Here, I will occasionally post challenge questions for the true maths brains to attempt.
Questions are not intended to reflect the scope of the difficulty in actual HSC questions, however may be completed using only knowledge taught in the course. Spoilers are intended to reveal what topics to draw knowledge from when a genuine, unaided attempt has been unsuccessful.
I invite everyone to also post their own questions at their own discretion, and for anyone who has completed 4U maths or equivalent to also answer. I invite collaboration, as from time to time, some questions may be, one will argue, ridiculous.
Spoiler
Required knowledge:
a) HSC 4U Complex Numbers
b) Preliminary 2U Tangent to a Curve and Derivative of a Function, HSC 4U Complex Numbers
c) Preliminary 2U Basic Arithmetic and Algebra, HSC 4U Complex Numbers, HSC 2U Trigonometric Functions
Edit: 3 years later I found a typo with part a). It has been fixed.
Here is one extremely brutal way of doing the above question. It overuses partial fractions and throws complex numbers into an integral - something not needed in the HSC, just like my very first question
I considered the hint, but it took me way too long to figure out what happens after the hint is applied, hence all of this.
I won't put up the solution to the 3x quicker method just yet. Unnecessarily over complicated non-HSC solution
(http://uploads.tapatalk-cdn.com/20161006/1273c354f7b7aa54fa08ea9caf390900.jpg)
Consider an alphabet with n different available letters.
Let P(k) be the number of ways you can use exactly k different letters in an n letter word.
i) Explain why
ii) Show that
Let the Score of a word, X, be defined as 1/(1+ρ(X)), where ρ(X) is the number of letters that were not used by the word X.
iii) Show that the sum of all the Scores, S, over all possible n letter words, is given by:
iv) Hence, evaluate S in closed form.
To help explain part (i) (because I took a while understanding what was going on)
Explanation
Consider a three letter alphabet: {a, b, c}
Then, P(1) is the number of ways we can make three letter words, out of just one letter of the alphabet. If that letter was a, then aaa is the only word.
P(2) is the number of ways we can make three letter words, out of any two letters of the alphabet. If the letters are a and b, then the words are:
aab, aba, baa, bba, bab, abb
P(3) is the number of ways we can make three letter words using all the letters. So that's the easy 3!
The point of introducing the nCk is to quantify the fact we could've chosen any k of the 3 letters. In P(1), we could've chosen a, b or c to be our letter. (And indeed, 3Ck=3 possible letters.)
And now for the question
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If a 4U student has not seen that Greek letter before, that is rho.
Explanation
Going back to the case n=3
If the words satisfied k=1 (e.g. aaa), then ρ(X) = 2 (Note: 1+2=3)
If the words satisfied k=2 (e.g. aab), then ρ(X) = 1 (Note: 2+1=3)
If the words satisfied k=3 (e.g. abc), then ρ(X) = 0 (Note: 3+0=3)
This can be seen by just realising how man letters we did not use, in each case.
The important thing to realise is that for each value of k, ρ(X) differs. As a matter of fact, ρ(X) always goes down by 1.
In fact, ρ(X) = 3 - k
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I don't fully trust what I say from here due to how P(k) has been defined.
Let be a root of the equation , where
The root locus of a complex quadratic is the set of all points on the Argand Diagram that could be a root of the quadratic.
Sketch the root locus of
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Note that through these computations, the points (1,0) and (-1,0) are technically excluded. Coincidentally, the other case(s) brings it back.
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Note that the above line was an abuse of notation. Infinity is not a number.
Proof of the limit used on g(p)
(Brief) Explanation as to why the functions decrease/increase from -1 respectively
In a similar way, the other case of \( p\le -2\) shows that \( y=0\) is a part of the locus for all \( x > 0\).
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A classic complex numbers question I just did a few minutes ago.
Hint
You only have to consider when \(|\alpha|=1\). After you prove it holds for \( |\alpha|=1\), the case \( |\alpha| < 1 \) falls out pretty easily from it
Spoiler
You need to know circle geometry as well as have a substantial understanding of complex number tricks.
Hint
Can be made much easier using one of the tricks used in the previous problem
Spoiler
Find a contradiction!
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.
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.
\[ \frac\pi{8072} \]
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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.
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.
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.
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.