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#### Leonard Blackburn

I'm a community college math professor. I teach algebra, geometry, trigonometry, precalculus, statistics, calculus, linear algebra, discrete math, and differential equations. I graduated from Knox College with a B.A. in mathematics and The University of Minnesota with an M.S. in mathematics. I enjoy mathematical logic, set theory, math contests, teaching math, disc golf, running, reading, chess, and spending time with my wife and daughter.

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## 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):

$xy = 75$

$y = \frac{75}{x}$

$P = 3x + y = 3x + \frac{75}{x} \geq 2 \sqrt{225} = 30$

$3x = \frac{75}{x}$

$x^2 = 75/3 = 25$

$x = 5, y = 15, P = 30.$

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 $a_1, a_2, \ldots, a_n$ are nonnegative real numbers.  Their arithmetic mean is

$\frac{a_1 + a_2 + \cdots + a_n}{n}$

and their geometric mean is

$\sqrt[n]{a_1 a_2 \cdots a_n}.$

The AM-GM inequality states that

$\sqrt[n]{a_1 a_2 \cdots a_n} \leq \frac{a_1 + a_2 + \cdots + a_n}{n}$

for all nonnegative real numbers $a_1, a_2, \ldots , a_n$, with equality if and only if $a_1 = a_2 = \cdots = a_n$.  This is easy to prove in the case $n = 2.$  Just note that the following inequalities are equivalent:

$\sqrt{ab} \leq \frac{a+b}{2}$

$2 \sqrt{ab} \leq a + b$

$4ab \leq a^2 + 2ab + b^2$

$0 \leq a^2 - 2ab + b^2$

$0 \leq (a - b)^2$

This last inequality is certainly true with equality if and only if $a = b.$  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 $n = 2, a = 3x$, and $b = 75/x$, which gives

$15 = \sqrt{3x \cdot (75/x)} \leq (3x + 75/x)/2$,

or multiplying by 2:

$30 \leq 3x + 75/x$

with equality if and only if $3x = 75/x$ or $x = 5.$

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

Problem 1

Find the minimum value of $f(x) = x^3 + \frac{16}{x} + \frac{32}{x^2}$ for $x > 0$.

Solution

By the AM-GM inequality, we have

$x^3 + \frac{16}{x} + \frac{32}{x^2} \geq 3 \cdot \sqrt[3]{x^3 \cdot \frac{16}{x} \cdot \frac{32}{x^2}} = 24$

with equality if and only if $x^3 = \frac{16}{x} = \frac{32}{x^2} =8$, which is achieved when $x = 2.$  So the minimum is 24.  That sure beats having to find the critical numbers for that function.

Problem 2

Find the minimum value of $f(x,y) = xy + \frac{8}{x} + \frac{1}{y}$ for $x > 0, y >0.$

Solution

$xy + \frac{8}{x} + \frac{1}{y} \geq 3 \cdot \sqrt[3]{xy \cdot \frac{8}{x} \cdot \frac{1}{y}} = 6$

with equality if and only if $xy = \frac{8}{x} = \frac{1}{y}=2$, which is achieved when $x = 4$ and $y = \frac{1}{2}$.  No partial derivatives required.

Problem 3

Find the maximum value of $f(x,y,z) = 160xyz - 5x^2yz - 8xy^2z-20xyz^2$ for $x > 0, y>0, z>0.$

Solution

First we write

$f(x,y,z) = xyz(160-5x-8y-20z).$

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

$f(x,y,z) = \frac{1}{800}(5x)(8y)(20z)(160-5x-8y-20z),$

and then note that $f(x,y,z)$ is maximized when $(5x)(8y)(20z)(160-5x-8y-20z)$ is.  But by the AM-GM inequality, we have

$(5x)(8y)(20z)(160-5x-8y-20z) \leq \left(\frac{5x+8y+20z+160-5x-8y-20z}{4}\right)^4 = 40^4,$

with equality if and only if $5x = 8y = 20z = 160 - 5x - 8y- 20z = 40,$ which is achieved when $x = 8, y = 5, z = 2.$  So, the maximum value of $f$ is $\frac{1}{800}(40)^4 = 3200$.

## The Fundamental Theorem of Calculus Is Obvious

Calculus students are often told that the equation

$\int_a^b f(x) dx = F(b) - F(a)$   (for $f$ continuous on $[a,b]$)

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 $[a,b]$ and some function $F(x)$ defined on that interval.  Suppose further that you were interested in computing the change of $F(x)$ over $[a,b]$, i.e. $F(b) - F(a)$, but all you know about $F(x)$ is its rate of change with respect to $x$ at each value of $x$ in $[a,b]$.  Let’s call this rate of change function $f(x)$ and suppose it is continuous.

