Each question is worth 10 points, for a maximum of 120 points. But the average (median) score is usually about 1 point, even though it is mostly taken by students who have prepared specifically for the exam and are specialized in math. Also one cannot guess or simply solve for an answer; the test requires writing proofs to justify each answer.

Today’s problem is adapted from the 2004 test, problem A1 (the easiest of the first set of problems).

**Casual statement of problem**

DeAndre has been working hard to improve his free throws. Early in the season his free throw percentage was below 75%, but by the end of the season his free throw percentage was above 75%.

Was there necessarily a point during the season where his free throw percentage exactly equaled 75%?

**More precise statement of problem**

DeAndre keeps track of *S*(*N*), the number of successful free throws in the first *N* attempts of the season. Early in the season, *S*(*N*) was less than 75% of *N*, but by the end of the season, *S*(*N*) was more than 75% of *N*.

Was there necessarily a point during the season where *S*(*N*) was exactly 75% of *N*?

Watch the video for a solution.

**Easy Problem From The Hardest Test**

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**Answer To Easy Problem From The Hardest Test**

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

As a fun fact, DeAndre Jordan has tremendously improved his free throw percentage. In 2015 he averaged about 45%, one of the worst all-time. But in 2018 he averaged about 75%, which is about league average. The lesson is that hard work and practice pay off in sports; and I would say the same for math class!

Anyway, back to the problem. At first it might seem the answer is “no” because percentages can jump. Consider someone that makes 1 of 2, and then makes 2 of 3, so that:

*S*(2)/2 = 1/2 = 50%

*S*(3)/3 = 2/3 = 66.6…%

Notice the free throw percentage was below and then above 60%, but it never was equal to 60%. So you might think you can jump over 75% as well.

But surprisingly you cannot! The answer is you must equal 75% if you were below and then above it. Let’s prove why.

**Proof by contradiction**

Suppose DeAndre is below and then above 75%. This means for some *n* the free throw percentage is below 75% and then on the next shot it is above 75%. This leads to the inequality:

*S*(*n*)/*n* < 75% < *S*(*n* + 1)/(*n* + 1)

Since the percentage increased, the *n* + 1 shot must have been a successful attempt. Hence *S*(*n* + 1) = *S*(*n*) + 1. So we have:

*S*(*n*)/*n* < 75% < (*S*(*n*) + 1)/(*n* + 1)

Now let’s write 75% = 3/4, and then work on each inequality by cross-multiplying:

*S*(*n*)/*n* < 3/4

4*S*(*n*) < 3*n*

3/4 < (*S*(*n*) + 1)/(*n* + 1)

3(*n* + 1) < 4(*S*(*n*) + 1)

3*n* < 4*S*(*n*) + 1

Combining the two results with 3*n* in the middle gives:

4*S*(*n*) < 3*n* < 4*S*(*n*) + 1

But this is impossible because it implies the integer 3*n* is strictly in between the consecutive integers 4*S*(*n*) and 4*S*(*n*) + 1. So we have a contradiction, and our original assumption was false.

DeAndre must have had *S*(*N*) = 75% for some value of *N*.

**Generalizing**

The proof works for any percentage whose fraction is (*k* – 1)/*k*.

If you assume otherwise, then like above, we would have:

*S*(*n*)/*n* < (*k* – 1)/*k* < (*S*(*n*) + 1)/(*n* + 1)

After some algebra we get:

*k**S*(*n*) < (*k* – 1)*n* < *k**S*(*n*) + 1

This is also impossible because the middle term is an integer in between the consecutive integers in the lower and upper bounds.

So the free throw percentage result also holds for 1/2 = 50%, 2/3 = 66.66%, 3/4 = 75%, 4/5 = 80%, …, 9/10 = 90%, etc…

It’s a fun and unexpected result! It’s sort of like a discrete intermediate value theorem for free throw percentages.

