HSC 2014 MX2 Marathon ADVANCED (archive) (2 Viewers)

FrankXie · Nov 10, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Chlee1998 said:
These problems belong in another thread, which is just called the 4u marathon. anyways, a^3 + b^3 can be factoed to (a+b)(a^2-ab + b^2) =2

But from AM GM, a^2 + b^2 > 2ab > ab (for a,b>0). Hence the second factor is positive.
Thus (a+b) multiplied by a positive number equals to 2, meaning that a+b is less than 2.

Are you sure? Your reasoning was quite wrong. (a+b) times a positive number=2, thus you deduce a+b less than 2?

FrankXie · Nov 10, 2014

Re: HSC 2014 4U Marathon - Advanced Level

counterexamples are everywhere, a third times (a+b)=2, then a+b=6 which is greater than 2

Sy123 · Nov 10, 2014

Re: HSC 2014 4U Marathon - Advanced Level

FrankXie said:
Try this one:
Given that $a>0, b>0$ and $a^3+b^3=2$. Without using calculus prove that $a+b\leq2$.

$\\ 2 = (a^3+b^3) = (a+b)(a^2-ab+b^2) = (a+b)((a+b)^2 - 3ab) \\ = (a+b)^3 + 3(a+b) (-ab) \geq (a+b)^3 + 3(a+b) \frac{-(a+b)^2}{4} = \frac{1}{4}(a+b)^3 \\ \\ 2 \geq \frac{1}{4}(a+b)^3 \Rightarrow \ 8 \geq (a+b)^3 \Rightarrow \ 2 \geq a+b$

FrankXie · Nov 10, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ 2 = (a^3+b^3) = (a+b)(a^2-ab+b^2) = (a+b)((a+b)^2 - 3ab) \\ = (a+b)^3 + 3(a+b) (-ab) \geq (a+b)^3 + 3(a+b) \frac{-(a+b)^2}{4} = \frac{1}{4}(a+b)^3 \\ \\ 2 \geq \frac{1}{4}(a+b)^3 \Rightarrow \ 8 \geq (a+b)^3 \Rightarrow \ 2 \geq a+b$

Nice job!

Chlee1998 · Nov 10, 2014

Re: HSC 2014 4U Marathon - Advanced Level

FrankXie said:
Are you sure? Your reasoning was quite wrong. (a+b) times a positive number=2, thus you deduce a+b less than 2?

Ah careless me. Sorry about that

Chlee1998 · Nov 10, 2014

Re: HSC 2014 4U Marathon - Advanced Level

H is the orthocenter of triangle ABC. Use AH as diameter to draw a circle which meets circle ABC at F. Prove that FH bisects BC

FrankXie · Nov 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Chlee1998 said:
H is the orthocenter of triangle ABC. Use AH as diameter to draw a circle which meets circle ABC at F. Prove that FH bisects BC

Well, a hard one, belongs to Olympiad question not HSC, although the proof is only a few lines.

Here is the outline of proof, to fully understand, you need draw a diagram.

Let feet of perpendicular on sides AB, AC are respectively D and E. Let FH meet the circle ABC again at G. I am going to prove BGCH is parallelogrm, thus FH bisects BC coz diagonals of a parallelogram bisects each other.

ABGF is cyclic, so $\angle{ABG}=90^\circ$ coz opposite angles are supplementary. Thus CH is parallel to BG.

ABGC is cyclic and $\angle{ABG}=90^\circ$ imply $\angle{ACG}=90^\circ$ . Thus BH is parallel to CG.

This proves BGCH is a parallelogram. Done!

Sy123 · Nov 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Find$ \ \sum_{n = 0}^{\infty} \frac{(-1)^n}{3n+1}$

FrankXie · Nov 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find$ \ \sum_{n = 0}^{\infty} \frac{(-1)^n}{3n+1}$

A method beyond HSC scope is:

Let
$f(x)=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{3n+1}$$ .
Find $f^\prime(x)$ , then integrate. Finally the desired sum is $-f(-1)$ .

