## The Inequality of Arithmetic and Geometric Means

A question on my calculus final exam stated:

Bobby wants to build a subdivided pen for his dog and rabbit. He plans on fencing in a total area of 75 square feet using his house as one side (see the diagram). Find the minimum length of fence needed to make the pen.

While grading the exams yesterday, I noticed that one of my students solved it in an unusual way. He wrote (after labeling the short side *x* and long side *y*):

He wasn’t being extremely clear about his solution, but after a moment’s thought I realized he was using the inequality of arithmetic and geometric means (AM-GM inequality) to arrive at the correct answer. I found this solution quite clever since we had never mentioned the AM-GM inequality in class before and we had always done problems like this with derivatives. (After all, it was a course in calculus.) Before I go any further, let’s state the AM-GM inequality.

Suppose are nonnegative real numbers. Their *arithmetic* *mean* is

and their *geometric mean* is

The AM-GM inequality states that

for all nonnegative real numbers , with equality if and only if . This is easy to prove in the case Just note that the following inequalities are equivalent:

This last inequality is certainly true with equality if and only if The AM-GM inequality can be proven for all values of *n* in a variety of ways, including induction. Proofs are given on the wikipedia page.

So, what my calculus student did was apply the AM-GM inequality to the case , and , which gives

,

or multiplying by 2:

with equality if and only if or

Intrigued by this, I invented some more examples of optimizing with the AM-GM inequality.

**Problem 1**

Find the minimum value of for .

**Solution**

By the AM-GM inequality, we have

with equality if and only if , which is achieved when So the minimum is 24. That sure beats having to find the critical numbers for that function.

**Problem 2**

Find the minimum value of for

**Solution**

with equality if and only if , which is achieved when and . No partial derivatives required.

**Problem 3**

Find the maximum value of for

**Solution**

First we write

In order to apply the AM-GM inequality with , we would like to have 4 factors with a constant sum. So we further write:

and then note that is maximized when is. But by the AM-GM inequality, we have

with equality if and only if which is achieved when So, the maximum value of is .

## The Fundamental Theorem of Calculus Is Obvious

Calculus students are often told that the equation

(for continuous on )

is a surprising and unexpected connection between area and slopes of tangent lines. The textbook I teach from calls it a “remarkable fact.” This theorem (known as the *fundamental theorem of calculus*) was discovered independently by Sir Isaac Newton (1642-1727) in England and by Gottfried Wilhelm Leibniz (1646-1716) in Germany, and this discovery is the primary reason that both men are credited with the invention of calculus. However, I find it odd that this fact needed to be *discovered* at all. The equality is intuitively and immediately obvious. I will give an intuitive explanation (not a proof) of this fundamental theorem below.

Suppose you had an interval and some function defined on that interval. Suppose further that you were interested in computing the change of over , i.e. , but all you know about is its rate of change with respect to at each value of in . Let’s call this rate of change function and suppose it is continuous.

Note that if (constant) on some interval , then (by definition of rate of change) the change in on is equal to . The problem is that is in general not constant on . So what we will do is partition into subintervals and choose a point in each subinterval . Then, if the subinterval is small, we can get a good approximation of the change in over that subinterval by assuming that (constant) over that subinterval. (Note that the value of does not change much over a small interval because is continuous.) So, the change in over is approximately equal to , and so the change in over is approximately equal to

where To make this approximation precise it seems that all we need to do is take a limit as the norm of the partition goes to zero:

But this is precisely the definition of .

So, by asking the question “What is the change of on given its continuous rate of change ?” it seems as though the definite integral is *designed* to compute making the fundamental theorem of calculus intuitively obvious and not at all surprising.

## Vieta’s Formulas

Suppose *a*, *b*, *c*, and *d* are complex numbers and consider the product

Let’s find *r*, *s*, *t*, and *u* in terms of *a*, *b*, *c*, and *d*. If we expand the product on the left, we will get terms. Each of these 16 terms can be created by choosing one of the two terms from each of the four factors and forming the product.

Let’s find *r*. from the four factors, we need to choose three *x*‘s and one constant in all possible ways. This gives

Likewise, we get

and

Note that the zeros of the polynomial in question are and So, we have

(sum of the zeros)

sum of all subproducts of zeros taken 2 at a time

(sum of all subproducts of zeros taken 3 at a time)

product of zeros

We can easily generalize this. Consider the polynomial

This polynomial of degree *n* has *n* zeros (not necessarily distinct). We have

the sum of the zeros

the sum of the subproducts of the zeros taken two at a time

the sum of the subproducts of the zeros taken three at a time

the product of the zeros

These formulas are known as Vieta’s formulas, after François Viète (1540-1603). As an application of these formulas, consider the polynomial

Check that this polynomial has no rational zeros. Thus it is difficult to find the zeros and perform calculations with them. However, call the zeros *a*, *b*, and *c*, and as an exercise, compute

and

(You should have gotten and 19, respectively.)

