The Extremal Principle
Example 1. Given are n red points and n blue points in the plane, no three collinear. Show
that we can draw n nonintersecting segments connecting the blue points to the red points.
Solution using monovariance: Consider any arbitrary pairing of the blue points with the
red points. We consider the following operation: If any two such segments cross, their
endpoints form a convex quadrilateral with vertices in the order R, B, B, R (clockwise or
counterclockwise).
Replace the two R−B diagonals (which the segments must be, since they intersect)
with the two R−B sides. It follows from the triangle inequality that the sum of the lengths of
these sides is less than the sum of the diagonals, so our operation decreases the total length
of the n segments, let’s call it ℓ. This total length, say ℓ, is our monovariant.
Since there are only finitely many possible pairings, iteration of the above operation
must eventually lead us somewhere from which we can proceed no further, i.e. no two
segments intersect, as desired.
Solution using extremal principle: Consider all possible pairings and for each pairing
consider the value of this total length ℓ of the n segments. Since there are only finitely
many possible pairings, there must a configuration with a minimal value of ℓ . In this
configuration no two segments must intersect, because otherwise we can perform the
above operation to decrease ℓ further, contradicting the minimality.
Example 2. The Parliament of Sikinia consists of one house. Every member has at most
three enemies among the remaining members. Note that a member cannot be his own
enemy, and enmity is mutual. Show that one can split the house into two houses so that
every member has one enemy at most in his house.
Solution. We consider all possible partitions of the house. For each partition, we consider
the sum of the number of enemies each member has in his own house, let’s call it E. We
claim that a partition with minimal E will have the desired property. Indeed, if in this
minimal partition one member has more than one enemy in his own house, then we can
switch him to the other house, which decreases E further, hence contradicting the
minimality of E for this partition.
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Extremal Principle: In a given setup, we consider an element/configuration with some
maximal or minimal property and show that it has/leads to the desired property.
Example 3. On each square of an 8×8 chessboard, a real number is written. Suppose that
each number equals the average (arithmetic mean) of its neighbours (above, below, left,
right). Show that all the numbers on the chessboard must be equal.
Solution. Among the numbers on the board, consider a number which is maximal, say m.
(Maximal in the sense that no number on the chessboard is strictly larger than this m. Why
maximal, not maximum? Because it may not be unique.). This number m is the average of
its neighbours. Now if any of its neighbours is strictly smaller than m, then their average
will be strictly smaller than m (since the other neighbours are ≤ m), which will contradict
the fact that their average is exactly m. This implies that each of its neighbours must be
equal to m. Now it is easy to see that this process spreads across the whole board.
Example 4. Let S be a set of points in the plane such that each point in S is a midpoint of
two other points in S. Show that S must be an infinite set.
Solution. Let, if possible, S be finite. Setup a coordinate system. Now consider the points
which are rightmost (i.e. those points which have a maximal x-coordinate). Among these,
consider the topmost point (which have the largest y-coordinate among these points), let
us call this P. Now P must be the midpoint of a segment joining two points, say A and B.
Since P has a maximal x-coordinate, it follows that none of A and B can lie on its right side.
This forces A and B to lie on the line x = c where c is the x-coordinate of P. But then one of
these points A and B will lie above P, contradicting the maximality of the y-coordinate of P
(among all points of S lying on that line x = c).
Example 5. Show that for every n > 1, the number 2^n -1 is not a multiple of n.
Solution. Let, if possible, let n > 1divide 2n − 1. Note that n cannot be even, since then it
won’t divide the odd number 2n − 1. Consider a prime p that divides n . Since p is odd,
Fermat’s theorem tells us that 2p−1 = 1 (mod p) . Thus, p | 2n − 1 and p | 2p−1 − 1 , which
implies p | gcd(2n − 1, 2p−1 − 1) = 2gcd(n,p−1) − 1. (Not familiar with this result? see below.)
So for any prime p dividing n , we have shown that p ∣ 2gcd(n,p−1) − 1 . Now among all
prime divisors of n, take p to be the minimum one. The minimality of p implies that gcd(n,
p − 1) = 1, because each prime that divides p − 1 is less than p and hence cannot divide n.
