SOLUTIONS TO PROBLEMS
MATH 397, WILLIAMS COLLEGE, FALL 2010
A BSTRACT. These are solutions to various interesting problems available on the web.
Our course homepage is
http://www.williams.edu/Mathematics/sjmiller/public html/161/index.htm
and we generously thank many institutions for posting and sharing their training material; we hope our information will be similarly useful. The solutions below are written
by the students before or after our discussion sessions; note that the write-up is not
necessarily by the presenter of the problem. Many of the solutions below have been
expanded by the instructor, Steven J. Miller. For more information, please contact him
at [email protected].
0. W EEK 0: I NDUCTION AND P IDGEONHOLE P RINCIPLE
0.1. Problem 5 (Liyang Zhang).
Problem 0.1. Let 𝑃 (𝑥) be a polynomial with integer coefficients at most 𝑛 and integer
degree at most 𝑛 but greater than 1. Suppose that ∣𝑃 (𝑥)∣ < 𝑛 for all ∣𝑛∣ < 𝑛2 . Show
that 𝑃 (𝑥) is constant.
Solution: Let’s first analyze what strategy we should use. Should we use induction
or the pidgeonhole principle (as we have the hint that this problem is from a section
on those two topics!). Induction at first seems promising; perhaps we can induct on
the degree 𝑛. It’s not a bad idea to say to yourself: let me try using technique 𝑋. In
this case, though, it seems like induction won’t work. The problem is that a general
polynomial need not factor into polynomials of smaller degree. Even assuming it did,
there is nothing that would ensure that the resulting coefficients of the polynomials of
smaller degree would satisfy the needed bounds. Thus, it is unlikely that induction is
the way to go.
One natural thing that comes to mind after reading the first sentence is a polynomial
with degree at most 𝑛 has at most 𝑛 solutions in the real numbers. We are asked to
prove that 𝑃 (𝑥) is constant. There are several ways to prove a polynomial is constant:
its derivative is 0, or it can be factored into polynomials that are constants, or maybe
the values we plug in are all roots to of polynomial. Let’s now consider some standard
strategies. We can try and factor polynomials to lower degrees, but as discussed above
there are numerous issues with such an approach. The two bounds in the problem must
suggest something. If we plug integers into 𝑃 (𝑥), we will get integer values. So for
𝑥 with ∣𝑥∣ within the bound, our polynomial takes on one of 2𝑛 − 1 values; we have
2𝑛2 − 1 choices for 𝑥. We now use the Pigeonhole principle, and see that at least one
of these 2𝑛 − 1 values in {−(𝑛 − 1), . . . , 𝑛 − 1} (say 𝑎) must be assumed by at least
2𝑛2 − 1
(2𝑛2 − 𝑛) + (𝑛 − 1)
𝑛−1
=
= 𝑛+
> 𝑛
2𝑛 − 1
2𝑛 − 1
2𝑛 − 1
1
(0.1)
2
MATH 397, WILLIAMS COLLEGE, FALL 2010
different 𝑥 with ∣𝑥∣ < 𝑛2 . Of course, the above is greater than 𝑛 only if 𝑛 > 1; the
problem as stated is actually wrong when 𝑛 = 1. So the related polynomial 𝑃 (𝑥) − 𝑎
has more than 𝑛 roots, which is impossible as 𝑃 is a polynomial of degree 𝑛 and thus
𝑃 (𝑥) and 𝑃 (𝑥) − 𝑎 can have at most 𝑛 roots unless they are constant.
0.2. Fibonacci Problem (David Thompson).
2
Problem 0.2. Prove 𝐹𝑛+1
+ 𝐹𝑛2 = 𝐹2𝑛+1 .
Proof : We proceed inductively. For 𝑛 = 1 we see:
𝐹22 + 𝐹12 = 12 + 12 = 2 = 𝐹3 .
