http://jipam.vu.edu.au/
Volume 7, Issue 5, Article 165, 2006
INEQUALITIES FOR INSCRIBED SIMPLEX AND APPLICATIONS
SHIGUO YANG AND SILONG CHENG DEPARTMENT OFMATHEMATICS
ANHUIINSTITUTE OFEDUCATION
HEFEI, 230061 P.R. CHINA
sxx@ahieedu.net.cn chenshilong2006@163.com
Received 18 January, 2005; accepted 06 June, 2005 Communicated by W.S. Cheung
ABSTRACT. In this paper, we study the problem of geometric inequalities for the inscribed simplex of an n-dimensional simplex. An inequality for the inscribed simplex of a simplex is established. Applying it we get a generalization ofn-dimensional Euler inequality and an inequality for the pedal simplex of a simplex.
Key words and phrases: Simplex, Inscribed simplex, Inradius, Circumradius, Inequality.
2000 Mathematics Subject Classification. 51K16, 52A40.
1. MAINRESULTS
Letσnbe ann-dimensional simplex in then-dimensional Euclidean spaceEn, V denote the volume ofσn, Randr the circumradius and inradius ofσn, respectively. Let A0, A1, . . . , An be the vertices of σn, aij = |AiAj|(0 ≤ i < j ≤ n), Fi denote the area of theith facefi = A0· · ·Ai−1Ai+1· · ·An((n−1)-dimensional simplex) ofσn, pointsOandGbe the circumcenter and barycenter ofσn, respectively. Fori= 0,1, . . ., n, letA0ibe an arbitrary interior point of the ith facefi ofσn. Then-dimensional simplexσn0 =A00A01· · ·A0nis called the inscribed simplex of the simplexσn. Leta0ij =|A0iA0j|(0≤i < j ≤n), R0 denote the circumradius ofσn0, P be an arbitrary interior point ofσn, Pibe the orthogonal projection of the pointP on theith facefi ofσn. Then-dimensional simplexσ00n =P0P1· · ·Pnis called the pedal simplex of the pointP with respect to the simplexσn[1] – [2], letV00 denote the volume ofσ00n, R00andr00 denote the circumradius and inradius ofσn00, respectively. We note that the pedal simplexσ00nis an inscribed simplex of the simplexσn. Our main results are following theorems.
Theorem 1.1. Letσ0nbe an inscribed simplex of the simplexσn, then we have (1.1) (R0)2(R2−OG2)n−1 ≥n2(n−1)r2n,
ISSN (electronic): 1443-5756
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019-05
with equality if the simplexσnis regular andσn0 is the tangent point simplex ofσn.
LetTibe the tangent point where the inscribed sphere of the simplexσntouches theith face fiofσn. The simplexσn=T0T1· · ·Tnis called the tangent point simplex ofσn[3]. If we take A0i ≡ Ti (i = 0,1, . . ., n)in Theorem 1.1, thenσn0 andσn are the same andR0 = r, we get a generalization of then-dimensional Euler inequality [4] as follows.
Corollary 1.2. For ann-dimensional simplexσn, we have
(1.2) R2 ≥n2r2+OG2,
with equality if the simplexσnis regular.
Inequality (1.2) improves then-dimensional Euler inequality [5] as follows.
(1.3) R ≥nr.
Theorem 1.3. Let P be an interior point of the simplex σn, and σn0 the pedal simplex of the pointP with respect toσn, then
(1.4) R00Rn−1 ≥n2n−1(r00)n,
with equality if the simplexσnis regular andσn00is the tangent point simplex ofσn.
2. SOME LEMMA ANDPROOFS OFTHEOREMS
To prove the theorems stated above, we need some lemmas as follows.
Lemma 2.1. Letσn0 be an inscribed simplex of then-dimensional simplexσn, then we have
(2.1) X
0≤i<j≤n
a0ij2
! n X
i=0
Fi2
!
≥n2(n+ 1)V2,
with equality if the simplexσnis regular andσn0 is the tangent point simplex ofσn.
Proof. Let B be an interior point of the simplex σn, and (λ0, λ1, . . ., λn) the barycentric co- ordinates of the point B with respect to coordinate simplex σn. Here λi = ViV−1(i = 0,1, . . ., n), Viis the volume of the simplexσn(i) =BA0· · ·Ai−1Ai+1· · ·AnandPn
i=0λi = 1.
LetQbe an arbitrary point inEn, then
−−→ QB =
n
X
i=0
λi−−→
QAi. From this we have
n
X
i=0
λi−−→
BAi =
n
X
i=0
λi−→
QAi−−−→ QB
=−→ 0,
n
X
i=0
λi−−→
QAi2
=
n
X
i=0
λi−−→
QB+−−→
BAi2
(2.2)
=
n
X
i=0
λi
−−→
QB2+ 2−−→ QB·
n
X
i=0
λi
−−→BAi+
n
X
i=0
λi
−−→
BAi
2
=−−→ QB2+
n
X
i=0
λi−−→
BAi2
.
