HSC 2016 MX2 Marathon ADVANCED (archive) (1 Viewer)

Sy123 · Nov 6, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
$\\ $On a perfect circular billiard table, the ball is hit at angle$ \ \theta \ $radians away from the line joining the starting point and the centre$$

$\\ $Find the average distance of the ball to the centre in terms of$ \ \theta$

Sy123 · Nov 6, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
$\\ $An$ \ n \ $sided regular polygon is such that no side is vertical in gradient$ \\ $Prove that if$ \ m_k \ $is the gradient of the$ \ k^{\text{th}} \ $side, then:$ \\\\ m_1 m_2 + m_2 m_3 + \dots + m_{n-1}m_n + m_n m_1 = - n$

$\\ $Place this polygon onto a cartesian plane, let the internal angle between each side be$ \ \theta$

$\\ $So, the angle between lines with gradient$ \ m_k \ $and$ \ m_{k+1} \ $is$ \ \tan \theta = \frac{m_k - m_{k+1}}{1 + m_k m_{k+1}}$

$\\ $So,$ \ m_k m_{k+1} = \frac{1}{\tan \theta} ( m_i - m_{i+1}) - 1$

$m_1m_2 + \dots + m_n m_1 = \frac{1}{\tan \theta} (m_1 - m_2 + m_2 - m_3 + \dots + m_n - m_1) -n$

$m_1m_2 + m_2m_3 + \dots + m_nm_1 = 0 - n = -n$

Sy123 · Nov 6, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Perhaps something easier (with a bit of guidance), this is possible with 3U knowledge and 4U polynomial knowledge:

$\\ $Note: one of the altitudes of a triangle, is the distance from one of its vertices to its opposite side$$

$\\ $A triangle has side lengths$ \ a,b,c \\ \\ $i) Show that the area of the triangle is$ \\ \sqrt{s(s-a)(s-b)(s-c)} \ $where$ \ s = \frac{a+b+c}{2} \\ \\ $ii) Show that if$ \ a,b,c \ $are roots of a cubic polynomial with rational coefficients, then the altitudes of the triangle are roots of a sixth degree equation with rational coefficients$$

Paradoxica · Nov 6, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:

Holy shit I was that close to the actual solution. I got the integral part but from there everything went haywire.

Sy123 · Nov 6, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
Holy shit I was that close to the actual solution. I got the integral part but from there everything went haywire.

In the future feel free to post your progress, in this case it was just a computation and the integral is not essential to understanding the problem

Sy123 · Nov 7, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Perhaps an interesting excercise:

$\\ $Prove that the function$ \ f(x) = \frac{2x}{1+x^2} \ $is monotone decreasing for$ \ x > 1 \ $without any recourse to calculus$$

Paradoxica · Nov 7, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
Perhaps an interesting excercise:

$\\ $Prove that the function$ \ f(x) = \frac{2x}{1+x^2} \ $is monotone decreasing for$ \ x > 1 \ $without any resort to calculus$$

$$Observe $(x-1)^2 \geq 0$

$Expanding and rearranging yields $1 \geq \frac{2x}{1+x^2}$ which is only true when $x$ is positive.$

$$Consider the points $P(x,f(x))$ and $Q(x+a, f(x+a))$ where $a>0$

$$Observe that the secant line between P and Q has a gradient of $ \frac{\frac{2(x+a)}{1+(x+a)^2} - \frac{2x}{1+x^2}}{x+a-x}$

$=\frac{2(x+a)(1+x^2) - (2x)(1+(x+a)^2)}{a(1+x^2)(1+(x+a)^2)} = \frac{2x^3 + 2x + 2ax^2 + 2a - 2x - 2x^3 - 4ax^2 - 2a^2x}{a(1+x^2)(1+(x+a)^2)}$

$=\frac{2(1-x^2-ax)}{(1+x^2)(1+(x+a)^2)}. $ This gradient is definitely negative, as $x>1$ and $a$ is positive$$

$$So any point on the curve with a larger $x$-value has a smaller $y$-value, and hence, the curve is monotonically decreasing for $x>1$

InteGrand · Nov 7, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
$$Observe $(x-1)^2 \geq 0$

$Expanding and rearranging yields $1 \geq \frac{2x}{1+x^2}$ which is only true when $x$ is positive.$

$Taking the limit as $x$ tends to infinity, the expression goes to zero. Q.E.D.$

$This isn't sufficient to prove it's monotone decreasing on $[1,\infty)$.$

Paradoxica · Nov 7, 2015

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
$This isn't sufficient to prove it's monotone decreasing on $[1,\infty)$.$

I'm not sure what I can argue that isn't calculus.

