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TANGENT SPHERE of EDGE in TETRAHEDRON 1. Introduction For TANGENT SPHERE OF EDGE IN TETRAHEDRON YU-DONG WU AND ZHI-HUA ZHANG Abstract. In this short note, using one of Yang’s Theorems, Lin-Zhu’s conjecture for a geometric inequality with respect to the tangent sphere of the edge in tetrahedron is proved. In the final, an open problem on the n¡dimension simplex is posed. 1. Introduction For the tetrahedron, there is the inscribed sphere and circumscribed sphere with similarly the incircle and circumscricle of a triangle. In 1996, Z.-Ch. Lin and H.-F. Zhu [1] studied a new sphere in tetrahedron, that is the tangent sphere of the edge. Definition 1.1. If a sphere is tangent with every edge of a tetrahedron, then this sphere is called the tangent sphere of the edge of this tetrahedron. However, it is not certain that there is a tangent sphere of the edge in every tetrahedron. In 1985, Zh. Yang given a sufficient and necessary condition to determine whether a tetrahedron has one tangent sphere of the edge in [2] (see also [1]). Let P denote tetrahedron P0P1P2P3 as follows. Theorem 1.1. P has the tangent sphere of the edge if and only if xi exist and satisfy aij = xi +xj for 0 · i < j · 3, where PiPj = aij. In [1], Lin and Zhu gave a formula to compute the radius and obtained several inequalities for the tangent sphere of the edge. Theorem 1.2. If l is the radius of the tangent sphere of the edge of P, then we have 2 2 (2x0x1x2x3) (1.1) l = P 2 2 2 2 2 2 2 2 2 2 2 2 : 2x0x1x2x3 xixj ¡ (x0x1x2 + x1x2x3 + x2x3x0 + x3x0x1) 0·i<j·3 Theorem 1.3. Let V be the volume, R the circumradius and r the inradius of P, respectively. Then for any P which has the tangent sphere of the edge, we have X X3 2 2 (1.2) 16l ¸ 2 xixj ¡ xi ; 0·i<j·3 i=0 (1.3) R2 ¸ 3l2; 1 2 (1.4) V l · (a a a a a a ) 3 ; 24 01 02 03 12 13 23 Date: September 21, 2005. 2000 Mathematics Subject Classification. 51M16, 52A40, 52B10. Key words and phrases. Geometric Inequality, tetrahedron, tangent sphere of the edge, inscribed sphere. The authors would like to thank Prof. Hanfang Zhang for his enthusiastic help. 1 2 Y.-D. WU AND ZH.-H. ZHANG p 8 8 3 (1.5) 8r2l · V · R2l · R3; 9 27 and 2 2 2 2 2 2 2 2 2 (1.6) 16R ¸ 15R + 3l ¸ a01 + a02 + a03 + a12 + a13 + a23: In the final of [1], the authors posed the following open problem (see also [3]). Conjecture 1.1. For any P which has the tangent sphere of the edge, prove or disprove that (1.7) l2 ¸ 3r2: The main purpose of this paper is to give a proof of this conjecture. 2. The Proof of Conjecture 1.1 In order to prove Conjecture 1.1, we require several lemmas. Lemma 2.1. For triangle ABC, if a = y + z, b = z + x and c = x + y, then xyz (2.1) r2 = : x + y + z Proof. Let p be the half perimeter, S the area, a; b; c the side-lengths of triangle ABC. Then we find that p (2.2) S = p(p ¡ a)(p ¡ b)(p ¡ c); and S (2.3) r = : p And for x = p ¡ a, y = p ¡ b, z = p ¡ c. Lemma 2.1 is proved. ¤ Lemma 2.2. If xi > 0 for 0 · i · 3, then µ ¶ x + x + x x + x + x x + x + x x + x + x 1 2 3 + 2 3 0 + 3 0 1 + 0 1 2 x1x2x3 x2x3x0 x3x0x1 x0x1x2 (2.4) 2 4(x0x1x2x3) ¢ P 2 2 2 2 2 2 2 2 2 2 2 2 ¸ 6: 2x0x1x2x3 xixj ¡ (x0x1x2 + x1x2x3 + x2x3x0 + x3x0x1) 0·i<j·3 Proof. In fact 2 2 2 2 2 2 2 2 2 x0x1(x2 ¡ x3) + x0x2(x1 ¡ x3) + x0x3(x1 ¡ x2) + 2 2 2 2 2 2 2 2 2 x1x2(x0 ¡ x3) + x1x3(x0 ¡ x2) + x2x3(x0 ¡ x1) ¸ 0; we have 2 X x2x2x2 + x2x2x2 + x2x2x2 + x2x2x2 ¸ x x x x x x : 0 1 2 1 2 3 2 3 0 3 0 1 3 0 1 2 3 i j 0·i<j·3 Therefor, we get X 4 X 2x x x x x x ¡ (x2x2x2 + x2x2x2 + x2x2x2 + x2x2x2) · x x x x x x ; 0 1 2 3 i j 0 1 2 1 2 3 2 3 0 3 0 1 3 0 1 2 3 i j 0·i<j·3 0·i<j·3 TANGENT SPHERE OF EDGE IN TETRAHEDRON 3 or 2 