Note that if $f(x) = c$ (constant) on some interval $[p,q]$, then (by definition of rate of change) the change in $F(x)$ on $[p,q]$ is equal to $c(q-p)$.  The problem is that $f(x)$ is in general not constant on $[a,b]$.  So what we will do is partition $[a,b]$ into subintervals $[a=x_0, x_1], [x_1, x_2], [x_2, x_3], \ldots, [x_{n-1}, x_n = b]$ and choose a point $w_k$ in each subinterval $[x_{k-1}, x_k]$.  Then, if the subinterval $[x_{k-1}, x_k]$ is small, we can get a good approximation of the change in $F(x)$ over that subinterval by assuming that $f(x) = f(w_k)$ (constant) over that subinterval.  (Note that the value of $f(x)$ does not change much over a small interval because $f(x)$ is continuous.)  So, the change in $F(x)$ over $[x_{k-1}, x_k]$ is approximately equal to $f(w_k)(x_k - x_{k-1})$, and so the change in $F(x)$ over $[a,b]$ is approximately equal to

$\sum_{k=1}^n f(w_k)(\Delta x_k)$

where $\Delta x_k = x_k - x_{k-1}.$  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:

$\lim_{\| P \| \rightarrow 0} \sum_{k=1}^n f(w_k)(\Delta x_k).$

But this is precisely the definition of $\int_a^b f(x) dx$.

So, by asking the question “What is the change of $F(x)$ on $[a,b]$ given its continuous rate of change $f(x)$?” it seems as though the definite integral $\int_a^b f(x) dx$ is designed to compute $F(b) - F(a)$ 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

$(x+a)(x+b)(x+c)(x+d) = x^4 + rx^3 + sx^2 + tx + u.$

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 $2^4 = 16$ 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

$r = a + b + c + d.$

Likewise, we get

$s = ab+ac+ad+bc+bd+cd,$

$t = abc + abd + acd + bcd,$ and

$u = abcd.$

Note that the zeros of the polynomial in question are $-a, -b, -c,$ and $-d.$  So, we have

$r=(-1)$(sum of the zeros)

$s =$ sum of all subproducts of zeros taken 2 at a time

$t =(-1)$(sum of all subproducts of zeros taken 3 at a time)

$u =$ product of zeros

We can easily generalize this.  Consider the polynomial

$a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_1 x + a_0.$

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

the sum of the zeros $= -\frac{a_{n-1}}{a_n}$

the sum of the subproducts of the zeros taken two at a time $= \frac{a_{n-2}}{a_n}$

the sum of the subproducts of the zeros taken three at a time $= -\frac{a_{n-3}}{a_n}$

$\vdots$

the product of the zeros $= (-1)^n \frac{a_0}{a_n}$

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

$x^3 - 5x^2 + 3x + 11.$

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

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ and $a^2 + b^2 + c^2.$

(You should have gotten $-3/11$ and 19, respectively.)

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

The product of two of the four roots of $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ is $-32.$  Determine k.

Solution

Call the roots a, b, c, and d with $ab = -32.$  Then, since $abcd = -1984,$ we have $cd = 62.$  We have to find

$k = ab + ac + ad + bc + bd + cd = -32 + ac + ad + bc + bd + 62 = 30 + (a+b)(c+d).$

Note that

$-200 = abc + abd + acd + bcd = 62(a+b) - 32(c+d)$, and

$18 = (a+b) + (c+d).$

Let $x = a+b$ and $y = c+d.$  Then solving $62x - 32y = -200$ and $x + y = 18$ gives $x = 4$ and $y = 14.$  So, $k = 30 + (4)(14) = 86,$ 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

$x^4 + 513x^2 + 7x + 18$

or

$3x^8 + 31x^7 + 314x^6 + 3141x^5 + 31415x^4 + 314159x^3 + 3141592x^2 + 31415926x + 314159265$

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 be positive integers.  Prove that for every $\epsilon > 0$, there are positive integers m and n such that $\epsilon < \lvert h \sqrt{m} - k \sqrt{n} \rvert < 2 \epsilon$.

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 $\epsilon < h \sqrt{m} - k \sqrt{n} < 2 \epsilon.$

Also, the expression $h \sqrt{m} - k \sqrt{n}$ 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 $m = k^2a$ and $n = h^2 b.$  Then,

$h \sqrt{m} - k \sqrt{n} = hk ( \sqrt{a} - \sqrt{b} )$.

So, letting $p = \frac{\epsilon}{hk}$, it suffices to show that there are positive integers a and b such that $p < \sqrt{a} - \sqrt{b} < 2p.$

There, that’s a simpler statement, and it seems reasonable enough.  In fact, I would conjecture that the set $\{ \sqrt{a} - \sqrt{b} \mid a, b$ 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 $\sqrt{b}$ to arrive at

$p + \sqrt{b} < \sqrt{a} < 2p + \sqrt{b}.$

Now, let’s square to arrive at

$p^2 + 2p\sqrt{b} + b < a < 4p^2 + 4p \sqrt{b} + b.$

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

$4p^2 + 4p\sqrt{b} + b - \left( p^2 + 2p\sqrt{b} + b \right) = 3p^2 + 2p\sqrt{b}.$