(0.2)
Assume Equation 0.2 holds for all 𝑛 up to some integer 𝑘, meaning
2
𝐹𝑘+1
+ 𝐹𝑘2 = 𝐹2𝑘+1 .
(0.3)
We show the 𝑘 + 1 case follows. Consider
2
2
2
𝐹𝑘+2
+ 𝐹𝑘+1
= (𝐹𝑘+1 + 𝐹𝑘 )2 + 𝐹𝑘+1
2
2
= 𝐹𝑘+1
+ 2𝐹𝑘+1 𝐹𝑘 + 𝐹𝑘2 + 𝐹𝑘+1
(0.4)
(0.5)
2
2
+
and 𝐹𝑘2 as 𝐹2𝑘+1 . Let 𝑆𝑘 = 𝐹𝑘+1
By our inductive hypothesis we can group 𝐹𝑘+1
2𝐹𝑘+1 𝐹𝑘 . If 𝑆𝑘 = 𝐹2𝑘+2 , we’re done. Let’s continue to expand 𝑆𝑘 . Substituting 𝐹𝑘+1 =
𝐹𝑘 + 𝐹𝑘−1 , we see:
2
2
𝑆𝑘 = 𝐹𝑘+1
+ 2𝐹𝑘+1 𝐹𝑘 = 𝐹𝑘2 + 2𝐹𝑘 𝐹𝑘−1 + 𝐹𝑘−1
+ 2𝐹𝑘2 + 2𝐹𝑘 𝐹𝑘−1
= 𝐹2𝑘−1 +
2(𝐹𝑘2
+ 𝐹𝑘 𝐹𝑘−1 ) = 𝐹2𝑘−1 + 2𝑆𝑘−1 .
(0.6)
(0.7)
We can analogously expand 𝑆𝑘−1 and we find 𝑆𝑘−1 = 𝐹2𝑘−3 + 2𝑆𝑘−2 . We can continue
this process all the way down to 𝑆1 = 𝐹22 + 2𝐹2 𝐹1 = 3. Therefore:
𝑆𝑘 = 𝐹2𝑘−1 + 2𝐹2𝑘−3 + 4𝐹2𝑘−5 + ... + 2𝑘−1 ⋅ 3
(0.8)
Since our goal is to show 𝑆𝑘 = 𝐹2𝑘+2 , let’s use induction again.
Claim: 𝑆𝑛 = 𝐹2𝑛+2 .
Proof: With 𝑛 = 2 we see 𝐹3 + 2 ∗ 3 = 8 = 𝐹6 . Assume our claim holds up to some
integer 𝑘. Then when 𝑛 = 𝑘 + 1 we see:
𝑆𝑘+1 = 𝐹2𝑘+1 + 2𝐹2𝑘−1 + ... + 2𝑘 ⋅ 3 = 𝐹2𝑘+1 + 2(𝐹2𝑘+2 ) = 𝐹2𝑘+4 ,
2
and our claim is proven. Thus 𝑆𝑛 = 𝐹2𝑛+2 , meaning 𝐹𝑛+1
+ 𝐹𝑛2 = 𝐹2𝑛+1 .
(0.9)
SOLUTIONS TO PROBLEMS
3
1. WEEK TWO
Problem 1.1. Find the smallest number with 28 divisors. (Solution by Liyang Zhang)
People can interpret this problem different ways and both ways are interesting!
Interpretation 1: Find the smallest number with exactly 28 divisors.