Therefore, p divides 2gcd(n,p−1) − 1 = 21 − 1 = 1, which is a contradiction.
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Why gcd(2a − 1, 2b − 1) = 2gcd(a,b) − 1?
Let gcd(2a − 1, 2b − 1) = d . Then d | 2a − 1, d | 2b − 1. Write b = a q + r . (say, b > a . )
Then d | (2b − 1) − (2a − 1) ⇒ d | 2b − 2a ⇒ d | 2b−a − 1 ⇒ d | 2b−a − 2a
… ⇒ d | 2b−aq − 1, i.e. d | 2r − 1.
Now repeat the above with d | 2a − 1 and d | 2r − 1, and so on. Notice that we have
Euclidean division algorithm running on the exponents of 2. It follows that at the last step
we would get d | 2gcd(a,b) − 1. On the other hand, since 2gcd(a,b) − 1 is a common divisor of
2a − 1 and 2b − 1, it must divide their gcd, d. Hence it follows that d = 2gcd(a,b) − 1.
Example 6. There are n points given in the plane. Any three of the points form a triangle of
area ≤ 1. Show that all n points lie in a triangle of area ≤ 4.
Solution. Consider all possible triangles that can be formed using these n points. Among
these triangles, consider the one with a maximal area, call it ABC. Draw lines through A,
B, C parallel to their opposite sides, intersecting at points D, E, F as shown. Since ABC is
the medial triangle of DEF, it follows that area(DEF) = 4 × area(ABC) ≤ 4.
Now we claim that all the n points must lie within this bigger triangle DEF. Suppose not,
say there is a point P outside DEF. Assume w.l.o.g that P lies on the opposite side of side
DF of the triangle. It is easy to see that area of triangle PAC is greater than the area of ABC,
thus contradicting the maximality of ABC.
Example 7. Rooks are placed on the n×n chessboard satisfying the following condition: If
the square (i, j) is free, then at least n rooks are on the i-th row and j-th column together.
Show that there are at least n 2 /2 rooks on the board.
Solution. The board has n rows and n columns. Among the rows and columns combined,
consider one with the minimal number of rooks. W.l.o.g. suppose that it is a row. Now if
this minimal row has at least n/2 rooks, then the conclusion follows (why?).
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Suppose not. Let, if possible, this row have k rooks where k < n /2. Consider the n − k
empty squares in this row. In each of their column, we must have at least n − k rooks (due
to the given condition). What about the remaining columns? Each of them must have at
least k rooks (since k is the minimal number of rooks in any row/column).
Therefore, the total number of rooks in the board is at least (n − k)2 + k 2 . It remains to
show that (n − k)2 + k 2 ≥ n 2 /2. It follows from the inequality 2(a 2 + b 2 ) ≥ (a + b)2 .
Example 8. We call a polynomial f (x) having integer coefficients to be primitive if the gcd
of the coefficients of f (x) is 1. Show that product of two primitive polynomials is also a
primitive polynomial.
Solution. Let f (x) = ∑k≥0 ak x k and g(x) = ∑k≥0 bk x k be two primitive polynomials. If the
product f (x)g(x) is not primitive, there must be a prime p which is a common divisor of all
its coefficients. But p can not divide all the coefficients of either f (x) or g(x) (otherwise they
would not be primitive). Let r be the minimum index such that ar is not divisible by p and
let s be the minimum index such that bs is not divisible by p. Now consider the term x r+s in
the product. Its coefficient (which can be 0 as well) must be of the form:
ar bs + ar+1bs−1 + ar+2 bs−2 + ⋯ + ar−1bs+1 + ar−2 bs+2 + ⋯ .
Note that the first term in the above sum is not divisible by p, while each of the remaining
terms are, so the entire sum cannot be divisible by p . But by our assumption all the
coefficients in the product f (x)g(x) must be divisible by p. Thus we get a contradiction.
In today’s class we saw various applications of the extremal principle, which is very
intuitive as a principle but widely applicable and tremendously useful for proof writing.
To master the usage of this principle, practice more problems from Problem Solving
Strategies (Arthur Engel) and other online resources.
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