Mathematics Extension 2 Challenge Marathon

[RuiAce]

We prove this, by demonstrating that in every set of \(n\) objects, all \(n\) of them have the same colour. Note that this is an equivalent statement, and thus is also valid to prove.

When \(n=1\) this is trivially true, because any set that has only has one object will only have one colour, so all objects must have the same colour.

Assume that every set with \(k\) objects have the same colour. Consider what happens when \(n=k+1\).

For any arbitrary set with \(k+1\) elements, we may label the elements in it, so that the set takes the form \( \{1,2,3,\dots,k,k+1\} \). A subset of this set would just be the first \(k\) elements, i.e. the set \( \{1,2,3,\dots,k\} \). Since this is one of the many sets that have \(k\) elements, from our inductive assumption all elements in this set must have the same colour.

Now, another subset of this set would be the last \(k\) elements, i.e. the set \( \{ 2,3,\dots,k,k+1\} \). Again, this is one of many sets that have \(k\) elements, so all the elements in this set must also have the same colour.

But these two sets have overlaps, so by consequence all elements in \( \{1,2,3,\dots,k,k+1\} \) must have the same colour. Hence, all elements in an arbitrary set with \(k+1\) elements all have the same colour.

So it follows by induction that in every set of \(n\) objects, all colours of the objects in said sets are the same, and hence every colour must be the same.

This is obviously wrong. But why?

[RuiAce]

This may be hard to believe, but I promise this question is doable via only 4U methods.

(Won't provide hints unless someone actually attempts it.)

[Opengangs]

About the Putnam Competition

The William Lowell Putnam Mathematical Competition (often just abbreviated to Putnam) is an annual Mathematics competition geared towards undergraduates, who all compete for prize money and scholarships to some of the most prestigious and highly regarded universities in the world. Examples include: Harvard University, MIT, Princeton, Carnegie Mellon, etc.

The test is divided up into halves (A and B). It's a typical 6 hour long paper (3 each), with only 12 questions. The maximum amount of marks awarded for each question is 10 marks. Even though there are 120 points to be awarded, the median is only around 1-5 points. The top ever score was 63/120 by students at Harvard University.

The very last question was a Question 6 from a previous Putnam paper, and the goal here is to walk you through the process by first depicting an easier way in approaching mathematical thinking.

[TheAspiringDoc]

This may be hard to believe, but I promise this question is doable via only 4U methods.

(Won't provide hints unless someone actually attempts it.)

I’m gonna guess the integrand is an odd function and therefore the definite integral is 0?

[RuiAce]

I’m gonna guess the integrated is an odd function and therefore the definite integral is 0?

You'll find that \(f(-x) \neq -f(x)\).

Edit: I had a go at this question again. I will post the answer, but not any solution yet.

\[ \frac\pi{8072} \]

________________

Seeing as though this is being used as a challenge, I'll just explain a bit about what's going on with the notation.

Not sure if this is a challenge question or not (depends on whether there's a simpler solution than mine which there probably is), but I'll post it here anyways.



I'm only like 85-90% sure my proof of part (i) is correct as it's quite lengthy and slightly complicated but I did put a lot of thought into it so hopefully it's fine.

\( \mathbb{N} \backslash \{ 2\} \) is just asking for all natural numbers, i.e. 0, 1, 2, 3, 4, 5, 6, ..., but except 2. So if you read on, we're really just assuming \(n\) is a prime number not equal to 2.

Part i is easy enough to visualise. But be careful how you argue it........that might be harder than actually doing parts ii and iii. In part ii, basically you know that \( \sum \) means to add all the terms. \( \prod\) means to multiply them instead.

[stolenclay]

This may be hard to believe, but I promise this question is doable via only 4U methods.

(Won't provide hints unless someone actually attempts it.)

It's been a while since I did maths on AN, but I spent too long on this to not cash in Hopefully there aren't any mistakes!