Alert: 1. sum of (functional) series 2. termwise differentiation applies 3. $f(x)$ only converges for $-1<x<1$ , but the series in question is convergent, so I can sub $x=-1$

Chlee1998 · Nov 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

FrankXie said:
Well, a hard one, belongs to Olympiad question not HSC, although the proof is only a few lines.

Here is the outline of proof, to fully understand, you need draw a diagram.

Let feet of perpendicular on sides AB, AC are respectively D and E. Let FH meet the circle ABC again at G. I am going to prove BGCH is parallelogrm, thus FH bisects BC coz diagonals of a parallelogram bisects each other.

ABGF is cyclic, so $\angle{ABG}=90^\circ$ coz opposite angles are supplementary. Thus CH is parallel to BG.

ABGC is cyclic and $\angle{ABG}=90^\circ$ imply $\angle{ACG}=90^\circ$ . Thus BH is parallel to CG.

This proves BGCH is a parallelogram. Done!

yeah good I showed that GHC is equal to BGH using basic angle chase to prove that CH is parallel to BG though. Other than that, we pretty much did the same thing. Oh and yeah I don't think this is HSC

FrankXie · Nov 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Show that for any real numbers $a_i, b_i, i=1,2,\cdots,n$ , where $n$ is a positive integer,

$\left(\sum_{k=1}^na_kb_k\right)^2\leq\left(\sum_{k=1}^na_k^2\right)\left(\sum_{k=1}^nb_k^2\right)$

Carrotsticks · Nov 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Haha a bit much to expect students to pull a proof of CS out of thin air.

Most accessible proof is probably the classic one using the sum of positive definite quadratics and setting a negative discriminant.

FrankXie · Nov 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Carrotsticks said:
Haha a bit much to expect students to pull a proof of CS out of thin air.

Most accessible proof is probably the classic one using the sum of positive definite quadratics and setting a negative discriminant.

quite so. and I was trying to say similarly prove that

$\left[\int_a^bf(x)g(x)dx\right]^2\leq\int_a^bf^2(x)dx\cdot\int_a^bg^2(x)dx$

where $a<b$ .

glittergal96 · Nov 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

FrankXie said:
Show that for any real numbers $a_i, b_i, i=1,2,\cdots,n$ , where $n$ is a positive integer,

$\left(\sum_{k=1}^na_kb_k\right)^2\leq\left(\sum_{k=1}^na_k^2\right)\left(\sum_{k=1}^nb_k^2\right)$

You can do it using a similar trick to my last inequality proof (last page depending on your forum formatting), we can scale so $\sum a_k^2=\sum b_k^2 =1.$ (By replacing each $a_i$ with $ta_i$ , both sides of the inequality are scaled by the same factor of $t^2$ . Similarly with b.)

Then $(\sum a_kb_k)^2\leq (\frac{1}{2}\sum (a_k^2 + b_k^2))^2=1=RHS.$

Exactly the same trick works for the integral version.

Sy123 · Nov 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

FrankXie said:
Show that for any real numbers $a_i, b_i, i=1,2,\cdots,n$ , where $n$ is a positive integer,

$\left(\sum_{k=1}^na_kb_k\right)^2\leq\left(\sum_{k=1}^na_k^2\right)\left(\sum_{k=1}^nb_k^2\right)$

Carrotsticks said:
Haha a bit much to expect students to pull a proof of CS out of thin air.

Most accessible proof is probably the classic one using the sum of positive definite quadratics and setting a negative discriminant.

Its just glorified AM-GM (if I didn't make a mistake)

$\\ \sum_{i < j} (a_i^2 b_j^2 + a_j^2 b_i^2) \geq \sum_{i <j} 2a_i b_i a_j b_j \ \ $(AM-GM)$$

$\sum_{k=1}^n a_k^2b_k^2 + \sum_{i < j} (a_i^2 b_j^2 + a_j^2 b_i^2) \geq \sum_{i <j} 2a_i b_i a_j b_j + \sum_{k=1}^{n} a_k^2b_k^2$

$\Rightarrow \ \left(\sum_{k=1}^na_kb_k\right)^2\leq\left(\sum_{k=1}^na_k^2\right)\left(\sum_{k=1}^nb_k^2\right)$

