Methods of Proof 355=cos∠DCA+cos∠BCA+cos∠CAD+cos∠CAB
=cos∠DCA+cos∠CAD+cos∠ADC+cos∠BCA+cos∠CAB+cos∠ABC,and we are done.
Remark.A more general theorem states that two triangulations of a polygonal surface
(not necessarily by diagonals) are related by the move from Figure 57 and the move from
Figure 58 or its inverse. These are usually called Pachner moves.
Figure 58(Indian Team Selection Test for the International Mathematical Olympiad, 2005,
second solution by A. Tripathy)
77.LetSbe the sum of the elements of the table. By performing moves on the rows or
columns with negative sum, we obtain a strictly increasing sequenceS 1 <S 2 <···.
BecauseScan take at most 2n
2
values (all possible sign choices for the entries of the table),
the sequence becomes stationary. At that time no row or column will have negative sum.
78.Skipping the first step, we may assume that the integers are nonnegative. The semi-
invariant isS(a, b, c, d)=max(a,b,c,d). Because for nonnegative numbersx, y,we
have|x−y|≤max(x, y),Sdoes not increase underT.IfSdecreases at every step,
then it eventually becomes 0, in which case the quadruple is( 0 , 0 , 0 , 0 ). Let us see in
what situationSis preserved byT.If
S(a, b, c, d)=S(T (a, b, c, d))=S(|a−b|,|b−c|,|c−d|,|d−a|),then next to some maximal entry there must be a zero. Without loss of generality, we
may assumea=S(a, b, c, d)andb=0. Then
(a, 0 ,c,d)
T
−→(a, c,|c−d|,|d−a|)
T
−→(|a−c|,|c−|c−d|,||c−d|−|d−a||,|a−|d−a||).CanSstay invariant in both these steps? If|a−c|=a, thenc=0. If|c−|c−d|| =a,
then sinceais the largest of the four numbers, eitherc=d=aor elsec=0,d=a.
The equality||c−d|−|d−a|| =acan hold only ifc=0,d=a,ord=0,c=a.
Finally,|a−|d−a||=aifd=a.SoSremains invariant in two consecutive steps only
for quadruples of the form