As a further application, consider the first problem from the 1984 USAMO:

The product of two of the four roots of is Determine *k*.

**Solution**

Call the roots *a*, *b*, *c*, and *d* with Then, since we have We have to find

Note that

, and

Let and Then solving and gives and So, and we’re done.

## Guessing a Polynomial

I’d like to play a guessing game with the readers of my blog. I want you to choose a polynomial with positive integer coefficients and write it down. Your polynomial can have any degree you want. You could choose

or

for example. Then I will try to guess your polynomial by asking you just two questions. Specifically, I will ask you to evaluate your polynomial at a particular value of *x* and tell me the answer. Then I will again ask you to evaluate your polynomial for a particular value of *x* and tell me the answer. Based on your two answers I will guess your polynomial.

If any reader would like to try this, simply write down your polynomial and then type “ready” or some such thing in the comments. Then I will respond in the comments with my first question.

## A Problem Concerning Equal Tangents

We review some basic geometry involving secants and tangents drawn to circles. Then, given two intersecting circles in a plane, we find all points P in that plane such that external tangents drawn from P to the two circles are equal.

## The William Lowell Putnam Mathematics Competition

On the first Saturday in December, the prestigious Putnam mathematics competition is held for undergraduates in North America. It is a notoriously very difficult exam consisting of two 3-hour sessions. In each session, students must work on six challenging problems in the areas of algebra, geometry, number theory, linear algebra, abstract algebra, calculus/analysis, and possibly other topics. Anything accessible to undergraduates is fair game (which may go well beyond standard course work).

Each problem is given an integer score from 0 to 10, so that a perfect score is 120 points (a feat rarely achieved). Solutions must be correct, complete, and well-written, and partial credit is hard to come by.

You can find an archive of old exams and solutions here.

I am eagerly awaiting this years problems. The problems, discussion, and proposed solutions typically start buzzing on the internet Sunday night after the contest. For example, keep your eye on this forum. Or probably soon you can start searching for the problems online and give them a try yourself before you read any discussion or solutions of the problems.

To whet your appetite, I decided to discuss a problem from last year’s exam. Judging by the scores, last year’s exam was particularly difficult. I was only able to solve a few of the problems (and not within the given time limit). Here is one of the easier problems:

**2011 Problem B1**

Let *h* and *k *be positive integers. Prove that for every , there are positive integers *m* and *n* such that .

Let’s talk our way through the solution, rather than presenting a polished proof. First, let’s try to simplify the problem. The absolute value is annoying. We’d rather not consider cases, so let’s try to prove the stronger statement that there are positive integers *m* and *n* such that

Also, the expression is hard to deal with because of the different coefficients in front of the square roots. So, let’s search for *m* and *n* of the form and Then,

.

So, letting , it suffices to show that there are positive integers *a* and *b* such that

There, that’s a simpler statement, and it seems reasonable enough. In fact, I would conjecture that the set are positive integers is dense in the set of real numbers (i.e., between any two real numbers, one can find a member of that set).

Ok, let’s add to arrive at

Now, let’s square to arrive at

Let’s find the length of that interval which must contain *a*:

All we need to do is choose *b* large enough so that the length of this interval is greater than 1 so that it will contain an integer we can choose for *a*. So, we need

or

So, if then we can choose . Otherwise, we can choose

Then there is a positive integer *a* in the interval and we’re done.

Now if you want a polished textbook-style solution, you basically just have to write down our discussion in reverse and hide all traces of your thinking process (smile).

Please feel free to use the comments to discuss this year’s Putnam once the problems are officially posted online.

## A Sea World Heptagon

Each natural number can be classified as interesting or dull (but not both).

**Theorem**

All natural numbers are interesting.

**Proof**

Suppose there are some dull numbers. Then there is a smallest dull number. But being the smallest of all dull numbers is pretty interesting. This is a contradiction. So there aren’t any dull numbers.

7 is a natural number. So, it’s interesting. Here are some of its interesting properties

- 7 is the only dimension other than 3 in which a vector cross product can be defined.
- Lagrange’s four square theorem states that every natural number can be expressed as the sum of the squares of four integers. 7 is the smallest natural number that cannot be expressed as the sum of three squares.
- When rolling two six-sided dice, 7 is the most frequently observed total.
- In 2000, the Clay Mathematics Institute announced 7 Millennium Prize Problems. There is a cash prize of $1,000,000 for a correct solution to any of the problems. (There are six left unsolved; the mathematician that solved one of the problems declined the cash prize.)

I thought of the number 7 recently while looking through vacation pictures. At Sea World in San Antonio, I took this picture of the ceiling in one of the stadiums (yellow highlight added):

When I looked up, I was surprised to see a heptagon prominently displayed in the center of an unusual tiling of the ceiling.

A heptagon is a polygon with 7 sides. Sometimes it’s called a septagon (which is funny because of the Latin prefix but Greek suffix.) One doesn’t get to see heptagons very frequently–unless you live in a place like Great Britain that has heptagonal coins.