Let \(k = 1009\). Then the integral becomes
\[
\int_{-\pi/(4k)}^{\pi/(4k)} \frac{1}{(2k)^{2kx}+1} \frac{\cos^{2k}(2kx)}{\sin^{2k}(2kx) + \cos^{2k}(2kx)} \, dx.
\]
Make the change of variables \(x = r/(2k)\). Then \(dx/dr = 1/(2k)\) and the integral becomes
\[
\int_{-\pi/2}^{\pi/2} \frac{1}{(2k)^{r}+1} \frac{\cos^{2k}r}{\sin^{2k}r + \cos^{2k}r} \frac{1}{2k} \, dr.
\]
Define the functions \(f\colon \mathbb{R} \to \mathbb{R}\) and \(g\colon \mathbb{R} \to \mathbb{R}\) by
\[
f(r) = \frac{1}{(2k)^r + 1}
\quad \text{and} \quad
g(r) = \frac{\cos^{2k}r}{\sin^{2k}r + \cos^{2k}r}
\quad \forall r \in \mathbb{R}.
\]
Then the integral we wish to compute is
\[
\frac{1}{2k} \int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx.
\]

Hint 1 (for my particular solution)

Try to use \(f(x) g(x) = [1 - f(x)] [1 - g(x)] + f(x) + g(x) - 1\).

Hint 2 (for my particular solution)

As is usual with these types of questions, we wish to make use of identities/symmetries satisfied by \(f\) and \(g\). For this particular problem, we will use the result that for every \(x \in \mathbb{R}\) we have

I leave it to the reader to verify that these hold. It is perhaps a bit arbitrary to begin with these, and in fact these are not identities that come out of thin air; rather they are motivated by the observation in Hint 1.

Solution

We integrate the equation in Hint 1 in \(x\) over the interval \([-\pi/2, \pi/2]\) to obtain
\[
\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \underbrace{\int_{-\pi/2}^{\pi/2} [1 - f(x)] [1 - g(x)] \, dx}_{I_1} + \int_{-\pi/2}^{\pi/2} f(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx - \pi.
\]
Let \(I_1\) be the integral as indicated above. We use a change of variables \(x = -r\). Then \(dx/dr = -1\) and \(I_1\) becomes
\[
I_1
= \int_{-\pi/2}^{\pi/2} [1 - f(x)] [1 - g(x)] \, dx
= \int_{\pi/2}^{-\pi/2} \underbrace{[1 - f(-r)]}_{{} = f(r)} [1 - \underbrace{g(-r)}_{{} = g(r)}] (-1) \, dr
= \int_{-\pi/2}^{\pi/2} f(r) [1 - g(r)] \, dr.
\]
Note that we have used some of the identities in Hint 2. We now have
\[
\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \int_{-\pi/2}^{\pi/2} f(x) [1 - g(x)] \, dx + \int_{-\pi/2}^{\pi/2} f(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx - \pi.
\]
Adding \(\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx\) to both sides, we obtain
\[
2\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= 2\int_{-\pi/2}^{\pi/2} f(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx - \pi
= 2\int_{-\pi/2}^{\pi/2} \underbrace{\bigg[f(x) - \frac{1}{2}\bigg]}_{\text{odd in \(x\)}} \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx
= \int_{-\pi/2}^{\pi/2} g(x) \, dx.
\]
Note that the fact that \(f(x) - 1/2\) is odd in \(x\) follows from an identity in Hint 2. We now wish to compute \(\int_{-\pi/2}^{\pi/2} g(x) \, dx\). By evenness in \(x\) of \(g(x)\), this is the same as \(2\int_{0}^{\pi/2} g(x) \, dx\), which is also the same as \(2\int_{-\pi/2}^{0} g(x) \, dx\). Since \(g(x) = 1 - g(x + \pi/2)\) for all real \(x\), making the substitution \(x = r - \pi/2\), we have
\[
\int_{-\pi/2}^{\pi/2} g(x) \, dx
= 2\int_{-\pi/2}^{0} g(x) \, dx
= 2\int_{-\pi/2}^{0} [1 - g(x + \pi/2)] \, dx
= 2\int_{0}^{\pi/2} [1 - g(r)] \, dr.
\]
We now write
\[
\int_{-\pi/2}^{\pi/2} g(x) \, dx
= \frac{1}{2} \Bigg[\int_{-\pi/2}^{\pi/2} g(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx\Bigg]
= \frac{1}{2} \Bigg[2\int_{0}^{\pi/2} g(x) \, dx + 2\int_{0}^{\pi/2} [1 - g(x)] \, dx\Bigg]
= \frac{\pi}{2}
\quad \implies \quad
2\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \frac{\pi}{2}.
\]
Dividing both sides by \(4k\), we obtain
\[
\frac{1}{2k} \int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \frac{\pi}{8k}.
\]
Recalling that \(k = 1009\), we have
\[
\int_{-\pi/4036}^{\pi/4036} \frac{1}{2018^{2018x}+1} \frac{\cos^{2018}(2018x)}{\sin^{2018}(2018x) + \cos^{2018}(2018x)} \, dx
= \frac{\pi}{8072}.
\]

Comments

This holds for all positive integers \(k\). Integrality of \(k\) is used when the identities regarding the function \(g\) in Hint 2 are derived.