Divisors of a number can always be related to its prime divisors. Two divisors are
different if they have different prime factorizations. So exactly 28 unique divisors mean
prime divisors of this number 𝑁 can be combined into exactly 28 unique numbers. So
let’s look at how a number factors:
𝑁 = 𝑝𝑟11 𝑝𝑟22 ⋅ ⋅ ⋅ 𝑝𝑟𝑛𝑛
(1.1)
with each 𝑝𝑖 distinct. We will have
𝑚 = (𝑟1 + 1)(𝑟2 + 1) ⋅ ⋅ ⋅ (𝑟𝑛 + 1)
(1.2)
distinct divisors since we can choose powers from 0 to 𝑟𝑖 for each prime 𝑝𝑖 . In this
problem, we know 𝑚 = 28; we just need to find each individual 𝑟𝑖 and 𝑝𝑖 to minimize
𝑁 . Now let’s worry about the choice for 𝑟𝑖 first. Note 28 can be factored as follows:
28 = 1 ⋅ 28 = 2 ⋅ 14 = 4 ⋅ 7 = 2 ⋅ 2 ⋅ 7.
(1.3)
Now we can assign 𝑝𝑖 to each 𝑟𝑖 . But we can eliminate two factorizations (1 ⋅ 28 and
2 ⋅ 14) immediately since any prime raised to the 27th or 13th power would be too big.
Now with the remaining two factorizations, we will assign the smallest prime 2 to the
power 6 since powers grow rapidly. Now we will either have 26 ⋅ 33 or 26 ⋅ 3 ⋅ 5. Since
we want the smallest one, 26 ⋅ 3 ⋅ 5 = 960 is the solution.
Interpretation 1: Find the smallest number with at least 28 divisors.
This is a harder problem if we ask the question this way. Suppose we have a factorization:
𝑁 = 𝑝𝑟11 𝑝𝑟22 ...𝑝𝑟𝑛𝑛 .
(1.4)
The number of divisors will be
𝑚 = (𝑟1 + 1)(𝑟2 + 1)...(𝑟𝑛 + 1).
(1.5)
Suppose the smallest number with at least 𝑚 divisors is 𝑁 . Now we ask: what is
smallest number with at least 𝑛 divisors for 𝑛 > 𝑚. We can do so by increasing some
𝑟𝑖 , adding a new prime 𝑝𝑛+1 , or doing the combination of reducing some 𝑟𝑖 and adding
a new prime. We don’t know which direction will give us a global minimum instead of
a local minimum. For now, we can repeat the process in Interpretation 1 for the next
several 𝑚 greater than 28 and find out that 𝑚 = 32 can give us a global minimum
with 𝑁 = 840. But this strategy is definitely not suitable for large 𝑚. (Possible future
research project?)
4
MATH 397, WILLIAMS COLLEGE, FALL 2010
Problem 1.2. Show for 𝑛 > 1 that 1 + 1/2 + 1/3 + ⋅ ⋅ ⋅ + 1/𝑛 is not an integer. (Solution
by David Thompson)
Proposition: For all 𝑘 > 1, the 𝑘 𝑡ℎ partial sum of the harmonic series,
𝐻𝑛 ≡
𝑘
∑
1
𝑖=1
𝑘
(1.6)
is not an integer.
Proof: We show that in lowest terms, 𝐻𝑛 = 𝐴𝑛 /𝐵𝑛 has an odd numerator and an even
denominator. We need two facts first.
Lemma 1: Let 𝑛𝑘 = [log2 (𝑘)]. Then 2𝑛𝑘 ∣ 𝐵𝑘 , but 2𝑛𝑘 +1 does not divide 𝐵𝑘 .
Proof: We proceed inductively. Notice that when 𝑘 = 1, 𝑛𝑘 = 0, so 2𝑛𝑘 = 1 which
certainly divides 𝐵𝑘 = 1, but 2𝑛𝑘 +1 = 2, which does not divide 𝐵𝑛 . Assume that for
some 𝑘, we know that 2𝑛𝑘 ∣ 𝐵𝑘 and 2𝑛𝑘 +1 does not divide 𝐵𝑛 . Then for the 𝑘 + 1𝑠𝑡 term
we have:
𝐴𝑘
1
𝐴𝑘 (𝑘 + 1) + 𝐵𝑘
+
=
.
