z4 = −2p4w3+z0,
where
u3(r1,r2,r3) = u1(r1,−r2,−r3),
v3(r1,r2,r3) = v1(r1,−r2,−r3),
w3(r1,r2,r3) = w1(r1,−r2,−r3).
x5 = 2p5u5+x0, y5 = 2p5v5+y0, z5 = 2p5w5+z0,
x6 = −2p6u5+x0, y6 = −2p6v5+y0, z6 = −2p6w5+z0,
RadiiofcirclesinApollonius’problem 365
where
u5(r1,r2,r3) = u1(r1,r2,−r3),
v5(r1,r2,r3) = v1(r1,r2,−r3),
w5(r1,r2,r3) = w1(r1,r2,−r3).
x7 = 2p7u7+x0, y7 = 2p7v7+y0, z7 = 2p7w7+z0,
x8 = −2p8u7+x0, y8 = −2p8v7+y0, z8 = −2p8w7+z0,
where
u7(r1,r2,r3) = u1(r1,−r2,r3),
v7(r1,r2,r3) = v1(r1,−r2,r3),
w7(r1,r2,r3) = w1(r1,−r2,r3).
(2)Therearecollineartripletsofpoints(S0,S1,S2),(S0,S3,S4),(S0,S5,S6)
and(S0,S7,S8),and
S0S1 ⊥ q1 : r1x+r2y+r3z = 0,
S0S3 ⊥ q3 : −r1x+r2y+r3z = 0,
S0S5 ⊥ q5 : r1x−r2y+r3z = 0,
S0S7 ⊥ q7 : r1x+r2y−r3z = 0,
where q1, q3, q5, q7 are the lines M12M22M32, M12M21M31, M11M22M31, and
M11M21M32 respectively.
(3)
1 1 2
− = , (23)
S0S1 S2S0 S0V1
where U1 = S0S1 ∩q1 and V1 and U1 are inversive to each other with respect to
theradicalcircle. Analogously,thisisassumedfortheothercentersS .
j
4. Three-dimensionalcase
Inthiscasewehavefourspheresandamaximumof16spheres,eachofwhich
beingtangenttoallofthefourgivenspheres. Analogousrelationsforradiiofthese
16 spheres will be found subsequently. Let us assume that the four spheres are
Φ1(O1,r1),Φ2(O2,r2),Φ3(O3,r3),Φ4(O4,r4). Wecantakethebasictetrahedron
ABCDtobetetrahedronO1O2O3O4 withO1 = A(1 : 0 : 0 : 0),O2 = B(0 : 1 :
0 : 0), O3 = C(0 : 0 : 1 : 0), O4 = D(0 : 0 : 0 : 1) given in the barycentric
coordinateswithmutualdistancesAB = c,AC = b,AD = d,BC = a,BD = e,
CD = f. AnimportantroleinfurtherinvestigationsisplayedbyCayley-Menger
determinant∆0 oftetrahedronABCD givenasfollows:
0 1 1 1 1
2 2 2
(cid:12)1 0 c b d (cid:12)
(cid:12) 2 2 2(cid:12)
∆0 = (cid:12)1 c 0 a e (cid:12). (24)
(cid:12) 2 2 2(cid:12)
(cid:12)1 b a 0 f (cid:12)
(cid:12) 2 2 2 (cid:12)
(cid:12)1 d e f 0(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
The known result is that the v(cid:12)olume V of tetrahedr(cid:12)on ABCD is given by for-
mula
2
∆0 = 288V .
366 M.R.Stevanovic´,P.B.Petrovic´,andM.M.Stevanovic´
Ifweapply∆ todenotethealgebraiccofactorofelementwithrow-columnposi-
ij
tion(i,j)incorrespondingCayley-Mengermatrix,weobtainthefollowingresult.
Proposition4. (1)ThecenterOofthecircumscribedsphereoftetrahedronABCD
(orcircumcenter)hasbarycentriccoordinates
O(∆12 : ∆13 : ∆14 : ∆15). (25)
(2)ThecircumradiusRoftheuppercircumsphereisgivenby
2 ∆11
R = − . (26)
2∆0
(3)IfforpointP(x : y : z : t)weintroducetworelevantexpressions
τ = τ(P) = x+y+z+t, (27)
2 2 2 2 2 2
T = T(P) = a yz+b zx+c xy+d xt+e yt+f zt, (28)
thenwehave
1
τ(O) = ∆0, T(O) = ∆0·∆11. (29)
2
LetusnowintroducetheradicalsphereΦ0(S0,r0),i.e.,thespherewithproperty
2 2 2 2 2 2 2 2 2
S0A −r1 = S0B −r2 = S0C −r3 = S0D −r4 = r0, (30)
orthespherewhichisorthogonaltothefourgivenspheresΦ1, Φ2, Φ3, Φ4. Then
wehave
Proposition5. (1)Thisradicalspherecorrespondstotheequation
2 2 2 2
T = τ(r x+r y+r z+r t). (31)
1 2 3 4
(2)ThecenterS0(x0 : y0 : z0 : t0)hascoordinates
2 2 2 2
x0 = ∆12+r1∆22+r2∆32+r3∆42+r4∆52, (32)
2 2 2 2
y0 = ∆13+r1∆23+r2∆33+r3∆43+r4∆53, (33)
2 2 2 2
z0 = ∆14+r1∆24+r2∆34+r3∆44+r4∆54, (34)
2 2 2 2
t0 = ∆15+r1∆25+r2∆35+r3∆45+r4∆55. (35)
(3)Fortheradiusr0,thefollowingformulaholds.
