770 Combinatorics and Probability
If the acute angle of them-gon isAkA 1 Ak+r, the condition that this angle is acute
translates intor≤n. The other vertices of them-gon lie betweenAkandAk+r; hence
m− 2 ≤r, and these vertices can be chosen in
(r− 1
m− 3)
ways. Note also that 1≤k≤ 2 n−r.
Thus the number ofm-gons with an acute angle atA 1 is
∑nr=m− 2(^2) ∑n−r
k= 1
(
r− 1
m− 3)
= 2 n∑nm− 2(
r− 1
m− 3)
−
∑nr=m− 2r(
r− 1
m− 3)
= 2 n(
n
m− 2)
−(m− 2 )(
n+ 1
m− 1)
.
There are as many polygons with an acute angle atA 2 ,A 3 ,...,A 2 n+ 1.
To count the number ofm-gons with two acute angles, let us first assume that these
acute angles areAsA 1 AkandA 1 AkAr. The other vertices lie betweenArandAs.We
have the restrictions 2≤k≤ 2 n−m+3,n+ 2 ≤r<s≤k+nifk≤nand no
restriction onrandsotherwise. The number of suchm-gons is
∑nk= 1(
k− 1
m− 2)
+
2 n+ (^1) ∑−(m− 2 )
k=n+ 1
(
2 n+ 1 −k
m− 2)
=
∑nk=m− 1(
k− 1
m− 2)
+
∑ns=m− 2(
s
m− 2)
=
(
n+ 1
m− 1)
+
(
n
m− 1)
.
This number has to be multiplied by 2n+1 to take into account that the first acute vertex
can be at any other vertex of the regularn-gon.
We conclude that the number ofm-gons with at least one acute angle is
( 2 n+ 1 )(
2 n(
n
m− 2)
−(m− 1 )(
n+ 1
m− 1)
−
(
n
m− 1))
.
903.Denote byUnthe set ofz∈S^1 such thatfn(z)=z. Becausefn(z)=zm
n
,Un
is the set of the roots of unity of ordermn−1. In our situationn=1989, and we are
looking for those elements ofU 1989 that do not have period less than 1989. The periods
of the elements ofU 1989 are divisors of 1989. Note that 1989= 32 × 13 ×17. The
elements we are looking for lie in the complement ofU 1989 / 3 ∪U 1989 / 13 ∪U 1989 / 17. Using
the inclusion–exclusion principle, we find that the answer to the problem is
|U 1989 |−|U 1989 / 3 |−|U 1989 / 13 |−|U 1989 / 17 |+|U 1989 / 3 ∩U 1898 / 13 |+|U 1989 / 3 ∩U 1989 / 17 |
+|U 1989 / 13 ∩U 1989 / 17 |+|U 1989 / 3 ∩U 1989 / 13 ∩U 1989 / 17 |,i.e.,
|U 1989 |−|U 663 |−|U 153 |−|U 117 |+|U 51 |+|U 39 |+|U 9 |−|U 3 |.