HSC 2016 MX2 Marathon ADVANCED (archive) (3 Viewers)

KingOfActing · Mar 11, 2016

Re: HSC 2016 4U Marathon - Advanced Level

agha said:
$\\ $Is$ \ 2^n + 3^n $, for positive integers$ \ n $, ever a square number?$$

$2^n + 3^n \equiv (-1)^n \mod{3}\\\text{But, note that } x^2 \equiv 0,1 \mod{3}\\\therefore n = 2m\\2^{2m} + 3^{2m} \equiv 4^m +9^m \equiv 2(-1)^m \mod{5}\\\text{However, } x^2 \equiv 0,1,-1 \mod{5}\\\pm 2 \notin \{0,1,-1\}\\\text{Hence }2^n+3^n \text{ is never a square for } n \in \mathbb{N}$

Paradoxica · Mar 11, 2016

Re: HSC 2016 4U Marathon - Advanced Level

$$\noindent Prove the irrationality of $\sqrt[n]{2} $ for $n \geq 2.$

KingOfActing · Mar 11, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
$$\noindent Prove the irrationality of $\sqrt[n]{2} $ for $n \geq 2.$

$\text{Assume } \sqrt[n]{2} = \frac{p}{q} $where $p$ and $q$ have no common factor. Then:$\\2 = \frac{p^n}{q^n}\\2q^n = p^n\\$So $ p = 2r\\2q^n = (2r)^n = 2^nr^n\\q^n = 2^{n-1}r^n\\$Since $ n \geq 2$:$\\q=2s\\$However, we stated that $p$ and $q$ may not have any common factors, so we have reached a contradiction. Hence, $ \sqrt[n]{2} \notin \mathbb{Q}\\\\$Alternatively (though overkill):$\\x = \sqrt[n]{2}\\x^n - 2 = 0\\$By the rational root theorem, the only possible rational roots are $ \pm 1, \pm 2, $ none of which are equal to $ \sqrt[n]{2}$

EDIT: SUPER OVERKILL

$Continuing from $ 2q^n = p^n\\q^n + q^n = r^n\\$However, by Fermat's Last Theorem, this has no integer solutions for $ n \geq 3 $. The case $ n = 2 $ is a trivial result.$

Drsoccerball · Mar 12, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Prove that the medians of a triangle meet at a point G which divides each median in the ratio 2:1 using vectors.

InteGrand · Mar 12, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
Prove that the medians of a triangle meet at a point G which divides each median in the ratio 2:1 using vectors.

I think a complex numbers vectors proof of this was in last year's Carrotsticks BOS 4U Trial. It's easy to prove by elementary geometry as well though. We'll give this proof below.

$\noindent Let the triangle have vertices $A, B, C$, with midpoints of $AB, BC, CA$ being $P, Q, R$ respectively. Join $AQ$ and $CP$, and let these medians intersect at point $G$.$

$\noindent Now, consider triangles $PQG$ and $CAG$. Note that $PQ$ is parallel to $CA$ (and half its length), using the HSC geometry result that the line segment joining midpoints of two sides of a triangle (the triangle $ABC$ here) is parallel to and half the length of the third side. Hence $\angle PQG = \angle CAG$ and $\angle QPG = \angle ACG$ (alternate angles in parallel lines $PQ$ and $CA$). So $\triangle PQG$ is similar to $\triangle CAG$ (equiangular).$

$\noindent Now, by corresponding sides of similar triangles having equal ratios of lengths, we know that as $PQ:CA =\frac{1}{2}$, we also have $PG:CG =\frac{1}{2} = QG:AG$. So the medians $CP$ and $AQ$ meet at the point $G$ that is divides each median in the ratio $2:1$. Repeating the argument with another pair of medians, say $AQ$ and $BR$ shows that the point $G^\prime$ where these two medians meet also divides each of these medians in the ratio $2:3$. But the point $G$ is the (obviously unique) point that divides $AQ$ in this ratio, so $G=G^\prime$. So all three medians go through this point $G$ and are divided in the ratio $2:1$ by it. Q.E.D.$

seanieg89 · Mar 17, 2016

Re: HSC 2016 4U Marathon - Advanced Level

A little sum I whipped up. For a fixed positive integer n, evaluate

$\sum_{k=1}^{2n}\frac{\sin(\frac{k\pi}{4n+2})\cos(\frac{2k\pi}{4n+2})}{\sin(\frac{2k\pi}{4n+2})\cos(\frac{k\pi}{4n+2})}$