(1.7)
𝐵𝑘 𝑘 + 1
(𝑘 + 1)𝐵𝑘
We need to reduce this to lowest terms. Let 𝑎 be the largest integer dividing 𝑘 + 1 and
𝐵𝑘 . Since 𝐴𝑘 and 𝐵𝑘 are already coprime, we have
𝐻𝑘+1 =
𝐴𝑘 (𝑘 + 1)/𝑎 + 𝐵𝑘 /𝑎
(1.8)
(𝑘 + 1)/𝑎 ⋅ 𝐵𝑘
in lowest terms. Notice that our new denominator, 𝐵𝑘+1 = (𝑘 + 1)/𝑎 ⋅ 𝐵𝑘 . Therefore,
if 𝑛𝑘 = 𝑛𝑘+1 , we’re done, since 2𝑛𝑘 ∣ 𝐵𝑘 (we note that 2𝑛𝑘 +1 cannot divide 𝐵𝑘+1 in this
case because 𝐵𝑘 has at least as many factors of 2 in it as 𝑘 + 1 does, meaning (𝑘 + 1)/𝑎
is odd). If 𝑛𝑘 ∕= 𝑛𝑘+1 , we know that 𝑘 + 1 is a power of 2, 𝑘 + 1 = 2𝑛𝑘 +1 . In this case
we must have 𝑎 = 2𝑛𝑘 since we are assuming 𝐵𝑛 is divisible by 2𝑛𝑘 but not by 2𝑛𝑘 +1 .
Therefore
𝐻𝑘+1 =
2𝐴𝑘 + 𝐵𝑘 /(2𝑛𝑘 )
,
(1.9)
2𝐵𝑘
= 2𝐵𝑘 . This shows that 2𝑛𝑘+1 does indeed divide 𝐵𝑘+1 , but that 2𝑛𝑘+1 +1
𝐻𝑘+1 =
implying 𝐵𝑘+1
does not.
Now we can prove the main result.
Theorem: 𝐻𝑘 is never an integer.
Proof: Again we proceed inductively. For all 𝑘 > 1 we claim that the numerator of
𝐻𝑘 is odd and the denominator is even, which implies 𝐻𝑘 is not an integer. For 𝑘 = 2
we see 𝐻𝑘 = 3/2. Assume that for some 𝑘, 𝐻𝑘 = 𝐴𝑘 /𝐵𝑘 , where 𝐴𝑘 is odd and 𝐵𝑘 is
even. Then:
1
𝐴𝑘 (𝑘 + 1) + 𝐵𝑘
𝐴𝑘
+
=
.
(1.10)
𝐵𝑘 𝑘 + 1
(𝑘 + 1)𝐵𝑘
If 𝑘 + 1 is odd, the result follows immediately. Assume 𝑘 + 1 is even, and factor out as
many powers of 2 as possible, writing 𝑘 + 1 = 2𝑚 ⋅ 𝑀 for 𝑀 odd. If 𝑚 < 𝑛𝑘 , we’re
done, since we can divide both sides by 2𝑚 . The numerator will be 𝐴𝑘 ⋅ 𝑀 + 𝐵𝑘 /2𝑚 ,
𝐻𝑘+1 =
SOLUTIONS TO PROBLEMS
5
which will be odd since 𝐵𝑘 /2𝑚 is even. Also, since 𝐵𝑘 is even, the denominator will be
even.
We don’t need to worry about the case when 𝑚 = 𝑛𝑘 . If 𝑘 + 1 = 2𝑛𝑘 ⋅ 𝑀 , then 𝑘 + 1
is equal to an odd number times the largest power of 2 smaller than 𝑘. Since we can’t
have 𝑀 = 1 (𝑘 + 1 can’t be smaller than 𝑘), we’re forced to conclude 𝑀 ≥ 3, which is
impossible since it would imply 𝑘 + 1 − 𝑘 > 2𝑛𝑘 +1 − 2𝑛𝑘 − (2𝑛𝑘 +1 − 1) > 1.
