HSC 2016 MX2 Marathon ADVANCED (archive) (4 Viewers)

InteGrand · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

RealiseNothing said:
That's the same instinct I had whilst thinking about why we get two different probabilities. Might look into it a bit more since it seems way more counter-intuitive than even the Monty Hall problem.

When I did this question I was assuming the temperatures were all fixed. Feel free to do it however you want though, I only have the answer (confirmed by other sources too so all good) using the assumption that the temperatures are fixed beforehand however.

And by the way, since the man (I think it was a man, forgot the name) is acting rationally, the answer depends on the man's knowledge of the whether the days' temperatures are fixed or not, right? If he thinks they are fixed, it matters not to the answer whether they actually are fixed or not, and similarly for if he thinks they are not fixed. So the key thing is not whether they are fixed, but what the man thinks/knows about them.

glittergal96 · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

With the probability question:

I think you need to be far more rigorous about how these days temperatures are "randomly" selected in order to have a well-posed question. Typically this involves a distribution.

Saying things like "the temperature is equally likely to be above the previous days temperature as below" does not really fit into any mathematical framework for randomness that I know of. Certainly no probability distribution that isn't trivial will satisfy this. (And our days are required to have distinct temperatures too!)

Without clearing this point up, the word "probability" is meaningless.

One way to do so is specify that the temperatures are selected uniformly from a given interval, say [0,1] for convenience. This makes the problem well-posed.

dan964 · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

I had a think about it. It is like having an infinite supply of numbered balls, with replenishment, where there is 4 draws.

Each day has a probability of 1/4 of being the hottest day; so does Day 1. That is simple to understand.
The 1/8 comes because of knowledge of the temperature of Day 1, meaning the other are thought of relative to the known i.e. Day 1.
So consider there are really only two probabilities; one that Day X (X from 2 to 4) is not hotter than Day 1 (i.e. Day 1 is hotter); or that it is.
Hence probability that Day 1 is hotter than Day X is P(E)=1/2.
Since the days are in an order: 1/2 * 1/2 * 1/2 = 1/8

fraserlawr · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

By using induction and contradiction, show that a^n + b^n ≠ c^n is true for all integers n>2 and a, b and c are positive integers

InteGrand · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

fraserlawr said:
By using induction and contradiction, show that a^n + b^n ≠ c^n is true for all integers n>2 and a, b and c are positive integers

There exists a truly marvellous proof that this post is too small to contain.

Paradoxica · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

fraserlawr said:
$By using induction and contradiction, show that $ a^n + b^n \neq c^n $ is true for all integers $n>2$ and $a, b$ and $c$ are positive integers$

$Um. If there were an elementary proof of Fermat's last theorem, surely someone would have discovered it in the 100+ years of head scratching.$

$$Find all polynomials $P(x)$ which satisfy $P(2) = 12, P(x^2) = x^2(x^2+1)P(x)$

glittergal96 · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
$Um. If there were an elementary proof of Fermat's last theorem, surely someone would have discovered it in the 100+ years of head scratching.$

$$Find all polynomials $P(x)$ which satisfy $P(2) = 12, P(x^2) = x^2(x^2+1)P(x)$

Comparing degrees, 2n=4+n, so all solutions must be quartic.

Take P=ax^4+bx^3+cx^2+dx+e and sub into the polynomial equation.

Considering the lowest order term on both sides we get d=e=0.

So
ax^6+bx^4+cx^2=(x^2+1)(ax^4+bx^3+cx^2)=ax^6+bx^5+(a+c)x^4+bx^3+cx^2

=> b=0.

=> a+c=0.

So P(x)=ax^2(x^2-1).

P(2)=12 => a=1.

So P(x)=x^2(x^2-1).

Carrotsticks · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

fraserlawr said:
By using induction and contradiction, show that a^n + b^n ≠ c^n is true for all integers n>2 and a, b and c are positive integers

There do exist integers for n=2 sastisfying the above (Pythagorean Triads). This establishes our base case.

The rest of the proof is left as an exercise for the reader.

