14.02.2013 Views

Gruber P. Convex and Discrete Geometry

Gruber P. Convex and Discrete Geometry

Gruber P. Convex and Discrete Geometry

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

302 <strong>Convex</strong> Polytopes<br />

Proof. (i)⇒(ii) The proof of the following simple elementary proposition is left to<br />

the reader; see also Alex<strong>and</strong>rov [16].<br />

(1) Let H = F + G, where F <strong>and</strong> G are proper convex polygons. Mark an<br />

edge of H by +, −, or leave it unmarked, if its corresponding side in F has<br />

length greater than, less than, or equal to its corresponding side in G. Ifno<br />

edge of H is marked, then F <strong>and</strong> G coincide up to translation. If at least<br />

one edge of H is marked then, omitting the unmarked edges, there are at<br />

least four changes of sign on each circuit of H.<br />

Consider the polytope<br />

R = P + Q.<br />

Each edge of R is the sum of corresponding sides in P <strong>and</strong> Q, respectively. At<br />

least one of these sides is an edge, <strong>and</strong> if both are edges, they are parallel in the<br />

ordinary sense. Mark an edge of R by +, −, or leave it unmarked, if the length of<br />

its corresponding side in P is greater than, less than or equal to the length of its<br />

corresponding side in Q.<br />

The main part of the proof is to show the following proposition.<br />

(2) Let H = F + G be a facet of R, where F <strong>and</strong> G are the corresponding<br />

faces in P <strong>and</strong> Q, respectively. If no edge of H is marked, then F <strong>and</strong> G<br />

both are facets which coincide up to translation. If at least one edge of H is<br />

marked then, omitting the unmarked edges, there are at least four changes<br />

of sign on each circuit of H.<br />

The following simple remark will be useful in the proof of (2).<br />

(3) For each edge of H the corresponding sides in P <strong>and</strong> Q actually are corresponding<br />

sides in F <strong>and</strong> G, respectively. Thus we may define the signs of<br />

the edges of H by means of F <strong>and</strong> G.<br />

To show (2), we distinguish four cases, according to the different possibilities for<br />

F <strong>and</strong> G.<br />

First, F <strong>and</strong> G are facets. Then take into account the assumption in statement (i)<br />

<strong>and</strong> apply (3) <strong>and</strong> (1).<br />

Second, F is a facet <strong>and</strong> G an edge (or vice versa). Since, by the assumption in<br />

statement (i), G cannot be embedded into F, the sides of F parallel to G have shorter<br />

length than G. These sides of F are separated by edges of F which are not parallel<br />

to G. Clearly, for these edges of F, the parallel sides in G are the endpoints of G <strong>and</strong><br />

thus have length 0. Noting (3), we see then that there are at least four changes of sign<br />

on each circuit of H.<br />

Third, F <strong>and</strong> G are non-parallel edges. Then H is a parallelogram <strong>and</strong>, noting<br />

(3), the edges of H have alternating sign.<br />

Fourth, F is a facet <strong>and</strong> G a vertex (or vice versa). Since then G can be embedded<br />

into F, this possibility is ruled out by the assumption in statement (i). The proof of<br />

(2) is complete.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!