HSC 2016 MX1 Marathon (archive) (4 Viewers)

KingOfActing · Oct 1, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
Magically complete some squares?

Nah I remember it starts with "Let a + b + c = S" but I can't be bothered looking it up.

Next question

$KingOfActing has done an amazing bottle flip. If KoA manages a successful bottle flip, the adrenaline rush makes him very focused, and so the probability that he'll manage to do it again in the next try is $0.9$. If he didn't manage to land a successful bottle flip, he feels deflated and the probability that he will be able to manage the next try it is $0.2\\\\$i) Let $ p_n$ be the probability of him successfully landing his $n$th bottle flip, with $p_0 = 1$. Find a relationship between $p_{n+1}$ and $p_n$.$\\\\$ii) Find the probability that flips his $100$th bottle successfully in exact form.$\\\\$iii) Suppose KingOfActing is a mythical being that will never die and will flip until time itself is over. What is the probability that he flips the last bottle in existence successfully?$

He-Mann · Oct 1, 2016

Re: HSC 2016 3U Marathon

KingOfActing said:
Nah I remember it starts with "Let a + b + c = S" but I can't be bothered looking it up.

Next question

$KingOfActing has done an amazing bottle flip. If KoA manages a successful bottle flip, the adrenaline rush makes him very focused, and so the probability that he'll manage to do it again in the next try is $0.9$. If he didn't manage to land a successful bottle flip, he feels deflated and the probability that he will be able to manage the next try it is $0.2\\\\$i) Let $ p_n$ be the probability of him successfully landing his $n$th bottle flip, with $p_0 = 1$. Find a relationship between $p_{n+1}$ and $p_n$.$\\\\$ii) Find the probability that flips his $100$th bottle successfully in exact form.$\\\\$iii) Suppose KingOfActing is a mythical being that will never die and will flip until time itself is over. What is the probability that he flips the last bottle in existence successfully?$

ANSWER BELOW! WATCH OUT!

You should give this question a try because it really helps logical thinking. I think 2U maths can do this, as well.

i) The probability of the future event is dependent on the present event, only. Either success or failure. So,

$p_{n+1} = 0.2 * (failed\_previous) + 0.9 * (succeeded\_previous) \\= 0.2 * (1 - p_n) + 0.9 * p_n \\= 0.2 + 0.7 * p_n$

ii) Expand the recursion and notice a pattern,

$\begin{align*}p_{n+1} &= 0.2 + 0.7p_n \\ &= 0.2 + 0.7(0.2 + 0.7 p_{n-1}) \\ &= 0.2 + 0.2(0.7)+ (0.7)^2 p_{n-1} \\ &\vdots \\ &= 0.2 + 0.2(0.7)+ 0.2(0.7)^2 + \dots + 0.2(0.7)^{n-1}p_{0}\end{align*}$

The pattern is the power of 0.7 is one more than the integer that n is getting subtracted by. Is there a more articulate way of putting it?

Anyway, let's attain a closed form before doing anything else.

$\begin{align*} p_{n+1} &= 0.2 + 0.2(0.7)+ 0.2(0.7)^2 + \dots + 0.2(0.7)^{n-1} \\ &= 0.2 \left(\frac{1- (0.7)^{n}}{1-0.7} \right )\end{align*}$

Let n = 99 for the 100-th success.

iii) As n increases without bound, we have a limiting sum:

$\lim_{n \to \infty} p_{n} = 0.2 \times \frac{1}{1 - 0.7} = \frac{2}{3}$

Drsoccerball · Oct 1, 2016

Re: HSC 2016 3U Marathon

KingOfActing said:
Nah I remember it starts with "Let a + b + c = S" but I can't be bothered looking it up.

Next question

$KingOfActing has done an amazing bottle flip. If KoA manages a successful bottle flip, the adrenaline rush makes him very focused, and so the probability that he'll manage to do it again in the next try is $0.9$. If he didn't manage to land a successful bottle flip, he feels deflated and the probability that he will be able to manage the next try it is $0.2\\\\$i) Let $ p_n$ be the probability of him successfully landing his $n$th bottle flip, with $p_0 = 1$. Find a relationship between $p_{n+1}$ and $p_n$.$\\\\$ii) Find the probability that flips his $100$th bottle successfully in exact form.$\\\\$iii) Suppose KingOfActing is a mythical being that will never die and will flip until time itself is over. What is the probability that he flips the last bottle in existence successfully?$

Have a video of KoA doing these bottle flips:

trecex1 · Oct 1, 2016

Re: HSC 2016 3U Marathon

KingOfActing said:
$\begin{align*}iii)\; \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} &=(a+b+c)\left(\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} \right ) - 3\\&=\frac{1}{2}\left((a+b) + (b + c) + (c + a)\right)\left(\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} \right ) - 3\\&\geq\frac{1}{2}\times 9 - 3\\&=\frac{3}{2}\end{align*}$

I sincerely remember there being a method to do this one by showing that equality like in the AM-GM inequality was when a=b=c, but I just can't remember how.