**Sources**

Putman problem

http://www.math.utah.edu/~palais/mst/frequency.pdf

http://faculty.sfasu.edu/judsontw/putnam/putnam-week06-solution.pdf

https://sites.math.northwestern.edu/~mlerma/problem_solving/putnam/easy_putnam_problems.pdf

DeAndre Jordan stats

https://www.theringer.com/nba/2018/12/5/18126305/deandre-jordan-free-throws-mavericks

https://www.basketball-reference.com/players/j/jordade01.html

It’s a fun mathematical exercise and is a good way to introduce a combinatorial identity. Watch the video for the calculation:

**What Are The Odds Of A Perfect NCAA Basketball Bracket?**

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**What Are The Odds Of A Perfect NCAA Basketball Bracket?**

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Let’s first count the number of games in the tournmanet. In the first round, there are 32 games (as the 64 teams are paired up). Since half of the teams are eliminated after each game, the number of games halves in each subsequent round. We keep doing this until the final championship game when the tournament ends. So the total number of games is:

32 + 16 + 8 + 4 + 2 + 1 = 63

Now I’ll mention that is the long way to calculate. The quicker way is to think logically. In each game there is exactly one loser. In the tournament, all teams except the champion ultimately lose. Hence 63 teams ultimately lose a game, which means there are 63 games in total.

We can extend this logic generally to derive the combinatorial identity:

2^{n} + 2^{n-1} + … + 1 = 2^{n+1} – 1

Pretty neat there is some advanced mathematics that we can teach using March Madness!

**The odds of a perfect bracket**

Suppose in each game you have a 50% chance, or a 1/2 probability, of predicting correctly. Then for 63 games, we would have a:

(1/2)^63 = 1 in about 9.2 quintillion chance (or to be exact, a 1 in 9,223,372,036,854,775,808 chance)

Most of us could never guess a coin toss 10 times in a row, and this is like guessing a coin toss 63 times in a row. It’s an absurdly low chance to get a perfect bracket by chance, which is part of the appeal of March Madess.

Good luck with your picks! You can get a few tips in the Bloomberg article How to Win Your Office NCAA Pool: Use Game Theory, Pick Duke. It’s a great article and I am quoted with a small tip too.

**References**

DePaul University mathematical professor Jeff Bergen calculates the odds

https://youtu.be/O6Smkv11Mj4

You might remember the terms *rational* and *irrational* numbers from math class. As a refresher, numbers like 3, 0.5, 0.333…., -10, -1/2, or 1/7 are known as rational numbers. All of these numbers can be written in fraction form as *a/b*, where the numbers *a* and *b* are integers. Rational numbers have the property that their decimal representation either terminates (like 2.2 or 1.41), or they will eventually repeat (like 0.3333… = 1/3 or 0.142857… = 1/7).

By contrast, an irrational number is a number where it is impossible to be expressed as a fraction *a/b*, where *a* and *b* are integers. Famous examples of irrational numbers are √2, the constant *e* = 2.71828…., and the constant π = 3.14159…

While it might seem intuitive or obvious that π is an irrational number, I was always curious how you would go about proving π is an irrational number.

Mathematically inclined readers can directly check out several proofs that pi is irrational. There is also a brief one-page proof plainly titled A simple proof that π is irrational from number theorist Ivan Niven, which is what is summarized in this post.

What mathematicians call “simple,” most of us find wildly complicated.

So below I offer a step-by-step guide to explain some of the steps more explicitly. I think this problem should be taught in calculus classrooms, as it utilizes the following mathematical concepts:

–Binomial Theorem

–Derivatives

–Chain Rule

–Integration

–Fundamental Theorem of Calculus

–Limits

–Squeeze Theorem

–Proof by Contradiction

All set mentally? Okay, now let’s get to proving that π is irrational.

Here’s a video with the main points. You may want to watch it and if you’re confused about any steps you can read the derivations in this blog post below.

**A Simple Proof Pi Is Irrational**

Details of the proof below…

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

**Preface: proving √2 is irrational**

Before we get to the matter of proving π is irrational, let us start out with a much, much easier proof. This will be an instructive example of *proof by contradiction*, which is the same method that will be used to show π is irrational.

The number √2 is either rational or it will be irrational. What we will do is the following: we will assume √2 is rational and then demonstrate that leads to a contradiction. Logically that means since √2 is not rational it must be irrational.

So suppose √2 = *a’*/*b’* for some integers *a’* and *b’*. One thing we will do is divide out any common factors between the two numbers. For instance, we know that 4 = 16/4, but we can divide the top and bottom by the common factor of 4 to get 4 = 4/1. Do this for √2 until the two numbers have no common factors (they are relatively prime). Then we can write √2 = *a*/*b* with *a* and *b* integers having no common factors.

We can multiply both sides by *b*, and then square both sides to get that *a*^{2} = 2*b*^{2}. This equation tells us that *a*^{2} is twice some other number, and therefore *a*^{2} must be an even number. In fact, this also means that *a* is an even number, as the square of an odd number is odd and the square of an even number is even.

