Radii of Circles in Apollonius Problem

Transcription

1 Forum Geometricorum Volume 7 ( FORUM GEOM ISSN Radii of ircles in pollonius Problem Milorad R. Stevanović, Predrag. Petrović, and Marina M. Stevanović bstract. The paper presents the relation for radii of the eight circles in pollonius problem for circles which are tangent to three given circles. nalogously, we derived the relations for radii of the 6 spheres which are tangent to four given spheres, with coordinates of their centers and with their radii.. Introduction It is well known that for three given circles generically there are eight different circles that are tangent to them. The problem of ruler and compass constructability of these eight circles is well-known. There are famous pollonius and Gergonne s solutions to this problem [3]. Special cases of the three given circles are considered 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 pollonius problem. The second case is when we have three excircles as a starting point. Three sides of the original triangle are three solutions to pollonius problem with infinite radii []. The nine-point circle is tangent externally to the three excircles, by Feuerbach theorem, and a relatively new object - the pollonius circle is tangent internally to three exircles (for some results about this circle see [4]-[7]. To these five circles we can add three Jenkins circles which are tangent to three excircles, by adding two of them externally and the third one internally. 2. Main result Let us assume that the three given circles arek (O,r,K 2 (O 2,r 2,K 3 (O 3,r 3, Figure, with distances between centers O 2 O 3 = a, O 3 O = b, O O 2 = c and with the area (O O 2 O 3 =, which is different from 0. The following theorem holds true: Theorem. Let us assume that the radii of the eight circles with centers S i given in Figure ((a, (b, (c and (d arep j, (j =,2,...,8. Then p p 2 p 3 p 4 p 5 p 6 p 7 p 8 = 0. ( Proof. Let us introduce the anglesϕ = O 2 SO 3,ϕ 2 = O 3 SO,ϕ 3 = O SO 2, S = S for Figure (a, so as to obtain cos 2 ϕ cos 2 ϕ 2 cos 2 ϕ 3 2cosϕ cosϕ 2 cosϕ 3 =. (2 Publication ate: June 25, 207. ommunicating Editor: Paul Yiu.

2 360 M. R. Stevanović, P.. Petrović, and M. M. Stevanović O 3 O 3 b a S 2 S b a S 4 c O O 2 S 3 O c O 2 O 3 b S 8 a O 3 c O O 2 S 6 b a S 7 O c S 5 O2 Figure. The eight different circles that are tangent to the three given circles. If we substitute t = sin 2 ϕ 2, t 2 = sin 2 ϕ 2 2, t 3 = sin 2 ϕ 3 2, from relation (2 we have t 2 t 2 2 t 2 3 2(t t 2 t 2 t 3 t 3 t 4t t 2 t 3 = 0. (3 If we generally denote the center of the circle and radius by S and p, then for the first unknown circle L (see Figure (a is SO = p r, SO 2 = p r 2,

3 Radii of circles in pollonius problem 36 SO 3 = pr 3 and cosϕ = (pr 2 2 (pr 3 2 a 2 2(pr 2 (pr 3 and analogously = t = a2 (r 2 r 3 2 4(pr 2 (pr 3, t 2 = b2 (r 3 r 2 4(pr 3 (pr, t 3 = c2 (r r 2 2 4(pr (pr 2. From relation (3 we now have relation (4: or in another form (a 2 (r 2 r (pr 2 (b 2 (r 3 r 2 2 (pr 2 2 (c 2 (r r (pr 3 2 2(a 2 (r 2 r 3 2 (b 2 (r 3 r 2 (pr (pr 2 2(a 2 (r 2 r 3 2 (c 2 (r r 2 2 (pr (pr 3 2(b 2 (r 3 r 2 (c 2 (r r 2 2 (pr 2 (pr 3 (a 2 (r 2 r 3 2 (b 2 (r 3 r 2 (c 2 (r r 2 2 = 0. (4 Equation (4 is of the second degree and is of the form F (p, r, r 2, r 3 = 0. (5 p 2 p = 0, (6 = 4(a 2 (r r 2 (r r 3 b 2 (r 2 r 3 (r 2 r c 2 (r 3 r (r 3 r = f(r,r 2,r 3, (7 = 2{r (a 2 (r 2 r r 2 (b 2 (r 3 r 2 2 r 3 (c 2 (r r (a 2 (r 2 r 3 2 (b 2 (r 3 r 2 (r r 2 (b 2 (r 3 r 2 (c 2 (r r 2 2 (r 2 r 3 (c 2 (r r 2 2 (a 2 (r 2 r 3 2 (r 3 r } = g(r,r 2,r 3, (8 = r 2 a 4 r 2 2b 4 r 2 3c 4 a 2 b 2 c 2 a 2 b 2 (r 2 r 2 2b 2 c 2 (r 2 2 r 2 3c 2 a 2 (r 2 3 r 2 a 2 (r 2 r 2 2(r 2 r 2 3b 2 (r 2 2 r 2 3(r 2 2 r 2 c 2 (r 2 3 r 2 (r 2 3 r 2 2 = h(r,r 2,r 3. (9

4 362 M. R. Stevanović, P.. Petrović, and M. M. Stevanović For the second unknown circle L 2 (see Figure (a we have SO = p r, SO 2 = pr 2,SO 3 = pr 3 and a corresponding equation in the form of equation (4, and with which implies that and F (p, r, r 2, r 3 = 0, (0 2 p 2 2 p 2 = 0, ( 2 = f (r, r 2, r 3 =, 2 = g (r, r 2, r 3 =, 2 = h (r, r 2, r 3 =, p 2 p = 0, p 2 2 p 2 = 0, p p 2 =. (2 For the third circlel 3 (see Figure (b we haveso = pr,so 2 = pr 2, SO 3 = pr 3 and we get F (p, r, r 2, r 3 = 0 with 3 p 2 3 p 3 = 0, 3 = f (r, r 2, r 3, 3 = g (r, r 2, r 3, 3 = h (r r 2, r 3 =. For the fourth circlel 4 (see Figure (b we haveso = pr,so 2 = pr 2, SO 3 = p r 3 and we get F (p, r, r 2, r 3 = 0 with 4 p 2 4 p 4 = 0, 4 = f (r, r 2, r 3 = 3, 4 = g (r, r 2, r 3 = 3, 4 = h (r,r 2,r 3 = and again we get nalogously, we have p 3 p 4 = 3. (3 = 5, p 5 p 6 (4 = 7, p 7 p 8 (5 5 = g (r, r 2, r 3, 7 = g (r, r 2,r 3. Formula ( follows from (2, (3, (4, (5 because g (r, r 2, r 3 g (r, r 2, r 3 g (r, r 2, r 3 g (r, r 2, r 3 = 0. Remark. If the indexj of the circle with radiusp j is an even (odd number, then /p j (in formula ( has the sign(.

5 Radii of circles in pollonius problem 363 Remark 2. In the first case of the three given circles mentioned in the introduction, we get the formula r = r r 2 r 3 r,r,r 2,r 3 are the inradius and exradii of triangle. Remark 3. In the second case we get p p 2 p 3 q = m, p,p 2,p 3 are the radii of Jenkins circles,q is the radius of pollonius circle andm = R/2 is the radius of Euler circle or nine-point circle. This formula can be proved independently since p = a bc q, p 2 = b ca q, p 3 = c ab q. Remark 4. In the same way, the same result can be proved for the three given circles, provided that two of them are inside of the third one. 3. Positions of8circles The radical circle of the three given circlesk (O,r,K 2 (O 2,r 2,K 3 (O 3,r 3, is the circle orthogonal to all of them, and pairs of circles (L,L 2, (L 3,L 4, (L 5,L 6, (L 7,L 8 Figure, are inversive with respect to the radical circle. For this radical circlek 0 (S 0,r 0, Figure 2, the following holds true. O 3 b S a S 0 S 2 c O O 2 Figure 2. The radical circle of the three given circles with the circlesl andl 2, inversive to the radical circle, based on Figure (a

6 364 M. R. Stevanović, P.. Petrović, and M. M. Stevanović Proposition 2. ( The center S 0 (x 0 : y 0 : z 0 has barycentric coordinates with respect to triangleo O 2 O 3 x 0 = a 2 (b 2 c 2 a 2 u 0, y 0 = b 2 (c 2 a 2 b 2 v 0, z 0 = c 2 (a 2 b 2 c 2 w 0, (6 u 0 = (r 2 2 r 2 3 2r 2 a 2 (r 2 r 2 3(b 2 c 2, (7 v 0 = (r 2 3 r 2 2r 2 2b 2 (r 3 r 2 (c 2 a 2, (8 w 0 = (r 2 r 2 2 2r 2 3a 2 (r r 2 2(a 2 b 2. (9 (2 The radius of the radical circle is given by formula r0 2 = 6 2. (20 (3 The coefficients, 3, 5, 7 are expressed in terms of the coordinates of S 0, i.e., = 2(r x 0 r 2 y 0 r 3 z 0, 3 = 2(r x 0 r 2 y 0 r 3 z 0, (2 5 = 2(r x 0 r 2 y 0 r 3 z 0, 7 = 2(r x 0 r 2 y 0 r 3 z 0. (22 (4 The line S 0 O 0, O 0 represents the circumcenter of triangle O O 2 O 3, is orthogonal to the line q : r 2x r2 2 y r2 3z = 0. This line passes through the midpoints of segments M M 2, M 2 M 22 and M 3 M 32, M and M 2 are the inner and outer centers of similarity of circles (K 2 and (K 3, which can analogously be applied to the other points. For the eight solutionsl j (S j,p j of pollonius problem, withs j (x j : y j : z j we have Proposition 3. ( The coordinates of the centerss j are as follows: x = 2p u x 0, y = 2p v y 0, z = 2p w z 0, x 2 = 2p 2 u x 0, y 2 = 2p 2 v y 0, z 2 = 2p 2 w z 0, u = (r 2 r 3 2r a 2 (r 2 r 3 (b 2 c 2 = u (r,r 2,r 3, v = (r 3 r 2r 2 b 2 (r 3 r (c 2 a 2 = v (r,r 2,r 3, w = (r r 2 2r 3 c 2 (r r 2 (a 2 b 2 = w (r,r 2,r 3. x 3 = 2p 3 u 3 x 0, y 3 = 2p 3 v 3 y 0, z 3 = 2p 3 w 3 z 0, x 4 = 2p 4 u 3 x 0, y 4 = 2p 4 v 3 y 0, z 4 = 2p 4 w 3 z 0, u 3 (r,r 2,r 3 = u (r,r 2,r 3, v 3 (r,r 2,r 3 = v (r,r 2,r 3, w 3 (r,r 2,r 3 = w (r,r 2,r 3. x 5 = 2p 5 u 5 x 0, y 5 = 2p 5 v 5 y 0, z 5 = 2p 5 w 5 z 0, x 6 = 2p 6 u 5 x 0, y 6 = 2p 6 v 5 y 0, z 6 = 2p 6 w 5 z 0,