The only other case we have to worry about, then, is when 𝑚 = 𝑛𝑘+1 , meaning
𝑘 + 1 = 2𝑛𝑘 +1 . In this case, divide both sides by 2𝑛𝑘 . This gives:
2𝐴𝑘 + 𝐵𝑘 /2𝑛𝑘
.
(1.11)
2𝐵𝑘
However, since 2𝑛𝑘 divides 𝐵𝑘 but 2𝑛𝑘 +1 does not, we know 𝐵𝑘 /2𝑛𝑘 is odd. Therefore
our numerator is odd, and our denominator is even. Thus for all 𝑘, 𝐻𝑘 is the ratio of an
odd number to an even number, and is therefore not an integer.
𝐻𝑘+1 =
2. J ENSEN ’ S I NEQUALITY: A RNOSTI
The following problems introduce (a specific form of) Jensen’s Inequality and explore its use in several problems.
Jensen’s Inequality states that if 𝑓 is convex (i.e. has positive second derivative) on
an interval 𝐼 of ℝ and 𝑥1 , . . . 𝑥𝑛 , ∈ 𝐼, then
𝑥1 + ⋅ ⋅ ⋅ + 𝑥𝑛
1
) ≤ (𝑓 (𝑥1 ) + ⋅ ⋅ ⋅ + 𝑓 (𝑥𝑛 )).
𝑛
𝑛
The case 𝑛 = 2 can be verified visually: it essentially states that if 𝑓 is concave-up,
then any chord connecting two points on the curve 𝑦 = 𝑓 (𝑥) lies above the function.
This can also be verified algebraically. Given any two points 𝑎, 𝑏 ∈ 𝐼 with 𝑎 ≤ 𝑏, note
that
∫ 𝑎
∫ 𝑏
𝑎+𝑏
𝑎+𝑏
′
𝑓 (𝑎) = 𝑓 (
)+
𝑓 (𝑥)𝑑𝑥,
𝑓 (𝑏) = 𝑓 (
)+
𝑓 ′ (𝑥)𝑑𝑥.
𝑎+𝑏
𝑎+𝑏
2
2
2
2
𝑓(
Then
𝑓 (𝑎) + 𝑓 (𝑏)
𝑎+𝑏
1
= 𝑓(
)+
2
2
2
(∫
𝑏
𝑓 ′ (𝑥)𝑑𝑥 −
𝑎+𝑏
2
∫
𝑎+𝑏
2
)
𝑓 ′ (𝑥)𝑑𝑥 .
(2.1)
𝑎
Note that the first of these integrals has value at least the minimum of 𝑓 ′ on [ 𝑎+𝑏
, 𝑏]
2
𝑏−𝑎
times the width of the interval, 2 . Similarly, the second of these integrals has value
at most the maximum of 𝑓 ′ on [𝑎, 𝑎+𝑏
] times the width of the interval (also 𝑏−𝑎
). Since
2
2
𝑎+𝑏
′′
′
′
𝑓 (𝑥) > 0 on 𝐼, 𝑓 is increasing on 𝐼, so the maximum of 𝑓 on [𝑎, 2 ] is at most its
minimum on [ 𝑎+𝑏
, 𝑏]. Thus, difference in the integrals in (2.1) is non-negative, proving
2
the claim.
Having established the case of 𝑛 = 2, we prove the full inequality by induction on 𝑛.
Rather than the standard induction technique of proving that if a property holds for 𝑘,
then it holds for 𝑘 + 1, we go backwards: if it holds for 𝑘, then it holds for 𝑘 − 1. We
then need only show that it holds for some family of values for 𝑘 which grows arbitrarily
large. In this case, I will take my family to be all powers of 2.
