HSC 2013 MX2 Marathon (archive) (5 Viewers)

Sy123 · Jul 25, 2013

Re: HSC 2013 4U Marathon

$$Prove for some positive$ \ n \ $and real$ \ k$

$e^{k\frac{n}{n+1}} < \left( 1+ \frac{k}{n} \right )^n < e^k$

$Hence evaluate the limit$

$\lim_{n \to \infty} \left( 1+ \frac{k}{n} \right )^n$

RealiseNothing · Jul 25, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Prove for some positive$ \ n \ $and real$ \ k$

$e^{k\frac{n}{n+1}} < \left( 1+ \frac{k}{n} \right )^n < e^k$

$Hence evaluate the limit$

$\lim_{n \to \infty} \left( 1+ \frac{k}{n} \right )^n$

Consider the curve $y=\frac{1}{x}$ from the x-values n to n+k. Construct an upper bound rectangle from n to n+k as the width, and f(n) as the height. This will give the inequality:

$\int_{n}^{n+k} \frac{1}{x} dx < [(n+k)-(n)]f(n)$

$\ln(n+k)-\ln(n) < \frac{k}{n}$

$\ln(\frac{n+k}{n}) < \frac{k}{n}$

$\ln(1+\frac{k}{n}) < \frac{k}{n}$

$n\ln(1+\frac{k}{n}) < k$

$\ln[(1+\frac{k}{n})^n] < k$

$(1+\frac{k}{n})^n < e^k$

Then do something similar for the lower bound, but I am still doing that so will edit it in.

Carrotsticks · Jul 25, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Very similar to last year's BOS trial question 15 by carrot.

In my paper, students were given the existence of the E-M constant. Sy's question is proving its existence.

Also Sy, how did you expect students to prove the existence of it without the monotone convergence theorem?

RealiseNothing · Jul 25, 2013

Re: HSC 2013 4U Marathon

By considering an isosceles triangle with one of it's angles 108 degrees, prove that:

$\sin(\frac{\pi}{10}) = \frac{1}{1+\sqrt{5}}$

Sy123 · Jul 25, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
In my paper, students were given the existence of the E-M constant. Sy's question is proving its existence.

Also Sy, how did you expect students to prove the existence of it without the monotone convergence theorem?

I expected students to do something similar to what Realise did. Does the existence of the limit need to be specified despite the upper bound that is independent of n?
The limit should be able to be done simply using squeeze theorem

RealiseNothing said:
By considering an isosceles triangle with one of it's angles 108 degrees, prove that:

$\sin(\frac{\pi}{10}) = \frac{1}{1+\sqrt{5}}$

Let the equal sides of the triangle be A and the side opposite 108 degrees be B.

$$Let$ \ \theta = \frac{\pi}{5}$

From the sine rule of a triangle:

$\frac{\sin(3\theta)}{\sin \theta} = \frac{B}{A}$

$4\cos^2 \theta - 1 = \frac{B}{A}$

Now, using the cosine rule on angle theta:

$\cos \theta = \frac{B^2+A^2-A^2}{2AB} = \frac{B}{2A}$

Now on the larger angel:

$\cos(3\theta) = \frac{2A^2-B^2}{2A^2} = 4\cos^3 \theta - 3\cos \theta = \cos \theta(4\cos^2 \theta - 1- 2)$

Now:

$\frac{2A^2-B^2}{2A^2} = \frac{B}{2A} \left(\frac{B}{A} - 1 \right )$

Simplifying:

$B^2 = AB+A^2$
$\left (\frac{B}{A} \right )^2 - \frac{B}{A} - 1= 0$

$\frac{B}{A} = \frac{1+\sqrt{5}}{2}$

Now, we know that:

$\sin(\frac{\theta}{2}) = \sqrt{\frac{1}{2} \left( 1 - \frac{B}{2A} \right)} = \frac{1}{2} \sqrt{2-\frac{B}{A}}$

Substituting in the value for B/A we get:

$\sin \left( \frac{\pi}{10} \right ) = \frac{1}{2} \sqrt{\frac{3-\sqrt{5}}{2}} = \frac{1}{2} \sqrt{\frac{(1-\sqrt{5})^2}{4}} = \frac{|1-\sqrt{5}|}{4} = \frac{1}{1+\sqrt{5}}$

I feel like I took a long approach

Carrotsticks · Jul 25, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
I expected students to do something similar to what Realise did. Does the existence of the limit need to be specified despite the upper bound that is independent of n?
The limit should be able to be done simply using squeeze theorem

1. Don't quite get what you mean by the upper bound of gamma being independent of n.

2. Squeeze Theorem is usually used to find the closed form for some sort of sequence/series/limit. Gamma, as we know so far, has no single closed form.

Sy123 · Jul 25, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
1. Don't quite get what you mean by the upper bound of gamma being independent of n.

2. Squeeze Theorem is usually used to find the closed form for some sort of sequence/series/limit. Gamma, as we know so far, has no single closed form.

Ah sorry I thought you meant the e^k question for some reason.

For that one, I aimed for students to first consider the upper rectangles of the graph y=1/x from x=1 to x=(n)

$H_n > \int_0^{n+1} \frac{dx}{x}$

$H_n - \ln(n+1) > 0$

$H_n - \ln(n) > 0$

Now the lower rectangles from x=2 to x=n

$(H_n - 1) < \int_0^{n} \frac{dx}{x}$

$H_n - \ln n < 1$

$\therefore \ 0 < H_n - \ln n < 1$

So no matter how n increases, it will always be less than 1, and greater than zero, thus converging to a finite limit.

Is this incorrect?

Carrotsticks · Jul 25, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Ah sorry I thought you meant the e^k question for some reason.

For that one, I aimed for students to first consider the upper rectangles of the graph y=1/x from x=1 to x=(n)

$H_n > \int_0^{n+1} \frac{dx}{x}$

$H_n - \ln(n+1) > 0$

$H_n - \ln(n) > 0$

Now the lower rectangles from x=2 to x=n

$(H_n - 1) < \int_0^{n} \frac{dx}{x}$

$H_n - \ln n < 1$

$\therefore \ 0 < H_n - \ln n < 1$

So no matter how n increases, it will always be less than 1, and greater than zero, thus converging to a finite limit.

Is this incorrect?

Yep, that is incorrect. A counter-example would be the curve y=sin(x). It is always less than 1 and greater than -1, but it most certainly does NOT converge to a finite limit.

Sy123 · Jul 25, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
Yep, that is incorrect. A trivial example is to consider the curve y=sin(x). It is always less than 1 and greater than -1, but it most certainly does NOT converge to a finite limit.

Ah yep that is true

Apologies

EDIT: What if I showed

$H_n - \ln (n) >H_{n+1} - \ln(n+1)$

And then made the argument that as n increases H_n - ln(n) keeps decreasing however it is bounded on the lower by 0, then it approaches some finite limit?

We can show the above by simply showing that, then subbing in x=1/n

$\ln(1+x) > \frac{x}{x+1}$

Which can be done through calculus.

Carrotsticks · Jul 25, 2013

Re: HSC 2013 4U Marathon

No need for apologies, this is how we learn things =) By making mistakes.

Carrotsticks · Jul 25, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Ah yep that is true

Apologies

EDIT: What if I showed

$H_n - \ln (n) >H_{n+1} - \ln(n+1)$

And then made the argument that as n increases H_n - ln(n) keeps decreasing however it is bounded on the lower by 0, then it approaches some finite limit?

That is the monotone convergence theorem, as I mentioned earlier =)

Sy123 · Jul 25, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
That is the monotone convergence theorem, as I mentioned earlier =)

Ah ok thanks for the clarification

RealiseNothing · Jul 25, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
I expected students to do something similar to what Realise did. Does the existence of the limit need to be specified despite the upper bound that is independent of n?
The limit should be able to be done simply using squeeze theorem

Let the equal sides of the triangle be A and the side opposite 108 degrees be B.

$$Let$ \ \theta = \frac{\pi}{5}$

From the sine rule of a triangle:

$\frac{\sin(3\theta)}{\sin \theta} = \frac{B}{A}$

$4\cos^2 \theta - 1 = \frac{B}{A}$

Now, using the cosine rule on angle theta:

$\cos \theta = \frac{B^2+A^2-A^2}{2AB} = \frac{B}{2A}$

Now on the larger angel:

$\cos(3\theta) = \frac{2A^2-B^2}{2A^2} = 4\cos^3 \theta - 3\cos \theta = \cos \theta(4\cos^2 \theta - 1- 2)$

Now:

$\frac{2A^2-B^2}{2A^2} = \frac{B}{2A} \left(\frac{B}{A} - 1 \right )$

Simplifying:

$B^2 = AB+A^2$
$\left (\frac{B}{A} \right )^2 - \frac{B}{A} - 1= 0$

$\frac{B}{A} = \frac{1+\sqrt{5}}{2}$

Now, we know that:

$\sin(\frac{\theta}{2}) = \sqrt{\frac{1}{2} \left( 1 - \frac{B}{2A} \right)} = \frac{1}{2} \sqrt{2-\frac{B}{A}}$

Substituting in the value for B/A we get:

$\sin \left( \frac{\pi}{10} \right ) = \frac{1}{2} \sqrt{\frac{3-\sqrt{5}}{2}} = \frac{1}{2} \sqrt{\frac{(1-\sqrt{5})^2}{4}} = \frac{|1-\sqrt{5}|}{4} = \frac{1}{1+\sqrt{5}}$

I feel like I took a long approach

My approach was to split the triangle into two smaller triangles by dividing the 108 angle into 72 and 36 angles and connecting that vertex to the opposite side. Let the equal sides of the larger triangle be 1, and the side opposite the 108 angle be some $x$

Now one of the smaller triangles has angles 36, 36, 108, and thus is similar to the larger triangle. We can also easily find the side lengths of the smaller triangle and use the ratio of the matching sides of the similar triangles to obtain:

$\frac{x}{1} = \frac{1}{x-1}$

Solving for $x$ gives:

$x^2 - x - 1 = 0$

$x = \frac{1+\sqrt{5}}{2}$

$2x = 1 + \sqrt{5}$

Now we use the sine rule on the larger triangle to obtain:

$\frac{\sin(36)}{1} = \frac{\sin(108)}{x}$

Now $\sin(36) = 2\sin(18)\cos(18)$ and $\sin(108) = \sin(72) = \cos(18)$

Substituting these in to what we have gives:

$2\sin(18)\cos(18) = \frac{\cos(18)}{x}$

Cancelling out the $\cos(18)$ and dividing both sides by 2 gives:

$\sin(18) = \frac{1}{2x}$

Substituting in our value for $2x$ gives the result:

$\sin(18) = \frac{1}{1+\sqrt{5}}$

RealiseNothing · Jul 25, 2013

Re: HSC 2013 4U Marathon

In the spirit of the AMC coming up next Thursday, here is a nice question from last year's paper:

$28 points are equally spaced around the circumference of a circle. What is the total number of triangles whose 3 vertices are from those 28 points and the size of one of the angles is twice the size of another?$

study1234 · Jul 26, 2013

Re: HSC 2013 4U Marathon

See question in image.

Sy123 · Jul 26, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
In the spirit of the AMC coming up next Thursday, here is a nice question from last year's paper:

$28 points are equally spaced around the circumference of a circle. What is the total number of triangles whose 3 vertices are from those 28 points and the size of one of the angles is twice the size of another?$

I got 840 and I think i got a more logical answer now:

My Solution

seanieg89 · Jul 26, 2013

Re: HSC 2013 4U Marathon

$1. For positive integers $d$ and $m$, evaluate:\\ $f(m,d)=\frac{1}{d}\sum_{j=1}^{d} \cos\left(\frac{2\pi jm}{d}\right).$\\ \\2. Use the above to find a function $g(n)$ defined in terms of MX2 functions such that $g(n)=1$ if $n$ is prime and $g(n)=0$ if $n$ is composite.\\ \\ 3. Use the function $g$ to construct a function $p(n)$ whose value at $n$ is the $n$-th prime number.$

Sy123 · Jul 28, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$1. For positive integers $d$ and $m$, evaluate:\\ $f(m,d)=\frac{1}{d}\sum_{j=1}^{d} \cos\left(\frac{2\pi jm}{d}\right).$\\ \\2. Use the above to find a function $g(n)$ defined in terms of MX2 functions such that $g(n)=1$ if $n$ is prime and $g(n)=0$ if $n$ is composite.\\ \\ 3. Use the function $g$ to construct a function $p(n)$ whose value at $n$ is the $n$-th prime number.$

1.

It is clear through Demoivre's theorem:

$\cos x + \cos(2x) + \cos(3x) + \dots + \cos (nx) = \frac{\sin \left(n+\frac{1}{2})x}{\sin\frac{x}{2}} - \frac{1}{2}$

Substituting in $x=\frac{2\pi m}{d}$

We get:

$f(m,d) = 0 \ \ $iff$ \ \frac{m}{d} \ $is not an integer$$

Taking the case of m/d being an integer, we look at the series version of f(m,d) it is clear that:

$f(m,d) = 1 \ \ $iff$ \ \frac{m}{d} \ $is an integer$$

There is no other case.

2.

I'm going to guess that

g(n) = f(x(n), y(n))

$\frac{x(n)}{y(n)} = $integer$ \ \ $for prime$ \ n$

And not an integer for composite n.

Now,from your earlier paper about the formula for primes lie, the property of prime numbers:

$p \ \ $is only a prime if and only if$ \ (p-1)! \equiv -1 \mod{p}$

I still don't know how to prove this but I'm going to take it as true

Thus, we can see that

$x(n) = (n-1)!+1$

$y(n) = n$

Therefore:

$g(n) = f((n-1)! + 1, n)$

3.

$p(n) = n \cdot g(n)$

===

Hopefully that is correct

Sy123 · Jul 28, 2013

Re: HSC 2013 4U Marathon

$An$ \ n \ $sided regular polygon is such that no side is vertical in gradient$

$Prove that if$ \ \ m_k \ \ $is the gradient of the$ \ k^{th} \ $ side of the polygon then$

$m_1m_2 + m_2m_3 + m_3 m_4 + \dots + m_{n-1}m_n + m_n m_1 = -n$

seanieg89 · Jul 29, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
1.

It is clear through Demoivre's theorem:

$\cos x + \cos(2x) + \cos(3x) + \dots + \cos (nx) = \frac{\sin \left(n+\frac{1}{2})x}{\sin\frac{x}{2}} - \frac{1}{2}$

Substituting in $x=\frac{2\pi m}{d}$

We get:

$f(m,d) = 0 \ \ $iff$ \ \frac{m}{d} \ $is not an integer$$

Taking the case of m/d being an integer, we look at the series version of f(m,d) it is clear that:

$f(m,d) = 1 \ \ $iff$ \ \frac{m}{d} \ $is an integer$$

There is no other case.

2.

I'm going to guess that

g(n) = f(x(n), y(n))

$\frac{x(n)}{y(n)} = $integer$ \ \ $for prime$ \ n$

And not an integer for composite n.

Now,from your earlier paper about the formula for primes lie, the property of prime numbers:

$p \ \ $is only a prime if and only if$ \ (p-1)! \equiv -1 \mod{p}$

I still don't know how to prove this but I'm going to take it as true

Thus, we can see that

$x(n) = (n-1)!+1$

$y(n) = n$

Therefore:

$g(n) = f((n-1)! + 1, n)$

3.

$p(n) = n \cdot g(n)$

===

Hopefully that is correct

Well you don't need to know how to prove the factorial thing, think about the definition of a prime number...it is convenient that we have found a function f that "checks divisibility".

Your answer is quite close, but your p(n) will actually spit out (1,2,3,0,5,0,7,0,0,0,11,etc) rather than (2,3,5,7,etc).

HSC 2013 MX2 Marathon (archive) (5 Viewers)

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