InteGrand · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Carrotsticks said:
There do exist integers for n=2 sastisfying the above (Pythagorean Triads). This establishes our base case.

The rest of the proof is left as an exercise for the reader.

The base case would be n = 3 lol. So the whole proof is an exercise for the reader.

Carrotsticks · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
The base case would be n = 3 lol. So the whole proof is an exercise for the reader.

If we prove it for n>2, I'm sure that includes n>3 =)

obliviousninja · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Carrotsticks said:
If we prove it for n>2, I'm sure that includes n>3 =)

Rektd

Paradoxica · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

glittergal96 said:
Comparing degrees, 2n=4+n, so all solutions must be quartic.

Take P=ax^4+bx^3+cx^2+dx+e and sub into the polynomial equation.

Considering the lowest order term on both sides we get d=e=0.

So
ax^6+bx^4+cx^2=(x^2+1)(ax^4+bx^3+cx^2)=ax^6+bx^5+(a+c)x^4+bx^3+cx^2

=> b=0.

=> a+c=0.

So P(x)=ax^2(x^2-1).

P(2)=12 => a=1.

So P(x)=x^2(x^2-1).

Could have gotten there faster by showing P(x) must be even.

InteGrand · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

obliviousninja said:
Rektd

The case of n = 2 is irrelevant.

glittergal96 · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
Could have gotten there faster by showing P(x) must be even.

Yep, of course. You can also very quickly show that P vanishes to second order at x=0, cutting down your work even more.

Just putting pen to paper immediately and solving it the way I did took way less time than it did to type it up though, so I didn't bother looking for a more elegant way.

Paradoxica · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$$a and b are rational numbers satisfying the equation $a^3 + 4a^2 b = 4a^2 + b^4$
$Prove $\sqrt{a} -1$ is a rational square.$

sy37 · Oct 31, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
$$a and b are rational numbers satisfying the equation $a^3 + 4a^2 b = 4a^2 + b^4$
$Prove $\sqrt{a} -1$ is a rational square.$

Where did you get this exactly?

kawaiipotato · Oct 31, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
$$a and b are rational numbers satisfying the equation $a^3 + 4a^2 b = 4a^2 + b^4$
$Prove $\sqrt{a} -1$ is a rational square.$

$a^3 + 4a^2 b = 4a^2 + b^4$

$a^2 + 4ab = 4a + \frac{b^{4}}{a} ($dividing by a since a $ \neq 0 )$

$a^2 + 4ab + 4b^2 = 4a + \frac{b^{4}}{a} + 4b^2$

$(a+2b)^2 = \frac{4a^2 + b^4 + 4ab^2 }{a}$

$(a+2b)^2 = \frac{(b^2 + 2a)^2 }{a}$

$a = \frac{(b^2 + 2a)^2 }{(a+2b)^2 }$

$\sqrt{a} = \frac{b^2 + 2a}{a+2b}$

$Since b and a are rational numbers then $ \sqrt{a} $ is rational$

$\sqrt{a} = \frac{a - 2b + b^2 }{a+2b}$

$a+2b = \frac{b^2 + 2a}{\sqrt{a}}$

$\sqrt{a}= \frac{2a - 2 \sqrt{a} b + b^2}{a}$

$\sqrt{a} - 1 = \frac{a - 2 \sqrt{a} b + b^2}{a}$

$= \frac{ ( \sqrt{a} - b)^2 }{a}$

$\therefore \sqrt{\sqrt{a} - 1} = \frac{\sqrt{a} - b}{\sqrt{a}}$

$and since RHS is rational then LHS is rational$

Paradoxica · Oct 31, 2015

Re: HSC 2016 4U Marathon - Advanced Level

sy37 said:
Where did you get this exactly?

Wouldn't you like to know.

sy37 · Nov 1, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
Wouldn't you like to know.

Polish MO Final 2010 - 2nd problem.

Paradoxica · Nov 1, 2015

Re: HSC 2016 4U Marathon - Advanced Level

sy37 said:
Polish MO Final 2010 - 2nd problem.

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