Paradoxica · Mar 19, 2016

Re: HSC 2016 4U Marathon - Advanced Level

$\noindent a) Let $m,n,$ be coprime naturals, and suppose $m>n$. The roots of $z^m = 1$ are used to construct a regular $m$-sided polygon on the unit circle. Let the variable $\pi$ be chosen such that none of the roots of $z^n = \cos{\pi}+i \sin{\pi}$ are equal to any of the $m^{\text{th}}$ roots of unity. Construct the $n$ sided regular polygon using the $n$ roots of the second equation.$\\$Select a vertex of the first polygon and a vertex of the second polygon and construct the corresponding chord through them. Repeat this construction for all remaining $mn -1$ pairs of vertices. Find, in simplest form, the expression for the product of all these chords.$

$\noindent b) Investigate the situation when $m$ and $n$ are not coprime.$

seanieg89 · Mar 22, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Easier than my unanswered summation problem:

Consider the solid of revolution S formed by rotating the disk $(x-R)^2+y^2\leq r^2 \quad (r < R)$ about the y-axis.

a) Find it's volume by standard MX2 methods.

b) Show that this volume is also equal to $Ad$ where A is the area of the disk used to construct S and d is the total distance travelled by the centre of the disk as we rotate this disk around the the y-axis.

c) Generalise the above to other solids of revolution.

Hint: The analogue of the centre of the circle in this case is the point on which the region would balance perfectly. First find an expression for this point using calculus or otherwise.

For the sake of simplicity, restrict your attention to solids of revolution generated by plane regions of the form:

$\{(x,y):a\leq x \leq b,f_1(x)\leq y \leq f_2(x)\}$

where a < b are positive reals and $f_1\leq f_2$ are continuous functions that are equal at a and b.

d) Inscribe and circumscribe regular 2n-gons about the unit circle centred at (2,0) (both oriented to have an edge parallel to the x-axis) and consider the plane region R bounded by these two polygons. What is the volume of the solid of revolution generated by R?

dan964 · Apr 7, 2016

Re: HSC 2016 4U Marathon - Advanced Level

seanieg89 said:
Easier than my unanswered summation problem:

Consider the solid of revolution S formed by rotating the disk $(x-R)^2+y^2\leq r^2 \quad (r < R)$ about the y-axis.

a) Find it's volume by standard MX2 methods.

b) Show that this volume is also equal to $Ad$ where A is the area of the disk used to construct S and d is the total distance travelled by the centre of the disk as we rotate this disk around the the y-axis.

c) Generalise the above to other solids of revolution.

Hint: The analogue of the centre of the circle in this case is the point on which the region would balance perfectly. First find an expression for this point using calculus or otherwise.

For the sake of simplicity, restrict your attention to solids of revolution generated by plane regions of the form:

$\{(x,y):a\leq x \leq b,f_1(x)\leq y \leq f_2(x)\}$

where a < b are positive reals and $f_1\leq f_2$ are continuous functions that are equal at a and b.

d) Inscribe and circumscribe regular 2n-gons about the unit circle centred at (2,0) (both oriented to have an edge parallel to the x-axis) and consider the plane region R bounded by these two polygons. What is the volume of the solid of revolution generated by R?

Someone is actually do a presentation on this next week in one of my subjects at university.

Paradoxica · Apr 8, 2016

Re: HSC 2016 4U Marathon - Advanced Level

seanieg89 said:
A little sum I whipped up. For a fixed positive integer n, evaluate

$\sum_{k=1}^{2n} \frac{\sin{\left(\frac{k\pi}{4n+2}\right)} \cos{\left(\frac{2k\pi}{4n+2}\right)}}{\sin{\left(\frac{2k\pi}{4n+2}\right)} \cos{\left(\frac{k\pi}{4n+2}\right)}}$