Nice, I did it the same way as well.

Prove
$\frac{a_1+a_2+a_3}{3}\geq \sqrt[3]{a_1a_2a_3}$

Using 3U inequalities only . (You can't assume Cauchy's Inequality obviously)

KingOfActing · Oct 1, 2016

Re: HSC 2016 3U Marathon

$Lemma: $ \frac{a_1+a_2+a_3+a_4}{4} \geq \sqrt[4]{a_1a_2a_3a_4}\qquad\qquad(*)\\\\$Proof:$

$\\\begin{align*} \frac{a_1+a_2 +a_3+a_4}{4} &\geq\frac{2\sqrt{a_1a_2}+2\sqrt{a_3a_4}}{4}\\&=\frac{\sqrt{a_1a_2}+\sqrt{a_3a_4}}{2}\\&\geq\sqrt{\sqrt{a_1a_2}\sqrt{a_3a_4}}\\&=\sqrt[4]{a_1a_2a_3a_4}\end{align*}$

$\\$Now, let $a_4 = \frac{a_1+a_2+a_3}{3}\\\begin{align*}\frac{a_1+a_2+a_3}{3} &\geq \left (a_1a_2a_3\left( \frac{a_1+a_2+a_3}{3}\right) \right )^\frac{1}{4}\\\left(\frac{a_1+a_2+a_3}{3} \right )^3 &\geq a_1a_2a_3\\\frac{a_1+a_2+a_3}{3} &\geq \sqrt[3]{a_1a_2a_3}\end{align*}$

...Or you know, just let n = 3 in the generalised case from before.

Paradoxica · Oct 1, 2016

Re: HSC 2016 3U Marathon

trecex1 said:
Nice, I did it the same way as well.

Prove
$\frac{a_1+a_2+a_3}{3}\geq \sqrt[3]{a_1a_2a_3}$

Using 3U inequalities only . (You can't assume Cauchy's Inequality obviously)

The inequality is only true for positive real numbers a₁,a₂,a₃

The inequality is obviously equivalent to:

$$\noindent$ (\sqrt[3]{a_1}+\sqrt[3]{a_2}+\sqrt[3]{a_3})\left(\frac{(\sqrt[3]{a_1}-\sqrt[3]{a_2})^2}{2}+\frac{(\sqrt[3]{a_2}-\sqrt[3]{a_3})^2}{2}+\frac{(\sqrt[3]{a_3}-\sqrt[3]{a_1})^2}{2}\right) \ge 0$

Which is trivially true.

davidgoes4wce · Oct 3, 2016

Re: HSC 2016 3U Marathon

$Cambridge Year 11 Exercise 9A Q 7 (a)$

$Use the distance formula to find the square of the distance between the point P(x,y) and each of the points A(0,4), B(0,-4) and C(6,3)$

In the back of the book there are no answers.

$$ My thinking is we have to find $ d_{PA}= \sqrt{(y-4)^2+x^2}$

or

$d_{PA}^2= {(y-4)^2+x^2}$

KingOfActing · Oct 3, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
$Cambridge Year 11 Exercise 9A Q 7 (a)$

$Use the distance formula to find the square of the distance between the point P(x,y) and each of the points A(0,4), B(0,-4) and C(6,3)$

In the back of the book there are no answers.

$$ My thinking is we have to find $ d_{PA}= \sqrt{(y-4)^2+x^2}$

or

$d_{PA}^2= {(y-4)^2+x^2}$

$PA^2 = x^2 + (y-4)^2\\PB^2 = x^2 + (y+4)^2\\PC^2 = (x-6)^2 + (y-3)^2$

davidgoes4wce · Oct 4, 2016

Re: HSC 2016 3U Marathon

This is from the 2014 HSC Extension 1 exam

Going through Q 13 (c), my initial thinking was the radius was constant from OQ and came across many different/varying solutions.

Would working out like this be sufficient for 3 marks? (this is the easiest understanding for me)

davidgoes4wce · Oct 4, 2016

Re: HSC 2016 3U Marathon

^^^ With that Q13 (c), I think I have seen 4 slightly different methods explained so far.

davidgoes4wce · Oct 4, 2016

Re: HSC 2016 3U Marathon

Was going through this binomial expansion tonight (from the 2013 HSC Exam):

Im aware that the coefficient

$^{{2n-k} \choose 0} =^{{2n-k} \choose {2n-k}}$

Aware that the 2nd term is equal to the 2nd last term , so on.