Since *a* is an even number, we can write *a* as two times some other number *c*, that is, *a* = 2*c*. We will substitute this back into *a*^{2} = 2*b*^{2} to find out that 2*b*^{2} = (2*c*)^{2} = 4*c*^{2}. This simplifies so that *b*^{2} = 2*c*^{2}. Now this equation tells us that *b* squared is twice some other number, and therefore we can conclude that *b* must be an even number.

But this is impossible! We cannot have *a* and *b* both being even numbers as we assumed *a* and *b* have no common factors.

Therefore, we know it’s impossible for √2 to be a rational number and it must be irrational.

**Sketch of proof that π is irrational**

The following proof is actually quite similar, except the steps involved require more complicated math.

There are four major steps in Niven’s proof that π is irrational. The steps are:

1. Assume π is rational, π =

a/bforaandbrelatively prime.2. Define a family of functions

f(x)depending on the constantsaandband an integernto be specified later.3. After much work, prove that integral of

f(x) sin(x)evaluated from 0 to π must be an integer, if π is rational.4. Simultaneously show that integral of

f(x) sin(x)evaluated from 0 to π will be positive but tend to 0 as the value ofngets arbitrarily large. This is the required contradiction: if the integral evaluates to an integer, it cannot also be equal to a value between 0 and 1.5. Conclude π is irrational.

Got all of that? Good. Now on to the details.

**Steps 1 and 2: create a function f(x)**

Assume that π = *a*/*b* for relatively prime integers *a* and *b*. We will create a polynomial function of power *2n* that depends on the constants *a* and *b*.

Define *f(x)*, for a constant integer *n*, as follows

This somewhat weirdly defined function will have some interesting properties that will be useful in this proof.

–First we have *f*(0) = 0

–Then we have *f*(*x*) = *f*(π –* x*), which can be verified

–The *k*^{th} derivative evaluated at 0, written as *f*^{(k)}(0), will always be an integer**

**This requires a little bit of verification. To see this, note that the polynomial *n!f*(*x*) = *x*^{n}(*a* – *bx*)^{n}. By the binomial theorem, the term (*a* – *bx*)^{n} can be expanded into a polynomial with integer coefficients ranging in powers from 0 to *n*. Each of these terms gets multiplied by the term *x*^{n}. Hence, the resulting polynomial *n!f(x)* is the sum of terms ranging in power from *n* to 2*n*. If we take the derivative *k* times and *k* is between 1 and less than *n*, then *f ^{(k)}(x)* has no constant terms so

–By the chain rule, *f'(x)* = –*f'(π – x)*, *f ^{(2)}(x)* =

–Since *f ^{(k)}(0)* is always an integer, and

We now have a bunch of properties of *f(x)*. Let’s see what we can do with this.

**Step 3: integral of f(x) sin(x) evaluated from 0 to π must be an integer**

If you thought *f(x)* was a weird function, the fun is only just beginning! We will need another, even stranger, function to help us. Let’s define *g(x)* as the sum of *f*(*x*) and the alternating sum of its even derivatives, as follows:

There are several properties about *g(x)* that we will find useful.

–The function *g*(*x*) evaluated at 0, *g*(0), will be an integer and so will *g*(π). This follows from the property that the derivatives of *f*(*x*) were integers when evaluated at 0 and π, and *g*(*x*) is a sum of these integers.

–We also have that *g*(*x*) + *g*^{(2)}(*x*) = *f*(*x*). We can verify this because *g ^{(2)}(x)* = f

–Now we will use a bit of MacGyver like instincts to come up with…another weird but useful function! Consider the function *g'(x)* sin *x* — *g(x)* cos *x*. What is its derivative? We can use the product rule (*a previous version I wrote “chain rule”) to figure it out.

Where the last step used the fact *g*^{(2)}*(x)* + *g(x)* = *f(x)*.

What’s the big deal of this? Since the derivative of *g'(x)* sin *x* — *g(x)* cos *x* results in *f(x)* sin*(x)*, that means the anti-derivative of *f(x)* sin*(x)* will be equal to *g'(x)* sin *x* — *g(x)* cos *x*.

–We can now evaluate the integral of *f(x)* sin*(x)* using the Fundamental Theorem of Calculus. We have

We already established that *g*(π) and *g*(0) are integers. Therefore, we have shown that if π is rational, the integral of *f(x)* sin*(x)* evaluated from 0 to π will be an integer.