$$\noindent The following identity is left as an exercise to the reader:$ \\ \tan{\theta} \cot{2\theta} \equiv \frac{1}{2} - \frac{1}{2}\tan^2{\theta}\\ \sum_{k=1}^{2n} \frac{\sin{\left(\frac{k\pi}{4n+2}\right)} \cos{\left(\frac{2k\pi}{4n+2}\right)}}{\sin{\left(\frac{2k\pi}{4n+2}\right)} \cos{\left(\frac{k\pi}{4n+2}\right)}} = \sum_{k=1}^{2n} \frac{1}{2} - \frac{1}{2}\tan^2{\frac{k\pi}{4n+2}} \\ $From De Moivre's Identity, consider the unsimplified quasi-binomial expansions of $\cos{(4n+2)\theta}$ and $\sin{(4n+2)\theta}\\ \cos{(4n+2)\theta} \equiv \cos^{4n+2}{\theta} - \binom{4n+2}{4n}\cos^{4n}{\theta}\sin^2{\theta} + \binom{4n+2}{4n-2}\cos^{4n-2}{\theta}\sin^4{\theta} -\cdots -\binom{4n+2}{4}\cos^4{\theta}\sin^{4n-2}\theta+\binom{4n+2}{2} \cos^2{\theta}\sin^{4n}{\theta}- \sin^{4n+2}{\theta}$

$$\noindent$ \sin{(4n+2)\theta} \equiv \binom{4n+2}{4n+1}\cos^{4n+1}{\theta}\sin{\theta} - \binom{4n+2}{4n-1}\cos^{4n-1}{\theta}\sin^3{\theta}+\cdots - \binom{4n+2}{3}\cos^3{\theta}\sin^{4n-1}{\theta} + \binom{4n+2}{1}\cos{\theta}\sin^{4n+1}{\theta} \\ $We can now obtain the rational expansion of $\tan{(4n+2)}\theta}$ by dividing the sine and cosine expansions together and then further dividing the numerator and denominator by $\cos^{4n+2}\theta$. Observe, that, however, the angles in the sum we are considering are exactly the angles of the specific solutions to $\tan{(4n+2)\theta} = 0$, which means we only need consider the sine expansion divided by $\cos^{4n+2}\theta$. In other words: $\\ \tan{(4n+2)\theta} = 0 \Leftrightarrow \binom{4n+2}{1}t^{4n+1} - \binom{4n+2}{3}t^{4n-1} + \cdots - \binom{4n+2}{4n-1}t^3 + \binom{4n+2}{4n+1}t = 0 $ where $ t = \tan{\theta}$

$$\noindent The desired sum does not have the angle zero, so $\tan{\theta} \neq 0$, so we can remove the factor of $t$ from the polynomial.$ \\ \binom{4n+1}{1}t^{4n} - \binom{4n+2}{3}t^{4n-2} + \cdots - \binom{4n+2}{4n-1}t^2 + \binom{4n+2}{4n+1} = 0\\$The roots of the current polynomial are $t = \tan{\frac{k\pi}{4n+2}}$, where $k$ ranges from 1 to $4n+1$, and excludes $2n+1$, as we cannot have an undefined root. The polynomial is also a polynomial in $t^2$, where the roots are $t^2 = \tan^2{\frac{k\pi}{4n+2}}$ as $k$ ranges from either 1 to $2n$, or $2n+2$ to $4n+1$. For convenience and coincidence with the sum, we pick the former range of $k$ values. \\By Vieta's Formulae for roots of Polynomials, we have the following: $\\ \sum_{k=1}^{2n} \tan^2{\frac{k\pi}{4n+2}} = \frac{\binom{4n+2}{3}}{\binom{4n+2}{1}} = \frac{2(4n+1)n}{3}$

$$\noindent We now have enough information required to finish off the original problem.$ \\ \sum_{k=1}^{2n} \frac{\sin{\left(\frac{k\pi}{4n+2}\right)} \cos{\left(\frac{2k\pi}{4n+2}\right)}}{\sin{\left(\frac{2k\pi}{4n+2}\right)} \cos{\left(\frac{k\pi}{4n+2}\right)}} = \sum_{k=1}^{2n} \frac{1}{2} - \frac{1}{2}\tan^2{\frac{k\pi}{4n+2}} = n - \frac{1}{2}\sum_{k=1}^{2n} \tan^2{\frac{k\pi}{4n+2}} = n - \frac{(4n+1)n}{3} = -\frac{4n^2+n-3n}{3}=-\frac{2n(2n-1)}{3} = -\frac{2}{3}\binom{2n}{2}$

Paradoxica · May 8, 2016

Re: HSC 2016 4U Marathon - Advanced Level

The odd natural numbers are partitioned as follows:

1
3,5
7,9,11
13,15,17,19
21,23,25,27,29
...
...
...

And so on, to infinity.

Prove the following statement:

The sum of all the numbers in the n^th row is equal to n³

KingOfActing · May 8, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
The odd natural numbers are partitioned as follows:

1
3,5
7,9,11
13,15,17,19
21,23,25,27,29
...
...
...

And so on, to infinity.

