On Dihedral Angles of a Simplex

For an n-simplex, let α, β denote the maximum, and the minimum dihedral angles of the simplex, respectively. It is proved that the inequality α ≤ arccos(1/n) ≤ β always holds, and either side equality implies that the nsimplex is a regular simplex. Similar inequalities are also given for a star-simplex, which is defined as a simplex that has a vertex (apex) such that the angles between distinct edges incident to the apex are all equal. Further, an explicit formula for the dihedral angle of a star-simplex between two distinct facets sharing the apex in common is presented in terms of the angle between two edges incident to the apex.


Introduction
Let σ be an n-dimensional simplex (n-simplex) in R n , n ≥ 2, and let f i , f j be two distinct facets of σ.The dihedral angle ∠( f i , f j ) between the facets f i , f j is defined as the supplement of the angle between the unit outer normal vectors of the facets f i , f j .For n ≥ 3, the sum of n+1 2 dihedral angles of an n-simplex is not constant.Indeed, the following holds (see Gaddum, 1952Gaddum, , 1956)).

£ The sum of the n+1
2 dihedral angles of an n-simplex lies between n 2 −1 4 π and n 2 π, and the sum can take any value in this range.
A dihedral angle is called acute (resp.nonobtuse) if the angle is less than (resp.not greater than) π/2.The next result seems first appeared in Fielder (1954), and rediscovered again in Leng (2003).
£ Every n-simplex has at least n acute dihedral angles.
For an n-simplex σ, let α = α(σ), β = β(σ) denote the minimum value and the maximum value of the dihedral angles in σ, respectively.If σ is a regular n-simplex, then α = β = arccos 1 n .In the 2-dimensional case n = 2, since the sum of the interior angles of a triangle is π, we have α ≤ π/3 ≤ β and α = π/3 ⇔ β = π/3.A similar assertion also holds in n ≥ 3, though the sum of dihedral angles are not constant.The next is the main result of this paper.
Theorem 1 For every n-simplex, α ≤ arccos 1 n ≤ β holds.Moreover, if either side equality holds, then the other side equality also holds and the simplex becomes a regular simplex.
We present a similar result for a family of star-simplexes.A star-simplex with vertex angle θ is defined to be a simplex that has a vertex v such that the plane angle between any two distinct edges incident to v is equal to θ.The vertex v is called the apex of the star-simplex.If those edges incident to the apex are of the same length, then the star-simplex is called a regular star-simplex.In a star-simplex, the dihedral angles between two distinct facets sharing the apex in common are all equal, and their common value is called the lateral angle of the star-simplex.

Proof of Theorem 1
Let S (O, x) ⊂ R n denote a sphere with center O and radius x.
OP n be the unit outer normal vectors of the facets of σ.Then, P 0 , . . ., P n span a simplex that contains O in its interior.
Proof.We may suppose that S (O, 1) is the inscribed sphere of σ, and the n + 1 points P 0 , . . ., P n are the contact points of S (O, 1) with the n + 1 facets of σ.Then, no closed hemisphere of S (O, 1) can contain these n + 1 points, for otherwise, the inscribed sphere S (O, 1) of σ can slip out of the simplex σ.Hence, O is an interior point of the simplex spanned by P 0 , . . ., P n .
Let us recall here some values concerning a regular simplex.If a regular n-simplex has unit circumradius, then • the radius of its inscribed sphere is equal to 1/n, • its edge-length is equal to l(n) := √ 2(n + 1)/n, and • its dihedral angle is equal to arccos 1 n .Note that l(n) is strictly monotone decreasing in n, and the edge-length of a regular n-simplex with circumradius R is given by R For a set V of n + 1 points on a unit sphere S (O, 1) ⊂ R n , let a(V), b(V) denote the minimum value and the maximum value of the Euclidean distance Lemma 2 Suppose that a set V of n + 1 points on S (O, 1) ⊂ R n spans an n-simplex V that contains O in its interior.Then (1) a(V) ≤ l(n), and if the equality holds, then V is a regular n-simplex.
(2) b(V) ≥ l(n), and if the equality holds, then V is a regular n-simplex.
Proof.(1) We use the following result in Deza and Maehara (1994): ( * ) For any m points P 1 , P 2 , . . ., P m on S (O, R), we have and the equality holds if and only if 1 Applying ( * ) to the point set V on S (O, 1), we have (n + 1) 2 ≥ n+1 2 a(V) 2 .This implies that a(V) ≤ l(n), and the equality holds only when |PQ| = l(n) for all P, Q ∈ V, P Q, which implies that V is a regular n-simplex.
(2) Proof is by induction on the dimension n.For n = 2, (2) can be easily seen.Suppose that (2) is true for every (n − 1)-simplex, and let us consider the n-dimensional case.We use the circumradius-inradius inequality for an n-simplex (see, e.g., Klamkin & Tsintsifas, 1979): ( * * ) The radius r of the inscribed sphere and the radius R of the circumscribed sphere of an n-simplex always satisfy r ≤ R/n.
Let τ = V .Since O is an interior point of τ, there is an x 0 > 0 such that S (O, x 0 ) is tangent to a facet f of τ and S (O, x 0 ) ⊂ τ.This x 0 must be smaller than or equal to the radius r 0 of the inscribed sphere of τ.Since r 0 ≤ 1/n by ( * * ), we have 1 The edge length of a regular (n − 1)-simplex with circumradius 1 − (1/n) 2 is given by 1 − (1/n) 2 • l(n − 1), which is equal to l(n) as easily verified.Note that the contact point of S (O, x 0 ) and the facet f is the circum-center of the facet f , and it is an interior point of f .Hence, we can apply the inductive hypothesis to the set V f of n vertices of the facet f on a sphere of radius 1 Then, for 0 < x < 1/n, no facet of τ can touch S (O, x), for otherwise, we have b(V f ) > l(n) for some facet f of τ, as easily seen.Hence we can deduce, from r ≤ 1/n, that the sphere S (O, 1/n) must be tangent to all facets of τ.In this case, b(V) = l(n) implies that all edge-lengths of τ are equal to l(n), and τ is a regular n-simplex.
Proof [Proof of Theorem 1].Let σ be an n-simplex and f 0 , f 1 , . . ., f n be the facets of σ.Let − − → OP i be the unit outer normal vectors of the facets f i .Then P i ∈ S (O, 1).Let V = {P 0 , . . ., P n }, and τ be the simplex spanned by V.By Lemma 1, O is an interior point of τ.The dihedral angle ∠( f i , f j ) of the facets f i , f j (i j) and the angle ∠P i OP j are related as By the cosine law, ∠P i OP j = arccos(1 − 1 2 |P i P j | 2 ), and |P i P j | = l(n) if and only if ∠P i OP j = arccos −1 n .Since arccos x is monotone decreasing for 0 < x < π, it follows from the inequality a(V) ≤ l(n) ≤ b(V) in Lemma 2 that the minimum value of ∠P i OP j is less than or equal to arccos −1 n and the maximum value of ∠P i OP j is greater than or equal to arccos −1 n .Now, since arccos 1 n = π − arccos −1 n , the theorem follows.