InteGrand · Nov 7, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
I'm not sure what I can argue that isn't calculus.

Try considering f(x+a) – f(x), where a is positive and x ≥ 1.

KingOfActing · Nov 7, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
Perhaps an interesting excercise:

$\\ $Prove that the function$ \ f(x) = \frac{2x}{1+x^2} \ $is monotone decreasing for$ \ x > 1 \ $without any recourse to calculus$$

$\begin{align*} \text{Let } y \geq x \\ \text{We would like to show } \frac{2x}{x^2 + 1} \geq \frac{2y}{y^2 + 1} \text{ for all } x > 1 \\ \frac{2x}{x^2 + 1} &\geq \frac{2y}{y^2 + 1} \\ 2x(y^2+1) - 2y(x^2 + 1) &\geq 0 \\x(y^2+1) - y(x^2 + 1) &\geq 0 \\xy^2+x - yx^2 - y &\geq 0 \\xy^2 - yx^2 + x - y &\geq 0 \\xy(y-x) + x - y &\geq 0 \\-xy(x - y) + 1(x - y) &\geq 0 \\(x - y)(1-xy) &\geq 0\end{align*}$

That's as far as I could get. I know that it must be shown to be always true given the conditions y >= x > 1, but I'm not sure how.

InteGrand · Nov 7, 2015

Re: HSC 2016 4U Marathon - Advanced Level

KingOfActing said:
$\begin{align*} \text{Let } y \geq x \\ \text{We would like to show } \frac{2x}{x^2 + 1} \geq \frac{2y}{y^2 + 1} \text{ for all } x > 1 \\ \frac{2x}{x^2 + 1} &\geq \frac{2y}{y^2 + 1} \\ 2x(y^2+1) - 2y(x^2 + 1) &\geq 0 \\x(y^2+1) - y(x^2 + 1) &\geq 0 \\xy^2+x - yx^2 - y &\geq 0 \\xy^2 - yx^2 + x - y &\geq 0 \\xy(y-x) + x - y &\geq 0 \\-xy(x - y) + 1(x - y) &\geq 0 \\(x - y)(1-xy) &\geq 0\end{align*}$

That's as far as I could get. I know that it must be shown to be always true given the conditions y >= x > 1, but I'm not sure how.

$Try substituting $y=x+\xi$ where $\xi >0$; then the desired result should fall out, recalling that $x\geq1$.$

$In fact, it is clear from your last line; we assume $x<y$, so $x-y<0$, and we can divide out the factor of $(x-y)$ from the last line and flip the inequality sign. Then we need to show $1-xy <0$, which is clearly true as $y>x\geq 1$. Since your steps were biconditional (based on my quick glance), the original inequality must also be true.$

KingOfActing · Nov 7, 2015

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
$Try substituting $y=x+\xi$ where $\xi >0$; then the desired result should fall out, recalling that $x\geq1$.$

$In fact, it is clear from your last line; we assume $x<y$, so $x-y<0$, and we can divide out the factor of $(x-y)$ from the last line and flip the inequality sign. Then we need to show $1-xy <0$, which is clearly true as $y>x\geq 1$. Since your steps were biconditional (based on my quick glance), the original inequality must also be true.$

I just realised it as I was posting. Great question!

lita1000 · Nov 7, 2015

Re: HSC 2016 4U Marathon - Advanced Level

does any 2016ers need solutions for sy's past 3 questions before his most recent? I can quickly type them up if requested!

lita1000 · Nov 7, 2015

Re: HSC 2016 4U Marathon - Advanced Level

oh nvm, just realised sy already posted solutions for the polygon and average distance question, no need to post those ones up then. If a 2016ers needs soln for the altitude q, I can type them up.

Paradoxica · Nov 7, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$$Prove by strong induction that $12|(n^4-n^2)$ for $n \in \mathbb N$

Paradoxica · Nov 7, 2015

Re: HSC 2016 4U Marathon - Advanced Level

lita1000 said:
oh nvm, just realised sy already posted solutions for the polygon and average distance question, no need to post those ones up then. If a 2016ers needs soln for the altitude q, I can type them up.

yes please.

lita1000 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
yes please.

Which part do you specifically need help with(as in first part or second part)? I'll type up as soon as I come back from work

dan964 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
$$Prove by strong induction that $12|(n^4-n^2)$ for $n \in \mathbb N$

First...
Case 0: trivial, Case 1: LHS = 6, which is divisible by 6,
Case 2: LHS = 30, which is divisible by 6
Assume formula is true up to n=k:
k(k+1)(2k+1)=6M, M is a natural number
For n=k+1
LHS = (k+1)(k+2)(2k+3)
= (k+1)(2k^2 + 7k + 6)
= k(k+1)(2k + 1+ 6) + 6(k+1)
= k(k+1)(2k + 1) + 6k(k+1)+6(k+1)
= 6M + 6(k+1)^2
= 6K, K is a natural number
Since true for first case, and also for two consecutive cases, it is therefore true for all natural n

am I right to assume that it is asking for 12 is a factor of the n^4-n^2
Case n=0 or n=1
n^4-n^2 = 0, trivial
Case n=2
n^4-n^2 = 4*3 = 12, which is divisible by 12

Assume formula to be true, up to some natural number 'k'
k^4-k^2=12N where N is a natural number (not 0 or 1)

For n=k+1
LHS = (k+1)^4 - (k+1)^2
= k^4 + 4k^3 + 6k^2 + 4k+1 - k^2 - 2k -1
= 12N + 4k^3 + 6k^2 + 2k
= 12N + 2k(2k^2 + 3k+1)
= 12N + 2k(2k+1)(k+1)
It remains to show that n(n+1)(2n+1) is a multiple of 6, which was conducted earlier
= 12N + 2k*6Z (Z a natural number)
= 12(N+kZ)
Hence since the first case, two consecutive cases are true, the statement is proved true for all natural n.

Paradoxica · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

lita1000 said:
Which part do you specifically need help with(as in first part or second part)? I'll type up as soon as I come back from work

second part

dan964 said:
First...
Case 0: trivial, Case 1: LHS = 6, which is divisible by 6,
Case 2: LHS = 30, which is divisible by 6
Assume formula is true up to n=k:
k(k+1)(2k+1)=6M, M is a natural number
For n=k+1
LHS = (k+1)(k+2)(2k+3)
= (k+1)(2k^2 + 7k + 6)
= k(k+1)(2k + 1+ 6) + 6(k+1)
= k(k+1)(2k + 1) + 6k(k+1)+6(k+1)
= 6M + 6(k+1)^2
= 6K, K is a natural number
Since true for first case, and also for two consecutive cases, it is therefore true for all natural n

am I right to assume that it is asking for 12 is a factor of the n^4-n^2
Case n=0 or n=1
n^4-n^2 = 0, trivial
Case n=2
n^4-n^2 = 4*3 = 12, which is divisible by 12

Assume formula to be true, up to some natural number 'k'
k^4-k^2=12N where N is a natural number (not 0 or 1)

For n=k+1
LHS = (k+1)^4 - (k+1)^2
= k^4 + 4k^3 + 6k^2 + 4k+1 - k^2 - 2k -1
= 12N + 4k^3 + 6k^2 + 2k
= 12N + 2k(2k^2 + 3k+1)
= 12N + 2k(2k+1)(k+1)
It remains to show that n(n+1)(2n+1) is a multiple of 6, which was conducted earlier
= 12N + 2k*6Z (Z a natural number)
= 12(N+kZ)
Hence since the first case, two consecutive cases are true, the statement is proved true for all natural n.

Alternatively, prove that if n is true, n+6 is true, then show trivially that it is true for the first six natural numbers and draw your conclusion from there.

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