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

$3p^2 + 2p \sqrt{b} > 1$

or

$\sqrt{b} > \frac{1 - 3p^2}{2p}.$

So, if $1 - 3p^2 \leq 0$ then we can choose $b = 1$.  Otherwise, we can choose

$b > \left( \frac{1 - 3p^2}{2p} \right)^2.$

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

1. 7 is the only dimension other than 3 in which a vector cross product can be defined.
2. 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.
3. When rolling two six-sided dice, 7 is the most frequently observed total.
4. 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 $2^{2^n} + 1.$  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 $360^{\circ}.$

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 $360^{\circ}.$

So, for a regular heptagon, each exterior angle measures $360/7 \approx 51.43^{\circ}$ and each interior angle measures $180 - 360/7 \approx 128.57^{\circ}.$

How many diagonals does an n-gon have?  Let’s count.  From each of the n vertices, draw a diagonal to each of the $n - 3$ non-adjacent vertices.  In this manner we will have drawn $n(n-3)$ diagonals, but we will have drawn each one twice (once from each of its endpoints).  So, there are

$\frac{n(n-3)}{2}$

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 $y = x(2-x)$ and $y = 0$.  I expected my students to use the method of cylindrical shells to arrive at

$\int_0^2 2\pi x \cdot x(2-x) dx.$

But several students erroneously wrote down

$\int_0^2 2\pi x(2-x) dx$

instead. However, I was surprised to find out that both integrals evaluated to $4/3$.  (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 $2\pi$‘s and interpreted the equality

$\int_0^2 x(2-x) dx = \int_0^2 x^2 (2-x) dx$

as a statement about areas.  Here are the graphs of $y = x(2-x)$ and $y = x^2(2-x)$:

I conjectured that if the graph of $y = f(x)$ is symmetric with respect to the line $x=1$, then the integrals would be equal.  Before I attempted a proof, I tried another example.  Note that $y = -1 + \cos (4x - 4)$ is symmetric with respect to $x = 1$ and you can check that

$\int_0^2 (-1 + \cos (4x - 4)) dx$  and  $\int_0^2 x(-1 + \cos (4x - 4)) dx$

are both equal to $-2 + 0.5 \sin 4$.  The graphs of the integrands are below.

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

Theorem

If $f(1-x) = f(1+x)$ for all $x \in [0,1]$ then $\int_0^2 f(x) dx = \int_0^2 x f(x) dx.$

Proof

Note that it suffices to prove that

$\int_0^2 (x-1) f(x) dx = 0.$

Writing

$\int_0^2 (x-1) f(x) dx = \int_0^1 (x-1) f(x) dx + \int_1^2 (x-1) f(x) dx$,

we see that it suffices to prove that

$\int_1^2 (x-1) f(x) dx = - \int_0^1 (x-1) f(x) dx$.

Letting $u = x - 1$ gives

$\int_1^2 (x-1) f(x) dx = \int_0^1 u f(u+1) du.$

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

$\int_0^1 u f(1-u) du.$

Now, letting $v = 1 - u$, this last integral equals

$\int_1^0 (1-v) f(v) (-dv) = \int_0^1 (1-v) f(v) dv = - \int_0^1 (v-1) f(v) dv$

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 $f(1) \neq f(3).$

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

$f'(x) = k(x-1)(x-2)(x-3) = k(x^3 - 6x^2 + 11x - 6)$

for some constant k. So, antidifferentiation gives:

$f(x) = k \left( \frac{1}{4}x^4 - 2x^3 + \frac{11}{2}x^2 - 6x \right) + C$

for some constant C.  Then I computed $f(1)$ and $f(3)$ 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 $2n + 2.$ Suppose that f  has $2n + 1$ local extrema at $x = a_1, a_2, ..., a_{2n+1}$ where $a_1 < a_2 < ... < a_{2n+1}$ is an increasing arithmetic progression. Then $f(a_j) = f(a_{2n+2-j})$ for $j=1,2,...,n.$

I decided to prove the stronger statement that $x = a_{n+1}$ is a vertical line of symmetry for the function f.  To do this, let’s define $g(x)=f(x+a_{n+1})$ and prove that $g(x)$ is an even function.  Let $d = a_2 - a_1.$ Then g has local extrema at $x = -nd, -(n-1)d, ..., -d, 0, d, ..., (n-1)d, nd.$ So,

$g'(x) = kx(x^2 - d^2)(x^2 - 4d^2) \cdots (x^2 - n^2d^2),$

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

$2\alpha, 2\beta, 2\gamma,$

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

Then,

$\gamma = \frac{\pi}{2} - (\alpha + \beta).$

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

$\frac{r}{c} = \tan(\frac{\pi}{2} - (\alpha + \beta)) = \cot(\alpha + \beta) = 1 / \tan(\alpha + \beta)$

and this equals

$\frac{1 - \tan(\alpha) \tan(\beta)}{\tan(\alpha) + \tan(\beta)} = \frac{1 - \frac{r^2}{ab}}{\frac{r}{a} + \frac{r}{b}} = \frac{ab - r^2}{r(a + b)}.$

Solving for c gives:

$c = \frac{r^2(a+b)}{ab - r^2}.$

Rearranging gives:

$r(a+b+c) = \frac{abc}{r}.$

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:

$A = \frac{abc}{r}.$