HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

RealiseNothing · Aug 27, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
For the probability question I got 3361/6561.

I just enumerated the colourings that don't contain a non-degenerate red triangle as follows:

(Colourings with less than 3 red points) + (Colourings with 3 red collinear points).

For Wilson's theorem, you could show that:

(~) If p is prime, then every nonzero a in {0,1,2,...,p-1} has a multiplicative inverse b (mod p), and that 1 and (p-1) are the only numbers that are self-inverse.

Once you have (~), the product (p-1)! cancels out in pairs except for the first and last term. This leaves a residue of 1*(p-1)=-1 mod p. This proves one direction of Wilson's and the converse is trivial.

To make this an extension two solution, I think we could just recast (~) as

For any prime p and any positive integer a there is a positive integer b such that ab-1 is divisible by p. Moreover, if b' is another such integer, then b-b' must be divisible by p. Also, if a-b is divisible by p and ab-1 is divisible by p, then either a-1 or a+1 is divisible by p.

Knowing about division/remainders is in the course, so this modified statement leads to a proof without mentioning anything out of syllabus I think.

Yer my approach was showing that:

$(p-1)! = (p-1)(a_0p+1)(a_1p+1)(a_2p+1)...(a_kp+1)$

Which becomes a polynomial of the form:

$b_np^n+b_{n-1}p^{n-1} + ... + b_1p - 1 = (p-1)!$

Take the constant term to the other side and we are done. Note $n,k$ were just arbitrarily chosen to represent whatever value it's meant to be as it really doesn't matter the amount of terms present.

The only thing you would really have to prove is the first line where we factorised $(p-1)!$ is only legit for primes. Though this is easily done by considering that if $a$ is an integer such that $1 \leq a < p$ then $aq$ is a multiple of $p$ IFF $p=q$ . Then it is forced that all preceding integers of $p$ must be a remainder of $\frac{aj}{p}$ for $j<p$ otherwise you end up stuck in a pattern of remainders that will never be 0 (hence contradiction).

So there exist some $b<p$ such that $ab = pt+1$ and the uniqueness of this is trivial.

RealiseNothing · Aug 27, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
For the probability question I got 3361/6561.

I just enumerated the colourings that don't contain a non-degenerate red triangle as follows:

(Colourings with less than 3 red points) + (Colourings with 3 red collinear points).

Exactly what I got which is equal to 51.2%

$1 - \frac{\binom{8}{0}2^8 + \binom{8}{1}2^7 + \binom{8}{2}2^6 + 4(2^5)}{3^8}$

glittergal96 · Aug 27, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
Exactly what I got which is equal to 51.2%

$1 - \frac{\binom{8}{0}2^8 + \binom{8}{1}2^7 + \binom{8}{2}2^6 + 4}{3^8}$

You probably just typo'd this, but you need to multiply that 4 by 2^5

.

RealiseNothing · Aug 27, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
You probably just typo'd this, but you need to multiply that 4 by 2^5 .

Cheers, forgot to add that on the end there.

sepseminar · Aug 27, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Yep, that's the way to do it. You consider the four cases in which you don't get a red triangle, which are 1) No reds, 2) One red, 3) Two reds, 4) Three reds in a line.

$Prove that $\\ |\mathrm{cis}(\alpha)-\mathrm{cis}(\beta)|\cdot|\mathrm{cis}(\gamma)-\mathrm{cis}(\delta)|+|\mathrm{cis}(\beta)-\mathrm{cis}(\gamma)|\cdot|\mathrm{cis}(\alpha)-\mathrm{cis}(\delta)|=|\mathrm{cis}(\alpha)-\mathrm{cis}(\gamma)|\cdot|\mathrm{cis}(\beta)-\mathrm{cis}(\delta)| $ where $ 0<\alpha<\beta<\gamma<\delta<2\pi\\$Interpret this result geometrically.$

RealiseNothing · Aug 27, 2014

Re: HSC 2014 4U Marathon - Advanced Level

sepseminar said:
Yep, that's the way to do it. You consider the four cases in which you don't get a red triangle, which are 1) No reds, 2) One red, 3) Two reds, 4) Three reds in a line.

$Prove that $\\ |\mathrm{cis}(\alpha)-\mathrm{cis}(\beta)|\cdot|\mathrm{cis}(\gamma)-\mathrm{cis}(\delta)|+|\mathrm{cis}(\beta)-\mathrm{cis}(\gamma)|\cdot|\mathrm{cis}(\alpha)-\mathrm{cis}(\delta)|=|\mathrm{cis}(\alpha)-\mathrm{cis}(\gamma)|\cdot|\mathrm{cis}(\beta)-\mathrm{cis}(\delta)| $ where $ 0<\alpha<\beta<\gamma<\delta<2\pi\\$Interpret this result geometrically.$

This is equivalent to proving Ptolemy's Theorem, but I'll leave this one for a current 4U student to have a crack at.

glittergal96 · Aug 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

sepseminar said:
Yep, that's the way to do it. You consider the four cases in which you don't get a red triangle, which are 1) No reds, 2) One red, 3) Two reds, 4) Three reds in a line.