The heptagon is a very interesting polygon. For example, did you know that regular (i.e. having equal sides and equal angles) heptagons can tile the hyperbolic plane? There’s a picture of such a tiling in the standard Poincare model here:

http://en.wikipedia.org/wiki/File:Uniform_tiling_73-t0.png

Also, the heptagon is the polygon with the fewest number of sides that is not constructible with the classical straightedge and compass instruments. Gauss and Wantzel proved that a regular polygon with *n* sides is constructible if and only if *n* is a product of a power of 2 and any number of distinct Fermat primes. Fermat primes are primes of the form The number 7 is not a Fermat prime, but 17 is (using *n* = 2), and Gauss once constructed a regular heptadecagon (17-sided polygon) at the age of 19 (or rather, he probably merely proved that one *could* be constructed). Gauss once asked that a regular heptadecagon be inscribed on his tombstone, but his wish was never carried out (probably because it was too difficult to inscribe).

Anyway, back to the heptagon. Let’s take this opportunity to review some basic geometric facts concerning polygons. The sum of the exterior angles (one at each vertex) of any convex polygon is

For an informal proof of this, start in the middle of one of the edges and walk once around the perimeter of the polygon, always facing the direction of motion. By the time you get back to your starting point, your head will have rotated a total of

So, for a regular heptagon, each exterior angle measures and each interior angle measures

How many diagonals does an *n*-gon have? Let’s count. From each of the *n* vertices, draw a diagonal to each of the non-adjacent vertices. In this manner we will have drawn diagonals, but we will have drawn each one twice (once from each of its endpoints). So, there are

diagonals. Thus, the heptagon has 14 diagonals.

Would anyone care to take a picture of a heptagon and link to it in the comments?

## A Curious Case of Equal Integrals

On a recent calculus test, I asked my students to find the volume of the solid obtained by revolving about the *y*-axis the region bounded by and . I expected my students to use the method of cylindrical shells to arrive at

But several students erroneously wrote down

instead. However, I was surprised to find out that both integrals evaluated to . (Check for yourself.) Sometimes one can chalk things like this up to coincidence, but I wondered if there was a deeper principle at work. So, I started to explore. First, I dropped the ‘s and interpreted the equality

as a statement about areas. Here are the graphs of and :

I conjectured that if the graph of is symmetric with respect to the line , then the integrals would be equal. Before I attempted a proof, I tried another example. Note that is symmetric with respect to and you can check that

and

are both equal to . The graphs of the integrands are below.

So, let’s attempt a proof of this phenomenon.

**Theorem**

If for all then

**Proof**

Note that it suffices to prove that

Writing

,

we see that it suffices to prove that

.

Letting gives

Then, by the symmetry hypothesis on , this last integral equals

Now, letting , this last integral equals

as desired. So, the proof is complete.

Can anyone find any generalizations?

## An Arithmetic Progression of Local Extrema

I was just trying to design a graph for an upcoming calculus exam. I wanted it to be a polynomial function *f* with local maxima at *x* = 1, 3 and a local minimum at *x* = 2. For variety I wanted

Since there are three local extrema, I knew I had to make the polynomial have degree at least 4. But after trying for a while, I couldn’t find a 4th degree polynomial to do the job. (I knew I was demanding too many conditions of a 4th degree polynomial.) Here’s how I proved it to myself:

For the sake of contradiction, suppose *f* is a 4th degree polynomial function with the given properties. Then

for some constant *k*. So, antidifferentiation gives:

for some constant *C*. Then I computed and and was slightly surprised to find that they were equal.

Intrigued by this, I sought a simple non-computational explanation. I decided to generalize, and conjectured this:

Suppose *n* is a positive integer and *f *is a polynomial function of degree Suppose that *f* has local extrema at where is an increasing arithmetic progression. Then for

I decided to prove the stronger statement that is a vertical line of symmetry for the function *f*. To do this, let’s define and prove that is an even function. Let Then *g* has local extrema at So,

which is an odd function. But an antiderivative of an odd polynomial is clearly an even polynomial. So, is even.

Satisfied, I can go back to work.

## A Thanksgiving Incircle

This past Thanksgiving, I was doodling on a napkin and drew this image (a little more crudely):

It’s a fun hobby to start with a geometric diagram and just start exploring. In this picture, we have a circle inscribed in triangle *ABC*. Radii are drawn to the points of tangency (and these radii must be perpendicular to the sides of the triangle). Lines are drawn from the center of the circle to the vertices of the triangle. These lines are the three angle bisectors of the triangle (easy exercise). Follow along on a napkin of you own:

Let’s call the angles at *A*, *B*, and *C*,

and let’s call the radius of the circle *r*.

Then,

Let’s take the tangent of both sides of this last equation to get:

and this equals

Solving for *c* gives:

Rearranging gives:

But, it is well-known (and easy to prove) that the area of a triangle equals the product of its semiperimeter and inradius. So, we’ve discovered an interesting formula for the area of a triangle, *A*: