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

seanieg89 · Apr 20, 2014

Re: HSC 2014 4U Marathon - Advanced Level

jyu said:
Thank you all for the thoughts.
I certainly have put time in analysing the idea.
The idea is not baseless. Well known geometric facts support that, such as the question I posted a+b+c>=abc in a unit circle; area of a rectangle; volume of a cuboid.
What I was not able to do was using the right name for the type of expressions generated. This allowed seanieg89 to construct the counterexamples.
'Well most homogeneous cyclic polynomials won't even have global extrema'- This is true unless there is a constraint placed on a homogeneous
cyclic function of the variables in the original polynomial, which I also forgot to mention previously.
For example, (x-2y)^2(2x-y)^2 is a homogeneous cyclic polynomial, now plus the constraint xy=4, the minimum is 16 when x=y >0.

Apologies for the delay. My family loves to drag on these holiday events.

Of course the idea is not baseless...but each of your successive guesses is. You successively strengthen your assumptions and weaken your conclusion hoping that the new claim is true, but each time you have not posted any evidence for why you think this slightly weaker claim is true, given that the previous claim wasn't.

Nor have you posted any proofs or attempts thereof.

I am not a machine for testing your claims one after another. To improve at mathematics you should learn to bash your head at things for longer...especially when you don't know if things are true or false...they are the best practice of all.

The promised counterexample is just a slightly modified version of the previous one, to make it homogeneous:

$P=\frac{(x-2y)^2(y-2x)^2}{x^4+y^4}.$

"Well most homogeneous cyclic polynomials won't even have global extrema- This is true unless there is a constraint placed on a homogeneous
cyclic function of the variables in the original polynomial, which I also forgot to mention previously.
For example, (x-2y)^2(2x-y)^2 is a homogeneous cyclic polynomial, now plus the constraint xy=4, the minimum is 16 when x=y >0."

Sure, before I thought you were only making claims about unconstrained optimisation. Then you introduced the positivity constraint, which as I pointed out does not change things.

Neither do such polynomial equation constraints.

Counterexample:
Minimise x+y subject to the constraint x^2 + y^2 >= 1 and x >= 0 and y >= 0. The extremal points are then (1,0) and (0,1) contrary to your claim.

(Neither do such constraints always imply existence of extrema, unless (for example) the function is continuous and the restricted domain of our function ends up being compact.)

seanieg89 · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Carrotsticks said:
Generally, I wouldn't tweak any assumptions or hypotheses under risk of being overly specific to say a class of functions and then mistaking the property to being applicable over a broader spectrum (which looks like what happened above)

Perhaps a difference in style, I wouldn't dedicate too much effort in an attempt to seek out a disgustingly pathological counter-example (if by pathological you mean something that really tests the boundaries of the current assumptions a la Riemann's Pathological function versus the former definition of Riemann integrability). I personally think that is a waste of time because they can be very difficult to think of sometimes (the 'disgustingly pathological' counterexamples)

However, I do see some merit in spending a couple of minutes thinking about a quick counter-example before jumping into proofs.

The tweaking I was objecting to was more the tweaking outlined in my second paragraph. He posted a "solution" to a question citing a result that is untrue. He has then successively weakened it many times (but not enough to make it true), and keeps asking me to find counterexamples. It is like a student of yours resorting to guessing after discovering only the first half of his argument was valid.

I don't mean any particularly outlandish counterexample. I just meant "pathological" to mean weird enough not to satisfy the theorem. (So any counterexample at all really.) As jyu said, the nicest objects like rectangles of fixed perimeter satisfy his claim. As it happens, so does the inequality he originally posted a solution to. But without isolating what it is about this inequality that makes the claim true, he hasn't proven anything at all.

I think it just depends on the starting point. Before I try to resolve a statement as true or false, I think about it a bit and see whether I THINK it is true or false.
I have been almost certain that each of jyu's statement's has been incorrect, hence me heading directly to counterexample hunts.
He seems to be sure of the opposite, (enough to keep bouncing slightly modified versions at me), so it seems logical to me that he should seek a proof and at least post his ideas on this.