7 Radii of circles in pollonius problem 365 u 5 (r,r 2,r 3 = u (r,r 2,r 3, v 5 (r,r 2,r 3 = v (r,r 2,r 3, w 5 (r,r 2,r 3 = w (r,r 2,r 3. x 7 = 2p 7 u 7 x 0, y 7 = 2p 7 v 7 y 0, z 7 = 2p 7 w 7 z 0, x 8 = 2p 8 u 7 x 0, y 8 = 2p 8 v 7 y 0, z 8 = 2p 8 w 7 z 0, u 7 (r,r 2,r 3 = u (r,r 2,r 3, v 7 (r,r 2,r 3 = v (r,r 2,r 3, w 7 (r,r 2,r 3 = w (r,r 2,r 3. (2 There are collinear triplets of points (S 0,S,S 2, (S 0,S 3,S 4, (S 0,S 5,S 6 and(s 0,S 7,S 8, and S 0 S q : r xr 2 y r 3 z = 0, S 0 S 3 q 3 : r xr 2 y r 3 z = 0, S 0 S 5 q 5 : r xr 2 y r 3 z = 0, S 0 S 7 q 7 : r xr 2 y r 3 z = 0, q, q 3, q 5, q 7 are the lines M 2 M 22 M 32, M 2 M 2 M 3, M M 22 M 3, and M M 2 M 32 respectively. (3 = 2, (23 S 0 S S 2 S 0 S 0 V U = S 0 S q and V and U are inversive to each other with respect to the radical circle. nalogously, this is assumed for the other centers S j. 4. Three-dimensional case In this case we have four spheres and a maximum of 6 spheres, each of which being tangent to all of the four given spheres. nalogous relations for radii of these 6 spheres will be found subsequently. Let us assume that the four spheres are Φ (O,r,Φ 2 (O 2,r 2,Φ 3 (O 3,r 3,Φ 4 (O 4,r 4. We can take the basic tetrahedron to be tetrahedron O O 2 O 3 O 4 with O = ( : 0 : 0 : 0, O 2 = (0 : : 0 : 0, O 3 = (0 : 0 : : 0, O 4 = (0 : 0 : 0 : given in the barycentric coordinates with mutual distances = c, = b, = d, = a, = e, = f. n important role in further investigations is played by ayley-menger determinant 0 of tetrahedron given as follows: 0 0 c 2 b 2 d 2 0 = c 2 0 a 2 e 2. (24 b 2 a 2 0 f 2 d 2 e 2 f 2 0 The known result is that the volume V of tetrahedron is given by formula 0 = 288V 2.

8 366 M. R. Stevanović, P.. Petrović, and M. M. Stevanović If we apply ij to denote the algebraic cofactor of element with row-column position(i,j in corresponding ayley-menger matrix, we obtain the following result. Proposition 4. ( The center O of the circumscribed sphere of tetrahedron (or circumcenter has barycentric coordinates O( 2 : 3 : 4 : 5. (25 (2 The circumradiusrof the upper circumsphere is given by R 2 = 2 0. (26 (3 If for point P(x : y : z : t we introduce two relevant expressions then we have τ = τ(p = xy z t, (27 T = T(P = a 2 yz b 2 zxc 2 xy d 2 xte 2 ytf 2 zt, (28 τ(o = 0, T(O = 2 0. (29 Let us now introduce the radical sphereφ 0 (S 0,r 0, i.e., the sphere with property S 0 2 r 2 = S 0 2 r 2 2 = S 0 2 r 2 3 = S 0 2 r 2 4 = r 2 0, (30 or the sphere which is orthogonal to the four given spheres Φ, Φ 2, Φ 3, Φ 4. Then we have Proposition 5. ( This radical sphere corresponds to the equation T = τ(r 2 xr 2 2y r 2 3z r 2 4t. (3 (2 The center S 0 (x 0 : y 0 : z 0 : t 0 has coordinates x 0 = 2 r 2 22 r r r , (32 y 0 = 3 r 2 23 r r r , (33 z 0 = 4 r 2 24 r r r , (34 t 0 = 5 r 2 25 r r r (35 (3 For the radiusr 0, the following formula holds. r 2 0 = R 2 ( r 2 x(mr 2 2y(Mr 2 3z(Mr 2 4t(M, (36 M is the midpoint of the segments 0 O. In Figure 3 we introduce corresponding ordered quadruplets. n appropriate number j in illustration (from (a to (p denotes that sphere L j is tangent to the four given spheres. The plus sign in the second position (given in brackets in each figure means that sphere with center is outside-externally tangent to spherel j, while the minus sign at the fourth position means that sphere with center is inside sphere L j internally tangent, and similarly in other cases. For each of the 6 possible layouts, corresponding signs are given immediately under the figure,

9 Radii of circles in pollonius problem 367 3a( 3b ( 3c( 3d( 3e( 3f( 3g( 3h( 3i( 3j( 3k( 3l( 3m( 3n ( 3o( 3p( Figure 3. The 6 possible quadruplets- spheres each of which being tangent to all of the four given spheres