I guess my way of expanding that binomial coefficient

for

${x^{2n-k}{(x+2)^{2n-k}}}= \binom{2n-k}{0} \ x^{2n-k} \ 2^0 \ x^{2n-k}+\binom{2n-k}{1} \ x^{2n-k-1} \ 2^1 \ x^{2n-k} + \binom{2n-k}{2} \ x^{2n-k-2} \ 2^2 \ x^{2n-k} +.................+\binom{2n-k}{2n-k} \ x^0 \ 2^{2n-k} \ x^{2n-k}$

I understand the solution has technically in a way reversed the order of the binomial expansion, something along the lines of :

$x^{2n-k}(2+x)^{2n-k}$

davidgoes4wce · Oct 4, 2016

Re: HSC 2016 3U Marathon

I also say now studying Binomials is the most exciting part of Maths Extension 1. (even though I don't necessarily see the link it has in real life applications from day to day).

Just think studying the patterns of Binomials is time consuming but the rewards you get from doing it makes it all the more worth it.

Paradoxica · Oct 4, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
I also say now studying Binomials is the most exciting part of Maths Extension 1. (even though I don't necessarily see the link it has in real life applications from day to day).

Just think studying the patterns of Binomials is time consuming but the rewards you get from doing it makes it all the more worth it.

Personally, the topic is a rung on a ladder that is part of a bigger picture...

penapplepineapplepen · Oct 4, 2016

Re: HSC 2016 3U Marathon

Can yous pliez anser this kweshtion yousing mathimatecal inducktion?

‘The journey, not the arrival, matters.’
Discuss this statement, focusing on how composers of texts represent the concept of the
journey.
In your answer, refer to your prescribed text, ONE text from the prescribed stimulus booklet,
Journeys, and at least ONE other related text of your own choosing.

zanking yous veri mutch

PPAP

DatAtarLyfe · Oct 5, 2016

Re: HSC 2016 3U Marathon

penapplepineapplepen said:
Can yous pliez anser this kweshtion yousing mathimatecal inducktion?

‘The journey, not the arrival, matters.’
Discuss this statement, focusing on how composers of texts represent the concept of the
journey.
In your answer, refer to your prescribed text, ONE text from the prescribed stimulus booklet,
Journeys, and at least ONE other related text of your own choosing.

zanking yous veri mutch

PPAP

By inspection, the statement does not hold true as "journeys" does not satisfy the domain of "AOS 2015-2020" for all real texts
Q.E.D

penapplepineapplepen · Oct 5, 2016

Re: HSC 2016 3U Marathon

DatAtarLyfe said:
By inspection, the statement does not hold true as "journeys" does not satisfy the domain of "AOS 2015-2020" for all real texts
Q.E.D

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youse halfwittt

do youse even understand nglish

DatAtarLyfe · Oct 5, 2016

Re: HSC 2016 3U Marathon

penapplepineapplepen said:
okoi dokie mayte

whi do yous youse inspecktion when i kleerly askd yous to youse mathimatemicalmolism inducktion

youse halfwittt

do youse even understand nglish

Inspection > MI

Paradoxica · Oct 5, 2016

Re: HSC 2016 3U Marathon

DatAtarLyfe said:
Inspection > MI

It's trivial, obvious and clearly true. There is no need for proof

#proofbyintimidation

#proofbyassertion

penapplepineapplepen · Oct 5, 2016

Re: HSC 2016 3U Marathon

whi youse have two b so meen two mee.

I asck won kwestion tat iz it

kan youse pliez anser it.

i giive youse halfwitts a hinty; my mathimethicalmolism teeacher told mee and mees frendz two youse an essaye tekniqqe

pliez just give moar then won lyin anser

youse halfwitt buffhed

Paradoxica · Oct 5, 2016

Re: HSC 2016 3U Marathon

The acute triangle ABC has sides a>b>c, where A is opposite a and likewise for B and C.

A square is inscribed with vertices on the edges of the triangle.

It is given that only three such squares exist, and that one side of each square is concurrent with the side of a triangle.

Suppose the side length of the squares are p,q,r (with any correspondence you want), and the altitudes of the triangle are a',b',c', which connect the vertices A,B,C with the sides a,b,c respectively.

i) Find expressions for each of the three squares in terms of all the variables described above.

ii) Determine which of the three squares has the largest area.

iii) Hence, or otherwise, find the smallest possible area of any such triangle ABC which encloses a square of unit area.

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