2 2 2 2 2 2
r = R − r x(M)+r y(M)+r z(M)+r t(M) , (36)
0 1 2 3 4
(cid:0) (cid:1)
whereM isthemidpointofthesegmentS0O.
In Figure 3 we introduce corresponding ordered quadruplets. An appropriate
number j in illustration (from (a) to (p)) denotes that sphere L is tangent to the
j
fourgivenspheres. Theplussigninthesecondposition(giveninbracketsineach
figure)meansthatspherewithcenterB isoutside-externallytangenttosphereL ,
j
while the minus sign at the fourth position means that sphere with center D is
insidesphereL –internallytangent,andsimilarlyinothercases. Foreachofthe
j
16 possible layouts, corresponding signs are given immediately under the figure,
RadiiofcirclesinApollonius’problem 367
C D C D
C D C D
A A
B B
B B
A A
3a(++++) 3b(−−−−) 3c(−+++) 3d(+−−−)
C D D C
C D C D
A A
B B
B B
A A
3e(+−++) 3f(−+−−) 3g(++−+) 3h(−−+−)
C D C D
D C C D
A A
B B
B B
A A
3i(+++−) 3j(−−−+) 3k(++−−) 3l(−−++)
D C D C
C D C D
A A
B B
B B
A A
3m(−+−+) 3n(+−+−) 3o(+−−+) 3p(−++−)
Figure3. The16possiblequadruplets-sphereseachofwhichbeingtangentto
allofthefourgivenspheres
368 M.R.Stevanovic´,P.B.Petrovic´,andM.M.Stevanovic´
related to the respective spheres with centers A, B, C and D, depending on their
positioninrelationtosphereL .
j
First of all, we will investigate sphere L1. If we generally denote the center of
the sphere and the radius by S and p, then for the first unknown sphere L1 (see
Figure3)is
SO1 = p+r1, SO2 = p+r2, SO3 = p+r3, SO4 = p+r4,
′ ′ ′ ′
andequationsofspheresΦ1(A,p+r1),Φ2(B,p+r2),Φ3(C,p+r3),Φ4(D,p+r4)
are
2 2 2 2 2 2 2
T = τ{−(p+r1) x+(c −(p+r1) )y+(b −(p+r1) )z+(d −(p+r1) )t},
(37)
2 2 2 2 2 2 2
T = τ{(c −(p+r2) )x−(p+r2) y+(a −(p+r2) )z+(e −(p+r2) )t},
(38)
2 2 2 2 2 2 2
T = τ{(b −(p+r3) )x+(a −(p+r3) )y−(p+r3) z+(f −(p+r3) )t},
(39)
2 2 2 2 2 2 2
T = τ{(d −(p+r4) )x+(e −(p+r4) )y+(f −(p+r4) )z−(p+r4) t}.
(40)
TheseequationsdeterminethepointS1(x1 : y1 : z1 : t1)whichisthecenterof
thesphereL1 andradiusp = p1 ofthatsphere. Forthemwehave
x1 = x0+2p(r1∆22+r2∆32+r3∆42+r4∆52), (41)
y1 = y0+2p(r1∆23+r2∆33+r3∆43+r4∆53), (42)
z1 = z0+2p(r1∆24+r2∆34+r3∆44+r4∆54), (43)
t1 = t0+2p(r1∆25+r2∆35+r3∆45+r4∆55), (44)
andtheseformulaeleadustotheequationforp = p1
2
F1(p, r1, r2, r3, r4) = A1p +B1p+C1 = 0, (45)
where
A1 = 2r1(r1∆22+r2∆32+r3∆42+r4∆52)
+2r2(r1∆23+r2∆33+r3∆43+r4∆53)
+2r3(r1∆24+r2∆34+r3∆44+r4∆54)
+2r4(r1∆25+r2∆35+r3∆45+r4∆55)+∆0
= f(r1,r2,r3,r4), (46)
2 2 2 2
B1 = 2r1(∆12+r1∆22+r2∆32+r3∆42+r4∆52)
2 2 2 2
+2r2(∆13+r1∆23+r2∆33+r3∆43+r4∆53)
2 2 2 2
+2r3(∆14+r1∆24+r2∆34+r3∆44+r4∆54)
2 2 2 2
+2r4(∆15+r1∆25+r2∆35+r3∆45+r4∆55)
= 2(r1x0+r2y0+r3z0+r4t0)
= g(r1,r2,r3,r4), (47)