Prove the following statement:

The sum of all the numbers in the n^th row is equal to n³

The first term of each row is given by $n^2 - n + 1$ . The term in row n and position m is given by $n^2-n - 1 + 2m$ .

This gives us that the sum of each row is:

$\sum_{m=1}^{n}n^2 - n - 1 + 2m = n^3 - n^2 - n + 2\frac{n(n+1)}{2} = n^3 - n^2 - n +n^2 + n = n^3$

abecina · Jun 22, 2016

Re: HSC 2016 4U Marathon - Advanced Level

$\textrm{The shaded (parallel) cross section is an isosceles trapezium with height } h \textrm{ and parallel sides equal to } y \textrm{ and } z \textrm{ as shown}. \\\textrm{The rear face } ABCD \textrm{ is a square with side length of } a. \textrm{ The base of the solid } CDEF \textrm{ is also a trapezium}\\ \textrm{ with height } a, \textrm{ and base angles equal to } $\pi/3$. \textrm{ The point } G \textrm{ is such that } AG=BG. \textrm{ Find the volume of the solid in terms of \textit{a}.}$

Paradoxica · Jul 4, 2016

Re: HSC 2016 4U Marathon - Advanced Level

abecina said:
$\textrm{The shaded (parallel) cross section is an isosceles trapezium with height } h \textrm{ and parallel sides equal to } y \textrm{ and } z \textrm{ as shown}. \\\textrm{The rear face } ABCD \textrm{ is a square with side length of } a. \textrm{ The base of the solid } CDEF \textrm{ is also a trapezium}\\ \textrm{ with height } a, \textrm{ and base angles equal to } $\pi/3$. \textrm{ The point } G \textrm{ is such that } AG=BG. \textrm{ Find the volume of the solid in terms of \textit{a}.}$ View attachment 33283

By similar triangles, the base of the triangle is a+2x/√3
The height is a-x since the diagonal which generates it is exactly 45 degrees inclined. (the base and height of the object have length a along the axis of generation)
The cross section along the middle is that of a pyramid, so by Cavalieri's Principle, it is a square, so the top length is a-x.

So the infinitesimal change in volume is equal to the area of the trapezium times the infinitesimal change in x

dV = (a-x)(a-x+a+2x/√3)/2 dx

Note that we should test the area function to make sure it is correct. Plugging in x=0 gives a² as expected, and plugging in x=a gives 0 as expected, as these correspond to the cross sections at the extreme ends of the solid.

x goes from 0 to a so the corresponding integral will go from 0 to a

$\int_0^a \frac{a-x}{2}\left ( a-x+a+\frac{2x}{\sqrt{3}} \right ) $d$x$

Use the basic border flip property on the integral.

$\int_0^a \frac{x}{2}\left (x+a+\frac{2(a-x)}{\sqrt{3}} \right ) $d$x$

After some tedious computation the volume turns out to be:

$\frac{a^3}{6}\left(\frac{5}{2} + \frac{1}{\sqrt{3}} \right)$

InteGrand · Jul 4, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
By similar triangles, the base of the triangle is a+2x/√3
The height is a-x since the diagonal which generates it is exactly 45 degrees inclined. (the base and height of the object have length a along the axis of generation)
The cross section along the middle is that of a pyramid, so by Cavalieri's Principle, it is a square, so the top length is a-x.

So the infinitesimal change in volume is equal to the area of the trapezium times the infinitesimal change in x

dV = (a-x)(a-x+a+2x/√3)/2 dx

Note that we should test the area function to make sure it is correct. Plugging in x=0 gives a² as expected, and plugging in x=a gives 0 as expected, as these correspond to the cross sections at the extreme ends of the solid.

x goes from 0 to a so the corresponding integral will go from 0 to a

$\int_0^a \frac{a-x}{2}\left ( a-x+a+\frac{2x}{\sqrt{3}} \right ) $d$x$

Use the basic border flip property on the integral.

$\int_0^a \frac{x}{2}\left (x+a+\frac{2(a-x)}{\sqrt{3}} \right ) $d$x$

After some tedious computation the volume turns out to be:

$\frac{a^3}{3}\left(\frac{1}{2} - \frac{1}{\sqrt{3}} \right) + \frac{a^2}{4}\left(1+\frac{2}{\sqrt{3}} \right)$

Now why is my volume negative.