6
MATH 397, WILLIAMS COLLEGE, FALL 2010
Suppose that Jensen’s Inequality holds for 𝑛 = 2𝑘 , 𝑘 > 0, and Suppose we have
values 𝑥1 , . . . , 𝑥2𝑘+1 ∈ 𝐼. Define
𝑥1 + ⋅ ⋅ ⋅ + 𝑥2𝑘
𝑥 𝑘 + ⋅ ⋅ ⋅ + 𝑥2𝑘+1
𝑦1 =
𝑦2 = 2 +1
.
𝑘
2
2𝑘
Note that because 𝑦1 and 𝑦2 are averages of points in 𝐼, they are also in 𝐼. Thus,
)
(
)
(
𝑦1 + 𝑦2
𝑓 (𝑦1 ) + 𝑓 (𝑦2 )
𝑥1 + . . . + 𝑥2𝑘+1
=𝑓
≤
.
𝑓
𝑘+1
2
2
2
But by our induction assumption,
𝑓 (𝑥1 ) + . . . + 𝑓 (𝑥2𝑘 )
𝑓 (𝑥2𝑘 +1 ) + . . . + 𝑓 (𝑥2𝑘+1 )
,
𝑓 (𝑦2 ) ≤
.
𝑘
2
2𝑘
Combining these inequalities yields the desired result, so Jensen’s Inequality holds for
all powers of 2.
To complete the proof, suppose that Jensen’s Inequality holds for 𝑛 = 𝑘 > 1. For
𝑘−1
arbitrary 𝑥1 , . . . , 𝑥𝑘−1 ∈ 𝐼, define 𝑥𝑘 = 𝑥1 +⋅⋅⋅+𝑥
, and note that 𝑥𝑘 ∈ 𝐼. By our
𝑘−1
assumption,
𝑓 (𝑦1 ) ≤
𝑥1 + ⋅ ⋅ ⋅ + 𝑥𝑘
1
) ≤ (𝑓 (𝑥1 ) + ⋅ ⋅ ⋅ + 𝑓 (𝑥𝑘 )).
𝑘
𝑘
from both sides yields
𝑓(
Subtracting
𝑓 (𝑥𝑘 )
𝑘
1
1
𝑥1 + ⋅ ⋅ ⋅ + 𝑥𝑘
) − 𝑓 (𝑥𝑘 ) ≤ (𝑓 (𝑥1 ) + ⋅ ⋅ ⋅ + 𝑓 (𝑥𝑘−1 )).
𝑘
𝑘
𝑘
Note that by our definition of 𝑥𝑘 ,
𝑓(
(2.2)
𝑥1 + ⋅ ⋅ ⋅ + 𝑥𝑘
1 𝑥1 + ⋅ ⋅ ⋅ + 𝑥𝑘−1
𝑥1 + ⋅ ⋅ ⋅ + 𝑥𝑘−1
= (1 +
)
=
= 𝑥𝑘 .
𝑘
𝑘−1
𝑘
𝑘−1
In fact, 𝑥𝑘 was carefully chosen to make this true. In general with this type of induction,
the challenge is making the right choice for 𝑥𝑘 . We see now that the left-hand-side of
(2.2) simplifies to
1
𝑘 − 1 𝑥1 + ⋅ ⋅ ⋅ + 𝑥𝑘−1
(1 − )𝑓 (𝑥𝑘 ) =
𝑓(
).
𝑘
𝑘
𝑘−1
The right-hand-side, meanwhile, can be rewritten
𝑘 − 1 𝑓 (𝑥1 ) + ⋅ ⋅ ⋅ + 𝑓 (𝑥𝑘−1 )
⋅
.
𝑘
𝑘−1
Multiplying both sides by
𝑘
𝑘−1
yields the desired inequality, completing our proof.