**Step 4: show that the integral of f(x) sin(x) cannot always be an integer for all values of n**

Consider a value of *x* that is between 0 and π. What can we say about the value of *f(x)* sin(*x*)? We know it will always be positive between 0 and π, because sin(*x*) is positive in this range, and the polynomial (*a* – *bx*) is positive in this range (recall π = *a*/*b* so then *a* – *bx* > *a* – *b*π = 0). The last step implies *f(x)* is positive in this range of *x*, and its product with sin(*x*) will also be positive in this range of *x*.

We now focus on the upper bound. One thing is that sin(*x*) is always between 0 and 1 when *x* is between 0 and π. Therefore, we can conclude *f(x)* sin(*x*) < *f(x)*.

Now we go back to the definition of *f(x)*, which is *f(x)* = *x*^{n}(*a* – *bx*)^{n}/*n!*. Notice that (*a* – *bx*) < *a* when *x* is between 0 and π. So we have (*a*–*bx*)^{n} must be less than (*a*)^{n}. Similarly, the term *x*^{n} will be between 0 and *π*^{n} when *x* is between 0 and π.

Putting this all together, we have that

If we integrate each of these functions, we can create a bound for the integral of *f(x)* sin*(x)* evaluated from 0 to π. We have:

We have so far not specified any value of *n*. What happens as *n* goes to infinity? We now employ the squeeze theorem to wedge the integral between 0 and the upper bound.

As *n* tends to infinity, the function 0 will remain 0, and the middle integral will go to some value–bounded by the fraction on the right hand side.

What happens to the upper bound? It turns out the fraction on the right will tend towards 0. One way to see that is π*e*^{aπ} has a Taylor series with terms of the form π^{n+1}*a*^{n}/*n!*. Since the series converges to π*e*^{aπ}, the limit of its terms must tend to 0. Hence we must have π^{n+1}*a*^{n}/*n!* tends to 0 as *n* tends to infinity. In other words, for some large value of *n*, the upper bound will be less than 1.

Since the integral is positive, but can be made arbitrarily small, the integral at some point will be a value between 0 and 1.

**Step 5: conclude π is irrational**

If π is rational, then step 4 says the integral can be between 0 and 1, but step 3 says the integral must always be an integer. This is a contradiction! There is no integer between 0 and 1, so we must have made a mistake in our original premise that π was rational.

Hence we can conclude π is irrational!

It took a little bit of work, but I think the steps are instructive and a beautiful demonstration of several principles of calculus.

Congratulations! You now have learned the proof of one of the most famous facts in mathematics.

**References**

Irrationality of pi at MathsChallenge.net

Niven, Ivan (1947), “A simple proof that pi is irrational” (PDF), Bulletin of the American Mathematical Society, 53 (6), p. 509, doi:10.1090/s0002-9904-1947-08821-2

http://www.ams.org/journals/bull/1947-53-06/S0002-9904-1947-08821-2/S0002-9904-1947-08821-2.pdf

(-1)^{π} = ?

It’s a neat little calculation that I wanted to share for March 14 = 3/14 = Pi Day. Watch the video to see the answer:

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**Answer To what is -1 to the power of pi equal to?**

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

We start with the formula:

*e*^{iθ} = cos θ + *i* sin θ

Substituting θ = π gives the famous formula:

*e*^{iπ} = cos π + *i* sin π = -1

Now we raise both sides to the power of π to get:

(-1)^{π}

= (*e*^{iπ})^{π}

= *e*^{iπ2}

Now we let θ = π^{2} to get:

*e*^{iπ2}

= cos π^{2} + *i* sin π^{2}

≈ -0.903 – (0.430)*i*

Hmm, that’s quite a *complex* answer for a seemingly simple problem. Happy Pi Day!

√(PI) + E = √(PIE)

Each letter is a different digit from 0 to 9, but P cannot be 0. Note PI is a two digit number and PIE is a three digit number.

No calculators or computer code allowed. Can you figure it out? Watch the video for a solution.

**Pi Day Puzzle: Solve For Each Letter**

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**Answer To Pi Day Puzzle – Solve For Each Letter**

There are many ways to solve this kind of problem. I will share how I solved it.

First, we want PI to be a perfect square, so we have the possibilities:

16

25

36

49

64

81

Now we need PIE to be a perfect square, so we consider which 3 digit numbers have the first two digits from the above list. There are three possibilities:

169

256

361

So now we check these three cases, and only the first possibility works. So we have:

P = 1, I = 6, E = 9

which works out to

√16 + 9 = 13 = √169

**Source**

Math Misery? by Manan Shah:

http://mathmisery.com/wp/2017/03/14/cryptarithmetic-puzzle-62/