In the best case scenario he finds a proof, and in the worst case scenario, he hits stumbling blocks that lead HIM to a counterexample without having to ask me.

jyu · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
Apologies for the delay. My family loves to drag on these holiday events.

Of course the idea is not baseless...but each of your successive guesses is. You successively strengthen your assumptions and weaken your conclusion hoping that the new claim is true, but each time you have not posted any evidence for why you think this slightly weaker claim is true, given that the previous claim wasn't.

Nor have you posted any proofs or attempts thereof.

I am not a machine for testing your claims one after another. To improve at mathematics you should learn to bash your head at things for longer...especially when you don't know if things are true or false...they are the best practice of all.

The promised counterexample is just a slightly modified version of the previous one, to make it homogeneous:

$P=\frac{(x-2y)^2(y-2x)^2}{x^4+y^4}.$

"Well most homogeneous cyclic polynomials won't even have global extrema- This is true unless there is a constraint placed on a homogeneous
cyclic function of the variables in the original polynomial, which I also forgot to mention previously.
For example, (x-2y)^2(2x-y)^2 is a homogeneous cyclic polynomial, now plus the constraint xy=4, the minimum is 16 when x=y >0."

Sure, before I thought you were only making claims about unconstrained optimisation. Then you introduced the positivity constraint, which as I pointed out does not change things.

Neither do such polynomial equation constraints.

Counterexample:
Minimise x+y subject to the constraint x^2 + y^2 >= 0 and x >= 0 and y >= 0. The extremal points are then (1,0) and (0,1) contrary to your claim.

(Neither do such constraints always imply existence of extrema, unless (for example) the function is continuous and the restricted domain of our function ends up being compact.)

Thanks again for another 'counterexample'.
Is 0 the minimum of x+y when x=y=0?

seanieg89 · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

jyu said:
Thanks again for another 'counterexample'.
Is 0 the minimum of x+y when x=y=0?

Sorry, my constraint was supposed to be x^2 + y^2 >= 1, not 0.

jyu · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
Sorry, my constraint was supposed to be x^2 + y^2 >= 1, not 0.

But this revised constraint is excluding x^2+y^2<1. I was thinking of all x and y > 0 including 0 if defined as pointed out in Sy's.

seanieg89 · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

jyu said:
But this revised constraint is excluding x^2+y^2<1. I was thinking of all x and y > 0 including 0 if defined as pointed out in Sy's.

I have already provided counterexamples for problems where you have allowed any positive x and y, you then brought up constrained optimisation in post #3067 (via the constraint xy=4), so I found one for that too.

jyu · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Thank you. Something wrong with my last post. I totally agree with you.

seanieg89 · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

No worries.

jyu · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sorry That I am doing it again.
If x^2 + y^2 = 1 in your last example?
Have you got a counterexample for zero degree cyclic rational functions?
Thanks for your patience.
To make clear, I wasn't trying to make any claim at the start, purely to do Sy's question the quickest way.

seanieg89 · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

1. I would have to cook up a different counterexample for a constraint like x^2+y^2=1 as we would then necessarily get a maximum. The claim is still not true though.

But as I said before, I am not going to bother to provide any more counterexamples for you without evidence on your behalf. These statements are all false until proven true, given how similar they are to things I have proven to be false in the earlier posts in this exchange.

2. I already gave you a counterexample for degree zero cyclic rational functions in post #3068.

3. Well if you are using things you aren't sure about the truth of, it is not really a solution.

TL1998 · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

'Prove that the tangents at oppositve vertices of a cyclic quadrilateral intersect on the secant through the other two vertcies IF AND ONLY IF the two products of opposite sides of the cyclic quadrilateral are equal

Davo_01 · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
'Prove that the tangents at oppositve vertices of a cyclic quadrilateral intersect on the secant through the other two vertcies IF AND ONLY IF the two products of opposite sides of the cyclic quadrilateral are equal

I wonder who can do this

Refer to attached diagram:

Some results we can prove using similarity:

