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ANSWERS,
H INTS &
SOLUTIONS
tUIallt
M1 16
Let n be the number of sides of
the given polygon iof course, n >- 1\.
Then the number of rts diagonals is
n(n - 3ll2 (because there are n - 3
diagonals issuing irom each oi its n
vertices, and this accounrs tr,r'ice for
each diagonal). So rre har.e to prove
that
sd
n n(n- 3r I
where s and d are the sums o{ the
iengths of all the pol,vgon's sides and
diagonals, respectir.elr-.
Consider two nonadiacent sides
AB and CD, and dragonals AC and
BD joining their endpoints and intersecting
at O 1iig. 1J. Since AB <
AO + OB, CD < CO + OD, we have
AB + CD. iAO + OC) + IBO + OD)
=AC+BD.
I{ u,e write out such inequalities
{or all pairs of nonadjacent sides and
add them up, we'l1 get (n - 3)s on the
le{t side (because each side enters r
- 3 pairs) andZd on the right side (because
each diagonal occurs in two
inequalities-it joins two pairs of
nonadjacent sides in the position of
figure 1). So (n - 3)s . 2d, which is
equivalent to what we set out to
prove.
Figure 1
M1 17
The answer isl-t. Consider any
of the permutations in question. Suppose
the number in the first place
(from the left) is k. If k > 1, then the
numbersk - l, k - 2,..., 1 will be met
just in this (decreasing) order as we
move from left to right. Indeed, the
number 1 must have 2 on its left side,
the number 2 must have 1 or 3-that
is, 3--on the left, 3 must have} or 4-
that is, 4-on the left, and so on.
Similarly, for k < n the numbers k + 1,
k + 2, ..., n must be arranged in increasing
order, becausen - 1 must be
to the left of n, then n - 2 to the left
of n - I, and so on. Therefore, arty of.
the permutations considered in the
problem is uniquely determined by
the set of places occupied by the
numbers l, 2, ...,k - 1 (there may be
no such places if k = l-that is, for
the identity permutation 1,2, ..., nl:
we have to affange these numbers in
these places in decreasing order and
the remaining numbers in the remaining
places in increasing order.
It's not hard to see that the number
of such sets is simply the number
of all subsets of the set of all the
n - 1 places except the first, and is
equal to 2" - I.
Ml18
We'Il use induction overm + fl to
prove a slightly generalized statement
with not Tess than k greatest
numbers underlined in each column,
andnot less thanl in each row
(this will be more convenient for
inductive reasoning).
Obviously, for m = n : k : I = I
the statement is true: ftJ: 1 number
(the only one) will be underlined
twice. Let's now show that the
statement for an m x n afiay can be
reduced to the case of an lm - Il x n
ormx(n-llaruay.
If all the underlined numbers in an
m x n afiay are underlined twice, then
there are no less thankJ of them. Otherwise,
let a be the greatest of the
numbers underlined once. It's either
one of the k greatest numbers in its
column or one of the I greatest numbers
in its row. Assume it was underlined
"along a column." Then the 1
greatest numbers in its row are
greater than a and so are underlined
twice. Cancel out this row. We get an
(m - ll x n arrayt in which at least 1
greatest numbers are underlined in
each line and at least k - 1 numbers
in each column. The induction hypothesis
implies that at least (k - 1)1
numbers in this reduced afiay are
underlined twice. The same numbers
are underlined twice in the big array;
together with at least I numbers
underlined twice in the line that was
deleted, this makes at least (k - lll + 1 :
ld numbers, completing the proof.
Ml19
Let's first prove that PAr = PA2 =
... : PAn.Suppose this is not true.
Choose the shortest and the longest
of these edges, PA, and PAr, respectively.
Now draw atangleBAC congruent
to angles PA.A,, ..., PAnAt,
and mark off the segmentAB : ArA,
: . . . = A,lron one of its sides, and segments
AC = Ar_rP and AD = Aj_P
on the other side. (Figure 2 shows the
Figure 2
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