4(x0x1x2x3) P 2 2 2 2 2 2 2 2 2 2 2 2 2x0x1x2x3 xixj ¡ (x0x1x2 + x1x2x3 + x2x3x0 + x3x0x1) 0·i<j·3 (2.5) 4(x x x x )2 3x x x x 0 1 2 3 P0 1 2 3 ¸ 4 P = : 3 x0x1x2x3 xixj xixj 0·i<j·3 0·i<j·3 It’s easy to find the following identity P 2 xixj x + x + x x + x + x x + x + x x + x + x 0·i<j·3 (2.6) 1 2 3 + 2 3 0 + 3 0 1 + 0 1 2 = : x1x2x3 x2x3x0 x3x0x1 x0x1x2 x0x1x2x3 From (2.5) and (2.6), we find immediately inequality (2.4). ¤ Lemma 2.3. Let V be the volume of P, r0, r1, r2 and r3 the inradius, S0, S1, S2 and S3 the areas of triangleP1P2P3, triangleP2P3P0, triangleP3P0P1 and triangleP0P1P2, respectively. Then we have 2 µ ¶ 2 9 V 1 1 1 1 (2.7) (S0 + S1 + S2 + S3) ¸ 2 + 2 + 2 + 2 ; 2 r0 r1 r2 r3 with the equality in (2.7) holds if and only if P is regular. Figure 1. Proof. As figure 1. Let ®, ¯ and γ be the plane angles of the dihedral angels PT0 ¡ P2P3 ¡ P1, P0 ¡P3P1 ¡P2 and P0 ¡P1P2 ¡PT3, respectively. We draw P0H?P1P2P3 from P0, P0 P1P2P3 = H, and P0Q?P2P3 from P0, P0Q P2P3 = Q, and link QH. Thus, the plane angles of the dihedral angels P0 ¡ P2P3 ¡ P1 is \P0QH and P0H = P0Q ¢ sin ®. Hence, p (2.8) P0H ¢ P2P3 = 2S1 sin ® = 2 (S1 + S1 cos ®)(S1 ¡ S1 cos ®): By the same argument as above, we also obtain p (2.9) P0H ¢ P3P1 = 2S2 sin ¯ = 2 (S2 + S2 cos ¯)(S2 ¡ S2 cos ¯); and p (2.10) P0H ¢ P1P2 = 2S3 sin γ = 2 (S3 + S3 cos γ)(S3 ¡ S3 cos γ): 4 Y.-D. WU AND ZH.-H. ZHANG From (2.8)–(2.10), together with P H = 3V and S0 = 1 (P P + P P + P P ), we get 0 S0 r0 2 1 2 2 3 3 1 3V p = (S + S cos ®)(S ¡ S cos ®) r 1 1 1 1 0 p (2.11) + (S2 + S2 cos ¯)(S2 ¡ S2 cos ¯) p + (S3 + S3 cos γ)(S3 ¡ S3 cos γ): By Cauchy’s inequality and S0 = S1 cos ® + S2 cos ¯ + S3 cos γ, we have µ ¶ 3V 2 ·(S1 + S1 cos ® + S2 + S2 cos ¯ + S3 + S3 cos γ) r0 (2.12) ¢(S1 ¡ S1 cos ® + S2 ¡ S2 cos ¯ + S3 ¡ S3 cos γ) =(S1 + S2 + S3 + S0)(S1 + S2 + S3 ¡ S0) 2 2 =(S1 + S2 + S3) ¡ S0 ; or µ ¶2 2 2 3V (2.13) (S1 + S2 + S3) ¡ S0 ¸ : r0 It’s easy to know that equality in (2.13) holds if and only if S1+S1 cos ® = S2+S2 cos ¯ = S3+S3 cos γ or S1¡S1 cos ® S2¡S2 cos ¯ S3¡S3 cos γ cos ® = cos ¯ = cos γ, also or ® = ¯ = γ. Similarly, yields µ ¶2 2 2 3 V (2.14) (S2 + S3 + S0) ¡ S1 ¸ ; r1 µ ¶2 2 2 3 V (2.15) (S3 + S0 + S1) ¡ S2 ¸ ; r2 and µ ¶2 2 2 3 V (2.16) (S0 + S1 + S2) ¡ S3 ¸ : r3 Summing (2.13) to (2.16), it follows that inequality (2.7) is true, and with the equality in (2.7) holds if and only if P is regular. Thus, the proof of Lemma 2.3 is completed. ¤ Remark 2.1. Inequality (2.7) is obtained by X.-Zh. Yang in [4]. Theorem 2.1. In P which has the tangent sphere of the edge, we have that inequality (1.7) holds. Proof. For P, from Lemma 2.1, Theorem 1.2 and Lemma 2.2, we have µ ¶ 1 1 1 1 2 (2.17) 2 + 2 + 2 + 2 l ¸ 6: r0 r1 r2 r3 By Lemma 2.3, and r = 3 V , it follows S0+S1+S2+S3 2 2 6 27 V 2 (2.18) l ¸ 1 1 1 1 ¸ 2 ¸ 3r : 2 + 2 + 2 + 2 (S0 + S1 + S2 + S3) r0 r1 r2 r3 Therefor, Theorem 2.1 is proved. ¤ Remark 2.2. Combining (1.3) and (1.7), we have (2.19) R2 ¸ 3l2 ¸ 9r2: TANGENT SPHERE OF EDGE IN TETRAHEDRON 5 3. An Open Problem As a generalization of the tetrahedron, we should consider the tangent sphere of the edge in n¡dimensional simplex in this section. The following definition and theorems of the n¡dimensional simplex are similar with the tetra- hedron in [5]. Definition 3.1. If a sphere is tangent with every edge of a n¡dimensional simplex, then this sphere is the tangent sphere of the edge of this n¡dimensional simplex. Theorem 3.1. Suppose that PiPj = aij is the edges of n¡dimensional simplex P0P1P2 ¢ ¢ ¢ Pn, then it has a tangent sphere of the edge if and only if xi exist and satisfy aij = xi + xj for 0 · i 6= j · n: Theorem 3.2.
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