Forj = 0,1, . . ., n, takingQ≡Aj in (2.2) we get (2.3)
n
X
i=0
λiλj
−−−→ AiAj
2
=λj
−−→
BAj
2
+λj n
X
i=0
λi
−−→
BAi
2
(j = 0,1, . . ., n).
Adding up these equalities in (2.3) and noting thatPn
j=0λj = 1, we get
(2.4) X
0≤i<j≤n
λiλj
−−−→ AiAj
2
=
n
X
i=0
λi
−−→
BAi
2
.
For any real numbers xi > 0 (i = 0,1, . . ., n)and an inscribed simplexσ0n = A00A01· · ·A0nof σn, we take an interior pointB0 ofσn0 such that (λ00, λ01, . . ., λ0n)is the barycentric coordinates of the pointB0with respect to coordinate simplexσ0n, hereλ0i =xi.
Pn
j=0xj (i= 0,1, . . ., n).
Using equality (2.4) we have X
0≤i<j≤n
λ0iλ0j a0ij2
=
n
X
i=0
λ0i−−→
B0A0i2
,
i.e.
(2.5) X
0≤i<j≤n
xixj a0ij2
=
n
X
i=0
xi
! n X
i=0
xi−−→
B0A0i2! .
SinceB0is an interior point ofσ0nandσn0 is an inscribed simplex ofσn, soB0is an interior point ofσn. Let the pointQibe the orthogonal projection of the pointB0 on theith facefiofσn, then
(2.6)
n
X
i=0
xi
−−→
B0A0i 2
≥
n
X
i=0
xi
−−→
B0Qi
2
.
Equality in (2.6) holds if and only ifQi ≡A0i(i= 0,1, . . ., n). In addition, we have (2.7)
n
X
i=0
−−→B0Qi
Fi =nV.
By the Cauchy’s inequality and (2.7) we have (2.8)
n
X
i=0
xi−−→
B0Qi2
! n X
i=0
x−1i Fi2
!
≥
n
X
i=0
−−→B0Qi ·Fi
!2
= (nV)2.
Using (2.5), (2.6) and (2.8), we get
(2.9) X
0≤i<j≤n
xixj a0ij2
! n X
i=0
x−1Fi2
!
≥n2
n
X
i=0
xi
! V2.
Takingx0 =x1 =· · ·=xn= 1in (2.9), we get inequality (2.1). It is easy to prove that equality in (2.1) holds if the simplexσnis regular andσn0 is the tangent point simplex ofσn. Lemma 2.2 ([1, 6]). For then-dimensional simplexσn, we have
(2.10)
n
X
i=0
Fi2 ≤[nn−4(n!)2(n+ 1)n−2]−1 X
0≤i<j≤n
a2ij
! , with equality if the simplexσnis regular.
Lemma 2.3 ([2]). LetP be an interior point of the simplexσ, σ00nthe pedal simplex of the point P with respect toσn, then
(2.11) V ≥nnV00,
with equality if the simplexσnis regular.
Lemma 2.4 ([1]). For then-dimensional simplexσn, we have
(2.12) V ≥ nn/2(n+ 1)(n+1)/2
n! rn, with equality if the simplexσnis regular.
Lemma 2.5 ([4]). For then-dimensional simplexσn, we have
(2.13) X
0≤i<j≤n
a2ij = (n+ 1)2
R2−OG2 .
HereOandGare the circumcenter and barycenter of the simplexσn, respectively.
Proof of Theorem 1.1. Using inequalities (2.1) and (2.10), we get
(2.14) X
0≤i<j≤n
a0ij2
! X
0≤i<j≤n
a2ij
!n−1
≥nn−2(n!)2(n+ 1)n−1V2.
By Lemma 2.5 we have
(2.15) X
0≤i<j≤n
a0ij2
≤(n+ 1)2(R0)2.
From (2.13), (2.14) and (2.15) we get
(2.16) (R0)2
R2−OG2n−1
≥ nn−1(n!)2 (n+ 1)n+1V2.
Using inequalities (2.16) and (2.12), we get inequality (1.1). It is easy to prove that equality in (1.1) holds if the simplexσnis regular andσ0nis the tangent point simplex ofσn. Proof of Theorem 1.3. Since the pedal simplex σn00 is an inscribed simplex of the simplexσn, thus inequality (2.16) holds for the pedal simplexσ00n, i.e.
(2.17) (R00)2
R2−OG2n−1
≥ nn−2(n!)2 (n+ 1)n+1V2. Using inequalities (2.17) and (2.11), we get
(2.18) (R00)2R2(n−1) ≥(R00)2
R2−OG2n−1
≥ n3n−2(n!)2 (n+ 1)n+1 (V00)2 By Lemma 2.4 we have
(2.19) V00 ≥ nn/2(n+ 1)(n+1)/2
n! (r00)n.
From (2.18) and (2.19) we obtain inequality (1.4). It is easy to prove that equality in (1.4) holds if the simplexσnis regular andσ00nis the tangent point simplex ofσn.
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