A Proof of Gibson s and Rodgers Problem

Transcription

1 A Proof of Gibson s and Rodgers Problem Nguyen Minh Ha February 4, 007 Abstract Let A 0 B 0 C 0 be a triangle with centroid G 0 inscribed in a circle Γ with center O. The lines A 0 G 0, B 0 G 0, C 0 G 0 intersect Γ again in A, B, C, respectively, and G is the centroid of triangle A B C. A triangle A B C with centroid G is obtained in a similar way from A B C, and the prodcedure is repeated indefinitely, producing triangles with centroids G 3, G 4,... Given g n = OG n, it is known that the sequence {g 0, g, g,,...} is decreasing and converges to zero. Peter M. Gibson and Michael H.Rodgers posed a question of proving or disproving in [] that a similar result holds for tetrahedron inscribed in a sphere, or, more generally, for an n-simplex inscribed in an n- sphere. This prolem is hitherto unsolved. We give a positive answer and a proof in this note. Notations and Premilinary Results In this note, we give a proof to problem proposed by Gibson and Rodgers. For convenience, we adopt the following notations: S A, S B, S C, S D are the areas of the faces opposite to the vertice A, B, C, D of tetrahedron ABCD; (XYZ) the plane through three points X, Y, Z; V(XYZT) the volume of tetrahedron XYZT. For simplicity, we shall use the following sum notation to represent some sums throughout the paper, say S A LA = S A LA + S BLB + S CLC + S DLD, AB = AB + AC + AD + BC + BD + CD,

2 Definition. Let ABCD be a tetrahedron. A plane through the edge AB and the midpoint of the edge CD is called the median plane of the tetrahedron through the edge AB of the tetrahedron. A bisecting plane of the dihedral angle at the edge AB of the tetrahedron is called the bisector plane of the tetrahedron through the edge AB. A plane that is reflection of the median plane through edge AB about the bisector plane through the edge AB is called the symmedian plane through the edge AB of the tetrahedron. Each tetrahedron has six edges, thus has six median planes, six bisector planes, six symmedian planes. In each tetrahedron, six median planes intersect at one common point, namely the centroid of the tetrahedron, six bisector planes meet at a common point called the center of the inscribed sphere. Six symmedian planes of a tetrahedron intersect at a common point and we call this point the Lemoine point of the tetrahedron (we will prove this later). Now, we rephrase the problem for tetrahedron. Theorem. Let A 0 B 0 C 0 D 0 be a tetrahedron with centroid G 0 inscribed in a sphere Γ with center O. The lines A 0 G 0, B 0 G 0, C 0 G 0, D 0 G 0 intersect Γ again in A, B, C, D, respectively, and G is the centroid of tetrahedron A B C D. A tetrahedron A B C D with centroid G is obtained in a similar way from A B C D, and the prodcedure is repeated indefinitely, producing tetrahedrons with centroids G 3, G 4,... Then, {OG n } is decreasing and convergent to zero. We need the following lemmas for the proof of theorem. Lemma. Six symmedian planes of a tetrahedron intersect at one common point L defined by S A LA = 0. Note. This point is uniquely defined by the above equaltiy and referred to as the Lemoine point as aforementioned. Proof. Let the median plane, the bisecting plane, and the symmedian plane through the edge AB of the tetrahedron meet the edge CD at M, N, P (see Figure ). Consider the plane (π) perpendicular to the line AB. Let A be the orthogonal projection of A, B on the (π); let C, D, M, N, P be the orthogonal projections of C, D, M, N, P on the plane (π), respectively.

3 It is evident that in triangle A C D, A M, A N, and A P are respectively the median, bisector, and the symmedian from the vertex A. By the symmedian property, we have P C ( A P D = C ), A D and since CC, AA, and PP are parallel, we have PC ( A PD = C ), A D Suppose that E, F are respectively the orthogonal projections of C, D on AB. It is easily seen that A C = EC; A D = FD. A D F E B C P M N D π A C M N P Thus, PC ( EC ) = PD = ( AB.EC) FD ( = S D AB.FD) SC, 3

4 which leads to SC PC + S DPD = 0. It follows from the equality S ALA = 0 that Thus, S A LA + S BLB + S C ( LP + PC) + S D ( LP + PD) = 0. S A LA + S BLB + (S C + S D ) LP + (SC PC + S DPD) = 0. Now we can conclude that S A LA + S BLB + (S C + S D ) LP = 0, so that L lies in the plane (ABP). Similarly, we have L lies simultaneously in the six symmedian planes of the tetrahedron ABCD. This completes the proof. Lemma. Let M be any point interior to the tetrahedron ABCD. Let H, K, I, J be the orthogonal projections of point M on the plane (BCD), (CDA), (DAB), (ABC). Then, M is the centroid of tetrahedron HKI J if and only if M is the Lemoine point of tetrahedron ABCD. Proof. Let us recall the following well-known results V(MBCD) MA = 0, S A MH = 0. MH Combining the two equalities and the lemma implies that the following statements are equivalent + M is the centroid of tetrahedron HKIJ. + MH + MK + MI + MJ = 0. + S A MH = S B MK = S C MI = S D MJ. + S A 3 MH.S A = S B 3 MK.S B = S C 3 MI.S C = S D 3 MJ.S. D + S A V(MBCD) = + S A MA = 0. S B V(MCDA) = S C V(MDAB) = 4 S D V(MABC).