10 368 M. R. Stevanović, P.. Petrović, and M. M. Stevanović related to the respective spheres with centers,, and, depending on their position in relation to sphere L j. First of all, we will investigate sphere L. If we generally denote the center of the sphere and the radius by S and p, then for the first unknown sphere L (see Figure 3 is SO = pr, SO 2 = pr 2, SO 3 = pr 3, SO 4 = pr 4, and equations of spheresφ (,pr,φ 2 (,pr 2,Φ 3 (,pr 3,Φ 4 (,pr 4 are T = τ{(pr 2 x(c 2 (pr 2 y (b 2 (pr 2 z (d 2 (pr 2 t}, (37 T = τ{(c 2 (pr 2 2 x(pr 2 2 y (a 2 (pr 2 2 z (e 2 (pr 2 2 t}, (38 T = τ{(b 2 (pr 3 2 x(a 2 (pr 3 2 y (pr 3 2 z (f 2 (pr 3 2 t}, (39 T = τ{(d 2 (pr 4 2 x(e 2 (pr 4 2 y (f 2 (pr 4 2 z (pr 4 2 t}. (40 These equations determine the point S (x : y : z : t which is the center of the spherel and radiusp = p of that sphere. For them we have x = x 0 2p(r 22 r 2 32 r 3 42 r 4 52, (4 y = y 0 2p(r 23 r 2 33 r 3 43 r 4 53, (42 z = z 0 2p(r 24 r 2 34 r 3 44 r 4 54, (43 t = t 0 2p(r 25 r 2 35 r 3 45 r 4 55, (44 and these formulae lead us to the equation for p = p F (p, r, r 2, r 3, r 4 = p 2 p = 0, (45 = 2r (r 22 r 2 32 r 3 42 r r 2 (r 23 r 2 33 r 3 43 r r 3 (r 24 r 2 34 r 3 44 r r 4 (r 25 r 2 35 r 3 45 r = f(r,r 2,r 3,r 4, (46 = 2r ( 2 r 2 22 r r r r 2 ( 3 r 2 23 r r r r 3 ( 4 r 2 24 r r r r 4 ( 5 r 2 25 r r r = 2(r x 0 r 2 y 0 r 3 z 0 r 4 t 0 = g(r,r 2,r 3,r 4, (47

11 Radii of circles in pollonius problem 369 = 2 { r 2 ( 2 r 2 22 r r r r 2 2( 3 r 2 23 r r r r 2 3( 4 r 2 24 r r r r 2 4( 5 r 2 25 r r r ( r 2 2 r r r } = h(r,r 2,r 3,r 4, (48 For the second unknown sphere L 2 (see Figure 3 we have SO = p r, SO 2 = pr 2,SO 3 = pr 3,SO 4 = pr 4 as illustrated by the equation F (p, r, r 2, r 3, r 4 2 p 2 2 p 2 p 2 p = 0. (49 Now we get p p 2 =. (50 This means that to the difference/p /p 2 we can relate the ordered quadruple (,,, related to (r,r 2,r 3,r 4 since is linear with respect to all r j. Since is the same for all 6 combinations (ε r, ε 2 r 2, ε 3 r 3, ε 4 r 4 for all ε {, }. When combining the signs of the ordered quadruplets, we obtain the following results given in the next theorem. Theorem 6. Let us assume that the radii of the sixteen spheres given in Figure 3 arep j (j =,2,...,6 and the volume V is different from0. Then = 0, (5 p p 2 p 3 p 4 p 5 p 6 p p 2 = 0, (52 p p 2 p 3 p 4 p 7 p 8 p 3 p 4 = 0, (53 p p 2 p 3 p 4 p 9 p 0 p 5 p 6 = 0, (54 p p 2 p 5 p 6 p 7 p 8 p 5 p 6 = 0, (55 p p 2 p 5 p 6 p 9 p 0 p 3 p 4 = 0, (56 p p 2 p 7 p 8 p 9 p 0 p p 2 ( ( ( ( p p 2 p 9 p 0 p 3 p 4 p p 2 p 5 p 6 p 3 p 4 p 7 p 8 p 5 p 6 = 0, (57