$\dfrac{AD}{DC}=\dfrac{FA}{FD}$

$\dfrac{BC}{AB}=\dfrac{FC}{FB}$

$FC \cdot FA = FD^2=FB^2$ $= FD \cdot FB$

$\therefore \dfrac{AD \cdot BC}{DC \cdot AB}=\dfrac{FC \cdot FA}{FD \cdot FB}=1$

$\therefore AD \cdot BC=DC \cdot AB$

Hence the product of opposite sides of a cyclic quadrilateral are equal when tangents at opposite vertices intersect on the secant through the other two vertices. (The converse is also true)

Note: If the above condition is not true, then one or more of the results proved using similarities cannot be true, implying A, C and F cannot be collinear, therefore the intersection can only occur if the result is true.

TL1998 · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
View attachment 30320 View attachment 30320

Refer to attached diagram:

Some results we can prove using similarity:

$\dfrac{AD}{DC}=\dfrac{FA}{FD}$

$\dfrac{BC}{AB}=\dfrac{FC}{FB}$

$FC \cdot FA = FD^2=FB^2$ $= FD \cdot FB$

$\therefore \dfrac{AD \cdot BC}{DC \cdot AB}=\dfrac{FC \cdot FA}{FD \cdot FB}=1$

$\therefore AD \cdot BC=DC \cdot AB$

Hence the product of opposite sides of a cyclic quadrilateral are equal when tangents at opposite vertices intersect on the secant through the other two vertices. (The converse is also true)

Note: If the above condition is not true, then one or more of the results proved using similarities cannot be true, implying A, C and F cannot be collinear, therefore the intersection can only occur if the result is true.

I was able to prove that if the intersection lies on the secant, the products would be equal.

But is the bolded part really enough to show the converse is true? I posted the question because i had difficulty with the converse.

Davo_01 · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
[/B]

I was able to prove that if the intersection lies on the secant, the products would be equal.

But is the bolded part really enough to show the converse is true? I posted the question because i had difficulty with the converse.

Thats pretty much what i did.

For the converse, I guess an alternate way of doing it is assuming the result holds, then proving A, C and F are collinear and if it isn't, then the intersection does not occur. Other than that, in my reasoning I try to explains that if the condition does not hold, then it must be due to one or more of the similarities we proved cannot be true and the only way it could be false if the points A, C and F are not collinear.

RealiseNothing · Apr 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Prove that the number $\frac{1}{n}$ will have at most $n-1$ decimal places before it repeats itself.

seanieg89 · Apr 22, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
Prove that the number $\frac{1}{n}$ will have at most $n-1$ decimal places before it repeats itself.

Haha that's a cute question. (Although I am not sure if you are are talking about a) the length of the block before the first copy of the repeating block, b) the length of the repeating block or c) the decimal at which the second copy of the repeating block begins). Here is my work anyway.

$By the pigeonhole principle, $\{10^k\}_{k=0}^{n-1}$ must contain at least two distinct elements $10^a < 10^b$ with equal remainders upon division by $n.$ We then have:$\\ \frac{10^b}{n}-\frac{10^a}{n}=m\in \mathbb{Z}.\\ \\ $But this just means that in a given repeating decimal expansion for $1/n$, if we kick the decimal point a times to the right or b times to the right, we get the same noninteger part. This means that the repeating block in this decimal expansion has size at most $b-a\leq n-1$ and the block prior to the repeating block has size at most $a<n-1$. The second copy of the repeating block begins at the $b\leq n-1$-th digit at the latest. Whatever the question actually is, I think we are done.$

seanieg89 · Apr 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$Let $P(x)$ be a polynomial with integer coefficients.\\ \\ Is it possible to find three distinct integers $a,b,c$ with $P(a)=b$, $P(b)=c$ and $P(c)=a$ ?\\ \\ Justify your answer.$

Sy123 · Apr 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
$Let $P(x)$ be a polynomial with integer coefficients.\\ \\ Is it possible to find three distinct integers $a,b,c$ with $P(a)=b$, $P(b)=c$ and $P(c)=a$ ?\\ \\ Justify your answer.$