5 + M is the Lemoine point of tetrahedron ABCD. Lemma 3. Let ABCD be a tetrahedron. Points X, Y, Z, T belong to the planes (BCD), (CDA), (DAB), (ABC) respectively. The sum XY is a minimum if and only if X, Y, Z, T are respectively the orthogonal projections of Lemoine point of ABCD onto the planes (BCD), (CDA), (DAB), (ABC). Proof. Let M be the centroid of the tetrahedron XYZT. Suppose that H, K, I, J are orthogonal projections of M on the planes (BCD), (CDA), (DAB), (ABC). We have XY = 4 MX 4 MH = 4 SA ( MH )( S A ) 4 ( S MH.S A ) A = 4 ( S 3V(MBCD)) 36 A S V (ABCD). A Therefore, XY 36 S V (ABCD), A with equality if and only if the following conditions are simultaneously satisfied: + X, Y, Z, T are respectively orthogonal projections of M on the planes (BCD), (CDA), (DAB), (ABC); + MH S A = MK S B = MI S C = MJ S D ; + M is interior to the tetrahedron ABCD. By Lemma, the equality holds if and only if X, Y, Z, T are respectively orthogonal projectios of Lemoine point of tetrahedron ABCD on the planes (BCD), (CDA), (DAB), (ABC). In conclusion, XY attains its minimum if and only if X, Y, Z, T are orthogonal projections of the Lemoine point of the tetrahedron ABCD on the planes (BCD), (CDA), (DAB), (ABC) respectively. Proof complete. 5

6 Proof of theorem Let (α), (β), (γ), (δ) be the planes through the points A 0, B 0, C 0, D 0 respectively and perpendicular to A 0 G 0, B 0 G 0, C 0 G 0, and D 0 G 0 in that order. Denote by A 0 = (β) (γ) (δ); B 0 = (γ) (δ) (α); C 0 = (δ) (α) (β); D 0 = (α) (β) (γ). Since G 0 is the centroid of tetrahedron A 0 B 0 C 0 D 0, by Lemma, G 0 is the Lemoine point of the tetrahedron A 0 B 0 C 0 D 0. Let A, B, C, D be the reflections of A, B, C, D across O. It is evident that A, B, C, D be on the planes (α), (β), (γ), (δ) respectively. By Lemma 3, we have and taking into account that we have A B A 0 B 0, A B = A B, A B A 0 B 0. Let R be the radius of Γ, we have the following well-known (and easy to prove) results. A 0 B 0 = 4R 4OG 0, A B = 4R 4OG. Thus, OG 0 OG. The equality holds if and only if OG 0 = OG, or A, B, C, D respectively coincide with A 0, B 0, C 0, D 0 (by Lemma 3). In other words, G 0 coincides with O, which is equivalent to A 0 B 0 C 0 D 0 is a quasi-regular tetrahedron. Repeating the procedure, we have OG 0 OG OG. It follows that OG n is a decreasing sequence and there exists lim OG n. () n Let Γ be a closed ball with boundary Γ. Since Γ is closed and bounded, there exist a subsequence {n k } of the sequence {n} such that the sequences {A nk }, {B nk }, {C nk }, {D nk }, {G nk }, {A nk +}, {B nk +}, {C nk +}, {D nk +}, {G nk +} are convergent in Γ. 6

7 Let A 0 = lim A n, A 0 = G0 = lim G n ; A = D = lim D n +; G = lim G n +. It is evident that lim B n, C0 = lim A n +; B = lim C n, D0 = lim B n +; C = lim D n ; lim C n +; OG 0 = lim n k OG n k ; It follows from () and () that OG = lim n k OG n k +. () OG 0 = OG. (3) On the other hand, since Γ is a closed and bounded, A 0, B 0, C 0, D 0, A, B, C, D belong to Γ. Since G n k, G nk + are centroids of tetrahedrons A nk B nk C nk D nk, A nk +B nk +C nk +D nk + respectively for all n k, we have that G0, G are respectively the centroids of A 0 B 0 C 0 D 0, A B C D. Since A n k +, B nk +, C nk +, D nk + are respectively the intersections of the lines A nk G nk, B nk G nk, C nk G nk, D nk G nk with Γ, so that A, B, C, D are intersections of A 0 G 0, B 0, G 0, C 0 G 0, D 0 G 0 with Γ, respectively. It follows from the above remarks that the relationship between tetrahedron A 0 B 0 C 0 D 0 and A B C D is the same as that between A 0B 0 C 0 D 0 and A B C D. By the same reasoning at the beginning of the proof, we have OG0 OG. (4) The equality (3) implies that equality in (4) occurs. Again, by analogous reasoning used in the begining of the proof, we can conclude that A 0 B 0 C 0 D 0 is a quasi-regular tetrahedron. This implies that G0 lies at the circumcenter O of the sphere. Then OG0 = 0, so that lim OG n = lim OG nk = OG n n 0 = 0. This completes the proof of Theorem for a tetrahedron inscribed in a sphere. The proof still holds for n simplex inscribed in n sphere. References [] Peter M.Gibson and Michael H. Rodger, Problem 884, Crux Math [] S. Rabinowitz, Index to Mathematical Problems , MathPro Press, Westford, Massachusetts USA 99, p

8 Nguyen Minh Ha, Hanoi University of Education, Hanoi, Vietnam address: 8