Proof of Theorems 2 and 3
Proof [Proof of Theorem 2].Let v 0 , v 1 , . . ., v n be an n-dimensional star-simplex with vertex angle θ, and suppose v 0 is the apex.Let To compute the lateral angle δ, we may suppose that |v 0 v i | = 1 for i = 1, 2, . . ., n.Let n i denote the unit outer normal vector of the facet opposite to the vertex v i .Then cos δ = −n 1 •n 2 .We can write n 1 as with some x 2 , . . ., x n , y ∈ R. Since a i • n 1 = 0 for i 1 and a i • a j = cos δ for i j, we have where Thus, we may put n 1 , and (by symmetry) n 2 as (3) Subtracting ( 2) × (n − 1)x from (3), we have On the other hand, Using (3), Hence Then, x = ty by ( 2), and substituting this in (4), we have Now, from (5) we have Simplifying this, we get Since cos δ = −n 1 • n 2 , we have the theorem.
Proof [Proof of Theorem 3].It is enough to show that an n-dimensional star-simplex of lateral angle δ exists if and only if arccos 1 n−1 < δ < π.The necessity of δ < π is obvious.Let σ be an n-dimensional star-simplex with lateral angle δ, and let f 0 , f 1 , . . ., f n be the facets of σ, f 0 be the facet opposite to the apex.Let − − → OP i be the unit outer normal vectors of the facets f i .Then P i ∈ S (O, 1), and the simplex P 1 , . . ., P n is a regular (n − 1)-simplex.By Lemma 1, the simplex P 0 , P 1 , . . ., P n contains O in its interior.Hence its facet P 1 , . . ., P n does not contain O. Therefore, the circumradius of the facet P 1 , . . ., P n is less than 1.Let Q be the circumcenter of P 1 , . . ., P n .Since this facet is a regular (n − 1)-simplex, we have Thus, δ > arccos 1 n−1 .Proof of the converse is easy now, and it is omitted.

Proof of Theorem 4
Lemma 3 Let τ denote a (variable) n-dimensional star-simplex with lateral angle δ, and let ω 1 , . . ., ω n be the dihedral angles between the facet opposite to the apex and other facets.Then (i) inf Proof.Let f 0 , f 1 , . . ., f n be the facets of τ, f 0 be the facet opposite to the apex, and let − − → OP i be the unit outer normal vector of f i .Then P i ∈ S (O, 1).Let P * i ∈ S (O, 1) denote the antipodal point of Since ∠( f i , f j ) = δ for 1 ≤ i < j ≤ n, we may put Since the simplex P 0 , P 1 , . . ., P n contains O in its interior by Lemma 1, OP * 0 passes through an interior point of the facet P 1 , . . ., P n .Hence, ω i < ∠P 1 OP 2 = π − δ for i = 1, 2, . . ., n.If P * 0 approaches P 1 , then ω 1 → 0, and ω 2 → ∠P 1 OP 2 = π − δ.Hence we have (i).Since the facet P 1 , . . ., P n is a regular (n − 1) simplex, the maximum value of min i ω i = min i ∠P i OP * 0 is attained when OP * 0 passes through the center Q := 1 n (P 1 + P 2 + • • • + P n ) of the facet P 1 , . . ., P n .Therefore, cos(max Proof [Proof of Theorem 4].Let τ be a (variable) n-dimensional star-simplex with lateral angle δ, and let f 0 , f 1 , . . ., f n be the facets of τ, f 0 opposite to the apex.Let ω i = ∠( f i , f 0 ).