$\\ $Using the theorem, that for integer polynomial$ \ P \ $and distinct integers$ \ a,b \ \ $then$ \ \ \frac{P(a) - P(b)}{a-b} \ \ $is an integer$$

$\\ $This is easily provable by showing that$ \\ \\ P(x) = \sum_0^n a_kx^k \ \Rightarrow \ P(a) - P(b) = \sum_1^n a_k (a^k - b^k) \\ $It is sufficient to show that$ \ \frac{a^k-b^k}{a-b} \ \ $is an integer$ \\ \\ \frac{a^k-b^k}{a-b} = b^{k-1} +b^{k-2}a + \dots + ba^{k-2} + a^{k-1} \in \mathbb{Z} \\ $Therefore the theorem is proven true$$

$\\ (1) \ \ \ \frac{P(a) - P(b)}{a-b} = \frac{b-c}{a-b} = k_1 \\ (2)\ \ \ \frac{P(b) - P(c)}{b-c} = \frac{c-a}{b-c} = k_2 \\ (3) \ \ \ \frac{P(c) - P(a)}{c-a} = \frac{a-b}{c-a} = k_3 \\ \\ (1) \times (2) \times (3) \Rightarrow \ k_1k_2k_3 = 1 \ \ \ $(for integers$ \ k_1,k_2,k_3)$ \\ \\ \therefore \ k_1 = k_2 = k_3 = 1 \\ \Rightarrow \ (b-c) = (a-b) \ , \ \ (c-a) = (b-c) \ , \ \ (a-b) = (c-a)$

$\\ \text{Without loss of generality, assume} \ \ a < b < c \\ \text{We see that} \ \ c = \frac{a+b}{2} \ , \ \ c = \frac{1}{2} b + \frac{1}{2}a < \frac{1}{2}b + \frac{1}{2}b = b \\ c < b \\ \\ \text{A contradiction!, therefore there is no distinct} \ a,b,c \ \text{so that the set of equations hold true}$

Sy123 · Apr 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Find$ \ \ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n + ik} \\ \\ $Where$ \ i = \sqrt{-1}$

seanieg89 · Apr 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Using the theorem, that for integer polynomial$ \ P \ $and distinct integers$ \ a,b \ \ $then$ \ \ \frac{P(a) - P(b)}{a-b} \ \ $is an integer$$

$\\ $This is easily provable by showing that$ \\ \\ P(x) = \sum_0^n a_kx^k \ \Rightarrow \ P(a) - P(b) = \sum_1^n a_k (a^k - b^k) \\ $It is sufficient to show that$ \ \frac{a^k-b^k}{a-b} \ \ $is an integer$ \\ \\ \frac{a^k-b^k}{a-b} = b^{k-1} +b^{k-2}a + \dots + ba^{k-2} + a^{k-1} \in \mathbb{Z} \\ $Therefore the theorem is proven true$$

$\\ (1) \ \ \ \frac{P(a) - P(b)}{a-b} = \frac{b-c}{a-b} = k_1 \\ (2)\ \ \ \frac{P(b) - P(c)}{b-c} = \frac{c-a}{b-c} = k_2 \\ (3) \ \ \ \frac{P(c) - P(a)}{c-a} = \frac{a-b}{c-a} = k_3 \\ \\ (1) \times (2) \times (3) \Rightarrow \ k_1k_2k_3 = 1 \ \ \ $(for integers$ \ k_1,k_2,k_3)$ \\ \\ \therefore \ k_1 = k_2 = k_3 = 1 \\ \Rightarrow \ (b-c) = (a-b) \ , \ \ (c-a) = (b-c) \ , \ \ (a-b) = (c-a)$

$\\ \text{Without loss of generality, assume} \ \ a < b < c \\ \text{We see that} \ \ c = \frac{a+b}{2} \ , \ \ c = \frac{1}{2} b + \frac{1}{2}a < \frac{1}{2}b + \frac{1}{2}b = b \\ c < b \\ \\ \text{A contradiction!, therefore there is no distinct} \ a,b,c \ \text{so that the set of equations hold true}$

. Right idea and pretty close to being done, but we can have a pair of the k's being -1 instead of all of them having to be 1.

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