I think you made a silly mistake in the computation of the integral at the end. (Also, we'd expect the final answer to only involve cubic terms in a as it's a volume. We can also see by inspection that the integral will result in only cubic terms in a.)

tywebb · Jul 4, 2016

Re: HSC 2016 4U Marathon - Advanced Level

2002 HSC Ext. 2 Q8a+solution by mojako attached. The question has four parts.

But the question can be extended with four more parts to prove

$\textstyle\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\dots=\frac{\pi^2}{6}$

as follows.

$\bf{Extension\ to\ 2002\ HSC\ Ext.\ 2\ Q8a}$

${\bf(v)}\text{ Use the identity }\cot^2\theta={\rm{cosec}}^2\ \!\theta-1 \text{ and the result of Question 8 (a) (iii) to prove that}$

${\rm{cosec}}^2\Big(\frac{\pi}{2m+1}\Big)+{\rm{cosec}}^2\Big(\frac{2\pi}{2m+1}\Big)+\dots+{\rm{cosec}}^2\Big(\frac{m\pi}{2m+1}\Big)=\frac{2m(m+1)}{3}$

${\bf(vi)} \text{ Use the fact that }\displaystyle {\rm{cosec}}\ \!\theta>\frac{1}{\theta}$ for $0<\displaystyle \theta<\frac{\pi}{2}$\text{ to show that}$

$\Big(\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\dots +\frac{1}{m^{2}}\Big)\frac{(2m+1)^{2}}{4m(m+1)}<\frac{\pi^{2}}{6}$

${\bf(vii)}\text{ Deduce that }\textstyle\frac{1}{(2m+1)^2}<1-\frac{6}{\pi^2}\big(\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\dots +\frac{1}{m^{2}}\big)<\frac{6m+1}{(2m+1)^2}$

${\bf(viii)}\text{ Taking the limit as }m\rightarrow\infty \text{ deduce that }\textstyle\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\dots=\frac{\pi^2}{6}$

Historical note: This was first proposed in 1644 by Pietro Mengoli and then solved by Euler in 1734. But Euler did not use this method. In fact there are several methods by which it can be proved. The method in this HSC exam together with the extension was first used by Augustin Louis Cauchy in 1821 in a book called Cours d'Analyse.

Nailgun · Jul 4, 2016

Re: HSC 2016 4U Marathon - Advanced Level

This thread makes me feel a little better for not attempting to do 4U.

InteGrand · Jul 4, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Nailgun said:
This thread makes me feel a little better for not attempting to do 4U.

The questions in HSC 4U (technically "Extension 2" of course) papers of today are overall much easier than most of the Q's in these 4U Advanced Marathons.

seanieg89 · Jul 23, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Here is a derivation of $\zeta(2)$ via a scenic route.

$$1. Prove that: $\int_0^\infty \frac{\sin(x)}{x}\, dx=\pi/2$ by considering the related integral $ \int_0^{\pi/2}\frac{\sin((2n+1)x)}{\sin(x)}\, dx$

$2. Find an integral expression for $\sum_{k=1}^n\frac{\sin(kx)}{k}$ where $x\in(0,2\pi)$.$

$3. Show that the the sum in 2. converges to $(\pi-x)/2$ for $x\in(0,2\pi)$. Justify your manipulations as best you can.$

$4. By integrating the sum in 2., find $\sum_{n=1}^\infty n^{-2}$. Justify your manipulations as best you can.$

$(*) 5. An interesting aspect of this method is how easily it generalises. Can you build on the work done in this question to compute $\sum_{n=1}^\infty n^{-4}$? Or even better, can you find a recursive relationship between $\sum_{n=1}^\infty n^{-2k}$ for positive integers $k$?$

Paradoxica · Jul 23, 2016

Re: HSC 2016 4U Marathon - Advanced Level

seanieg89 said:
Here is a derivation of $\zeta(2)$ via a scenic route.

$$1. Prove that: $\int_0^\infty \frac{\sin(x)}{x}\, dx=\pi/2$ by considering the related integral $ \int_0^{\pi/2}\frac{\sin((2n+1)x)}{\sin(x)}\, dx$

$2. Find an integral expression for $\sum_{k=1}^n\frac{\sin(kx)}{k}$ where $x\in(0,2\pi)$.$

$3. Show that the the sum in 2. converges to $(\pi-x)/2$ for $x\in(0,2\pi)$. Justify your manipulations as best you can.$

$4. By integrating the sum in 2., find $\sum_{n=1}^\infty n^{-2}$. Justify your manipulations as best you can.$

Should that be ∞ instead of n? Or perhaps the limit?

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