I changed the integration variable from \(r\) to \(x\) multiple times without comment. I'm not sure if it's possible for marks to be lost for this under a HSC marking scheme, so be careful.

[Ali_Abbas]

This may be hard to believe, but I promise this question is doable via only 4U methods.

(Won't provide hints unless someone actually attempts it.)

Alright well I have an alternative solution to that of stolenclay's and is as follows:

[RuiAce]

It is true to say that the 2018 could've been replaced with a lot of different numbers (in fact, as alluded to, the even numbers), and the integral ultimately boils down to the famous \( \int^{\pi/2}_{0} \frac{1}{1+\tan^\alpha x} dx\), where \(\alpha\) can be any real number. Many of these ideas were successfully deduced in these solutions.
The intended method basically involved three substitutions, the first one being the obvious \( u = 2018x\)

[Opengangs]

\[ \text{Let } I = \int_0^{\frac{\pi}{2}} \ln (\sin(x))\,dx.\]
\[ \text{(i) Show that:} \int_0^{\frac{\pi}{2}}\ln(\sin(x))\,dx = \int_0^{\frac{\pi}{2}}\ln(\cos(x))\,dx. \]\[ \text{(ii) By considering } 2I = \int_0^{\frac{\pi}{2}} \ln(\sin(x)\cos(x))\,dx, \text{ show that: } 2I = -\frac{\pi}{2}\ln2 + I.\]\[ \text{And hence, evaluate } \int_0^{\frac{\pi}{2}}\ln(\sin(x))\,dx.\]

[Opengangs]

Prove that among any ten points located on a circle with diameter 5, there exist at least two at a distance less than 2 from each other.

[RuiAce]

Prove that among any ten points located on a circle with diameter 5, there exist at least two at a distance less than 2 from each other.

Spoiler

Note that the radius of the circle is \(r = \frac{5}{2}\).

Partition the circle into nine congruent sectors. Each of these sectors will be formed by an angle of \( \frac{2\pi}{9}\) at the circle's centre.

By the pigeonhole principle, at least two such points must lie on one of the nine arcs corresponding to the sectors. Consider any such arc that has two points lying on it.

Form a triangle between these two points on the arc and the centre of the circle. Note that two of the triangles' edges will be radii of the circle, and the third is the distance between the two points.

Let the angle made at the centre of the circle be \(\theta\). On one hand, it now follows that \( \theta \in \left[0, \frac{2\pi}{9}\right]\). On the other hand, from the cosine rule,
\begin{align*}
D &= \sqrt{\left( \frac52 \right)^2 + \left( \frac52\right)^2 - 2 \left( \frac52\right) \left( \frac52\right) \cos \theta}\\
&= \frac{5}{2} \sqrt{2} \sqrt{1-\cos\theta}
\end{align*}
where \(D\) is the distance between the two points.

Since \(\theta\in \left[ 0,\frac{2\pi}{9}\right] \), we can simply plug in the values \(\theta = 0\) and \(\theta = \frac{2\pi}{9}\) to find that the boundary values of \(D\) are \(0\) and \(1.7101\), rounded to 4 decimal places. Note that plugging in was all that was required, because \(f(x) = \cos x\) is monotone on the open interval \( \left( 0, \frac{2\pi}{9}\right) \). (In particular, monotonic decreasing.)

But in any case, we see that the possible values for \(D\) are all strictly less than \(2\). So indeed, these two points of interest are always at a distance of less than 2 from each other.

[Opengangs]

A nice and simple pigeonhole principle (but not really pigeonhole principle) question:
For any positive integer \(a\) and \(b\), if \(ab + 1\) or more objects are placed into \(b\) boxes, then prove that one box will contain more than \(a\) objects.

Hint

Look for a contradiction.