12 370 M. R. Stevanović, P.. Petrović, and M. M. Stevanović ( ( ( ( p p 2 p 9 p 0 p p 2 p 9 p 0 p p 2 p 9 p 0 p 3 p 4 p 3 p 4 p 3 p 4 p p 2 p p 2 p p 2 p 5 p 6 p 5 p 6 p 5 p 6 p 3 p 4 p 3 p 4 p 3 p 4 p 7 p 8 p 7 p 8 p 7 p 8 p 5 p 6 p 5 p 6 p 5 p 6 = 0, = 0, = 0. (58 (59 (60 orollary 7. From the above formulae we can also obtain = 0, (6 p 3 p 4 p 5 p 6 p 3 p 4 p 5 p 6 = 0, (62 p 3 p 4 p 7 p 8 p p 2 p 5 p 6 = 0, (63 p 3 p 4 p 9 p 0 p p 2 p 3 p 4 = 0, (64 p 5 p 6 p 7 p 8 p p 2 p 3 p 4 = 0, (65 p 5 p 6 p 9 p 0 p p 2 p 5 p 6 = 0. (66 p 7 p 8 p 9 p 0 p 3 p 4 p 5 p 6 These formulae and formulae listed in Theorem 6 are all possible formulae of this type. orollary 8. For the radiusr 0 of the radical circle, Proof. From formula (37 we have Since r 2 0 = 0. (67 T = τ [ (c 2 y b 2 z d 2 t = (p r 2 τ ]. τ = τ(s = τ(s 0 = 0,

13 Radii of circles in pollonius problem 37 for the coefficient without o, we have Now the desired formula follows from = T 0 0 (c 2 y 0 b 2 z 0 d 2 t 0 r = 0, r0 2 = ( (c 2 y 0 b 2 z 0 d 2 t 0 r τ τ 2 0 T 0 0 τ 0 = ( (c 2 y 0 b 2 z 0 d 2 t 0 r 2 0 T Remark 5. If in the two-dimensional case, we introduce the determinant 0 0 = 0 c 2 b 2 c 2 0 a 2, b 2 a 2 0 then we have 0 = 6 2. onsequently, r 2 0 = / 0 in both cases. orollary 9. The centers S ands 2 have coordinates x = x 0 2p u, y = y 0 2p v, z = z 0 2p w, t = t 0 2p η, x 2 = x 0 2p 2 u, y 2 = y 0 2p 2 v, z 2 = z 0 2p 2 w, t 2 = t 0 2p 2 η, with the property u = r 22 r 2 32 r 3 42 r 4 52, v = r 23 r 2 33 r 3 43 r 4 53, w = r 24 r 2 34 r 3 44 r 4 54, η = r 25 r 2 35 r 3 45 r 4 55, u v w η = 0. nalogously to the planar case we can obtain coordinates of centers for all 6 spheres. There are eight planes, and each of them passes through six of the twelve inner or outer centers of mutual similarity of the given four spheres. s earlier, point S 0 with two centers is perpendicular to one of these eight planes. Theorem can be proved by the same technique used in the proof of Theorem 6. References [] H. S. M. oxeter, The Problem of pollonius, mer. Math. Monthly, 75 ( [2] F. G.-M., Exercices de géométrie, Tours, France: Maison Mame, 8-20 and 663, 92. [3] M. Gergonne, Recherche du cercle qui en touche trois autres sur une sphere, nn. math. pures appl., 4 ( [4]. Grinberg and P. Yiu, The pollonius circle as a Tucker circle, Forum Geom., 2 ( [5] J.. Lagarias,. L. Mallows, and. R. Wilks, eyond the escartes ircle Theorem, mer. Math. Monthly 09 (

14 372 M. R. Stevanović, P.. Petrović, and M. M. Stevanović [6]. Pedoe, On a theorem in geometry, mer. Math. Monthly, 74 ( [7] M. R. Stevanović, The pollonius circle and related triangle centers, Forum Geom. 3 ( Milorad R. Stevanović: University of Kragujevac, Faculty of Technical Sciences Čačak, Svetog Save 65, Čačak, Serbia Predrag. Petrović: University of Kragujevac, Faculty of Technical Sciences Čačak, Svetog Save 65, Čačak, Serbia address: predrag.petrovic@ftn.kg.ac.rs Marina M. Stevanović: University of elgrade, Faculty of Mathematics, Studentski trg 6, 000 eograd, Serbia address: marina.stevanovic42@gmail.com