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Gruber P. Convex and Discrete Geometry

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16 Volume of Polytopes <strong>and</strong> Hilbert’s Third Problem 291<br />

(3) Let P = P1 ˙∪··· ˙∪Pk, where P1,...,Pk ∈ Pp, <strong>and</strong> extend f to a Hamel<br />

quasi-function g compatible with P1,...,Pk. Then<br />

D f (P) = Dg(P) = Dg(P1) +···+ Dg(Pk).<br />

The set of all vertices of P, P1,...,Pk dissect the edges of these polytopes into<br />

smaller line segments which we call links. The dihedral angle of one of the polytopes<br />

P, P1,...,Pk at a link is the dihedral angle of this polytope at the edge containing<br />

this link.<br />

If a link is contained in an edge of P, then, for the dihedral angles ϑ, ϑ1,...,ϑk<br />

of P, P1,...,Pk at this link, we have, ϑ = ϑ1 +···+ϑk. Thus<br />

f (ϑ) = g(ϑ) = g(ϑ1) +···+g(ϑk).<br />

If a link is contained in the relative interior of a facet of P, then, for the dihedral<br />

angles π, ϑ1,...,ϑk of P, P1,...,Pk at this link, the equality π = ϑ1 +···+ϑk<br />

holds. Thus<br />

f (π) = 0 = g(π) = g(ϑ1) +···+g(ϑk).<br />

If a link is in the interior of P, then, for the dihedral angles 2π, ϑ1,...,ϑk of<br />

P, P1,...,Pk at this link, we have 2π = ϑ1 +···+ϑk <strong>and</strong> thus<br />

f (2π) = 0 = g(2π) = g(ϑ1) +···+g(ϑk).<br />

Multiplying these equalities by the lengths of the corresponding links <strong>and</strong> summing<br />

over all links yields the equality in (3). ⊓⊔<br />

Corollary 16.3. Let S be a regular simplex <strong>and</strong> K a cube in E 3 with V (S) = V (K ).<br />

Then S <strong>and</strong> K are not equidissectable.<br />

Proof. The dihedral angle ϑ of S at any of its edges satisfies the equation cos ϑ =<br />

1/3. Apply the formula cos(α + β) = cos α cos β − sin α sin β with α = nϑ <strong>and</strong><br />

β =±ϑ to express cos(n + 1)ϑ <strong>and</strong> cos(n − 1)ϑ. Addition then yields<br />

cos(n + 1)ϑ = 2 cos ϑ cos nϑ − cos(n − 1)ϑ for n ∈ N.<br />

This, in turn, implies, by a simple induction argument, that<br />

cos nϑ = an<br />

3 n for n ∈ N, where an ∈ Z, 3 � | an.<br />

Since an/3 n �=±1, it follows that nϑ is not an integer multiple of π for any n ∈ N.<br />

Hence ϑ <strong>and</strong> π are rationally independent. Let f be a Hamel quasi-function such<br />

that f (ϑ) = 1, f (π) = 0 <strong>and</strong> let l be the edge-length of S. Then<br />

D f (S) = 6 l �= 0 = D f (K ).<br />

Thus S <strong>and</strong> K are not equidissectable by Dehn’s theorem 16.4. ⊓⊔

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