From Jensen’s Inequality, we can derive the famed AM-GM inequality, which states
that if 𝑎1 . . . 𝑎𝑛 are positive real numbers, then their arithmetic mean is at least as large
as their geometric mean. How can we show this? We need to convert Jensen’s Inequality
(which involves only sums) to
𝑎1 + ⋅ ⋅ ⋅ + 𝑎𝑛
(𝑎1 ⋅ ⋅ ⋅ 𝑎𝑛 )1/𝑛 ≤
,
𝑛
SOLUTIONS TO PROBLEMS
7
which involves products. Therefore, it makes sense to consider the logarithm, which
converts products to sums. Let 𝑥𝑖 = ln(𝑎𝑖 ), and 𝑓 (𝑥) = 𝑒𝑥 . Note that the second derivative of 𝑓 is positive on (−∞, ∞), so we may apply Jensen’s Inequality. When we do
so, AM-GM pops out immediately!
)𝑥+1 for 𝑥 > 0. Instead
We will now use Jensen’s Inequality to prove that 𝑥𝑥 ≥ ( 𝑥+1
2
of proving this directly, we will take the natural logarithm of both sides and prove that
for positive 𝑥, 𝑥 ln(𝑥) ≥ (𝑥 + 1) ln( 𝑥+1
) (this is equivalent because 𝑔(𝑥) = ln(𝑥) in2
creases monotonically). Consider the function 𝑓 (𝑥) = 𝑥 ln(𝑥). Note that 𝑓 ′′ (𝑥) = 1/𝑥,
(1)
which is positive on (0, ∞). Hence, 𝑓 (𝑥)+𝑓
≥ 𝑓 ( 𝑥+1
). Since 𝑓 (1) = 0, this gives us
2
2
𝑥+1
𝑓 (𝑥) ≥ 2𝑓 ( 2 ), as desired.
We will conclude with a problem that uses the same idea, but in the opposite direction. If our function of interest is concave (has negative second derivative) on the
interval 𝐼, then Jensen’s Inequality can be reversed (feel free to prove this).
√
√
√
We
use
this
fact
to
prove
that
for
real
numbers
𝑎,
𝑏,
𝑐
≥
0,
𝑎
+
𝑏
+
𝑐 ≤
√
√
−𝑥−3/2
′′
3(𝑎 + 𝑏 + 𝑐). Consider the function 𝑓 (𝑥) = 𝑥. Then 𝑓 (𝑥) =
< 0 for
4
all positive 𝑥. Thus, we can apply our reversed form of Jensen’s Inequality, giving us
(𝑐)
that 𝑓 (𝑎)+𝑓 (𝑏)+𝑓
≤ 𝑓 ( 𝑎+𝑏+𝑐
). Multiplying both sides by 3 yields the desired inequality.
3
3
3. I NVARIANTS : W EN H AN
Sample 2 from the Ravi Vakil worksheet on invariants and monovariants.
Problem 1. Given n red points and n blue points in the plane, show that we can
draw n non-intersecting line segments, each having one red endpoint and one blue
endpoint.
Proof. So the goal here is to find a quantity that is monotone. Let’s consider an
arbitrary pairing of the blue points with the red points. We consider this operation: If
any two such segments cross, their endpoints form a quadrilateral with vertices in the
order R1;B2;B1;R2. Replace the two R–B diagonals with the two R–B sides. Call
the intersection point 𝑂, then because of the triangular inequality, ∣𝑅1𝐵2∣ < ∣𝑅1𝑂∣ +
∣𝐵2𝑂∣, and ∣𝑅2𝐵1∣ < ∣𝑅2𝑂∣ + ∣𝐵1𝑂∣. Thus
∣𝑅1𝐵2∣ + ∣𝑅2𝐵1∣ < (∣𝑅1𝑂∣ + ∣𝐵2𝑂∣) + (∣𝑅2𝑂∣ + ∣𝐵1𝑂∣) = ∣𝑅1𝐵1∣ + ∣𝑅2𝐵2∣.
Therefore, our operation decreases the total length of the n segments, which is our
mono- variant. 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.