$Prove that $\\ |\mathrm{cis}(\alpha)-\mathrm{cis}(\beta)|\cdot|\mathrm{cis}(\gamma)-\mathrm{cis}(\delta)|+|\mathrm{cis}(\beta)-\mathrm{cis}(\gamma)|\cdot|\mathrm{cis}(\alpha)-\mathrm{cis}(\delta)|=|\mathrm{cis}(\alpha)-\mathrm{cis}(\gamma)|\cdot|\mathrm{cis}(\beta)-\mathrm{cis}(\delta)| $ where $ 0<\alpha<\beta<\gamma<\delta<2\pi\\$Interpret this result geometrically.$

Call the points a,b,c,d in order of increasing argument in [0,2pi).

By the triangle inequality:

|(a-b)(c-d)|+|(a-d)(b-c)| >= |(a-c)(b-d)|.

with equality if and only if the quotient of the two LHS terms (without the modulus signs) is a positive real.

We have

arg{((a-b)(c-d))/((a-d)(b-c))} = arg (a-b)/(a-d) + arg (c-d)/(c-b)+pi = 0 (mod 2pi)

as opposite angles in a cyclic quadrilateral are supplementary. So we are done.

Edit: Oh, and the geometric conclusion is that the sum of the products of opposite side lengths is at least the product of diagonal lengths in any quadrilateral Q, with equality if and only if Q is cyclic. (This is more than the question asked for, but it was the same effort so why not.)

dunjaaa · Aug 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Nice question! Yeah, it's Ptolemy's theorem in circle geometry translated into complex numbers vector representation

RealiseNothing · Aug 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

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

abecina · Sep 5, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
$\sum_{k=1}^n \frac{n-k+1}{\sin^2(\frac{k\pi}{2n+1})}$

OK, I'm going to need a hint

RealiseNothing · Sep 5, 2014

Re: HSC 2014 4U Marathon - Advanced Level

abecina said:
OK, I'm going to need a hint

2002 HSC question 8 is similar.

Sy123 · Sep 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
2002 HSC question 8 is similar.

The problem comes with the k in the numerator

Can you post your solution?

glittergal96 · Sep 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Yeah, I couldn't solve that one either. (Although I could solve several slightly tweaked versions of it using similar techniques as in that HSC question.)

Could you either post a solution, or at least the expression we should be aiming for please?

glittergal96 · Sep 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

?

RealiseNothing · Sep 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
?

I did the question like a year ago, I'll try get the solution out again and post it when I'm bothered. Too much uni stuff atm.

glittergal96 · Sep 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
I did the question like a year ago, I'll try get the solution out again and post it when I'm bothered. Too much uni stuff atm.

Where is it from though? I want to try more ways of doing it myself but only if I know for sure that it does have a nice closed form.

RealiseNothing · Sep 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
Where is it from though? I want to try more ways of doing it myself but only if I know for sure that it does have a nice closed form.

I summed it myself. Tested it for the first few values of n and my closed expression worked so I think it was right. What did you get so far?

glittergal96 · Sep 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
I summed it myself. Tested it for the first few values of n and my closed expression worked so I think it was right. What did you get so far?

Nothing really for the sum as it's written (if you change the summations upper limit to 2n it is easy though), and I tried things along the same lines as in that HSC question.

I can't see any reason why there should be a closed expression for that sum, but maybe I'm missing something obvious. Anyway, I just asked because I didn't want to spend any more time on it if it can't be done. (I prefer "prove that X" questions for this reason lol).

TL1998 · Sep 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 · Sep 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
View attachment 30903

By scaling, we may as well assume the large circle has radius 1.

So the area of the shape AQBP (which is the union of two circular segments) is $\pi/3.$

Triangles AOP, BOP are isosceles so have equal base angles equal to $(\pi-\theta)/2$ . So $\angle APB = \pi-\theta.$

Also, the cosine rule in triangle AOP tells us that $AP^2=2(1-\cos(\theta))$ .

So, putting this all together using the formula for the areas of circular segments ( $r^2(\phi-\sin(\phi))/2$ ), we get:

$\frac{\pi}{3}= (\theta-\sin(\theta)\cos(\theta)) + (1-\cos(\theta))(\pi-\theta-\sin(\theta))=\pi-sin(\theta)-(\pi-\theta)\cos(\theta).$

Moving the theta dependent terms to one side and the constants to the other completes the proof.

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