$\\ $Set$ \ a_k = f \left(a + \frac{b-a}{n}\cdot k \right) \ , \ b_k = g \left(a + \frac{b-a}{n} \cdot k \right)$

$\\ \Rightarrow \ \left(\sum_{k=1}^n f\left(a + \frac{k(b-a)}{n} \right) g\left(a + \frac{k(b-a)}{n} \right)\right)^2\leq\left(\sum_{k=1}^n f^2\left(a + \frac{k(b-a)}{n} \right)\right)\left(\sum_{k=1}^ng^2\left(a + \frac{k(b-a)}{n} \right)\right)$

$\\ \Rightarrow \ \left( \frac{b-a}{n}\sum_{k=1}^n f\left(a + \frac{k(b-a)}{n} \right) g\left(a + \frac{k(b-a)}{n} \right)\right)^2\leq \\ \frac{b-a}{n}\left(\sum_{k=1}^n f^2\left(a + \frac{k(b-a)}{n} \right)\right) \cdot \frac{b-a}{n}\left(\sum_{k=1}^ng^2\left(a + \frac{k(b-a)}{n} \right)\right)$

$\\ $Take$ \ n \rightarrow \infty$

$\\ \Rightarrow \ \left( \int_a^b f(x) \cdot g(x) \ dx \right)^2 \leq \left(\int_a^b f(x) \ dx \right) \cdot \left( \int_a^b g(x) \ dx \right)$

FrankXie · Nov 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Its just glorified AM-GM (if I didn't make a mistake)

$\\ \sum_{i < j} (a_i^2 b_j^2 + a_j^2 b_i^2) \geq \sum_{i <j} 2a_i b_i a_j b_j \ \ $(AM-GM)$$

$\sum_{k=1}^n a_k^2b_k^2 + \sum_{i < j} (a_i^2 b_j^2 + a_j^2 b_i^2) \geq \sum_{i <j} 2a_i b_i a_j b_j + \sum_{k=1}^{n} a_k^2b_k^2$

$\Rightarrow \ \left(\sum_{k=1}^na_kb_k\right)^2\leq\left(\sum_{k=1}^na_k^2\right)\left(\sum_{k=1}^nb_k^2\right)$

$\\ $Set$ \ a_k = f \left(a + \frac{b-a}{n}\cdot k \right) \ , \ b_k = g \left(a + \frac{b-a}{n} \cdot k \right)$

$\\ \Rightarrow \ \left(\sum_{k=1}^n f\left(a + \frac{k(b-a)}{n} \right) g\left(a + \frac{k(b-a)}{n} \right)\right)^2\leq\left(\sum_{k=1}^n f^2\left(a + \frac{k(b-a)}{n} \right)\right)\left(\sum_{k=1}^ng^2\left(a + \frac{k(b-a)}{n} \right)\right)$

$\\ \Rightarrow \ \left( \frac{b-a}{n}\sum_{k=1}^n f\left(a + \frac{k(b-a)}{n} \right) g\left(a + \frac{k(b-a)}{n} \right)\right)^2\leq \\ \frac{b-a}{n}\left(\sum_{k=1}^n f^2\left(a + \frac{k(b-a)}{n} \right)\right) \cdot \frac{b-a}{n}\left(\sum_{k=1}^ng^2\left(a + \frac{k(b-a)}{n} \right)\right)$

$\\ $Take$ \ n \rightarrow \infty$

$\\ \Rightarrow \ \left( \int_a^b f(x) \cdot g(x) \ dx \right) \leq \left(\int_a^b f(x) \ dx \right) \cdot \left( \int_a^b g(x) \ dx \right)$

wow, very nice, especially the way you prove continuous C-S inequality by using discrete C-S. I would use positive definite quadratic and negative discrimint

FrankXie · Nov 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

dan964 said:
View attachment 31287
hmm can someone check the "legitness" while you do it...

so confusing a question. I believe $\theta$ should not be constant, is that correct? and launching from the equator, isn't the gravity pointing to the centre of the planet? if yes, it is not downward.

or maybe I totally understand the question wrong?

dan964 · Nov 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Amended question
View attachment Mechanics Question.pdf
View attachment Answers to Mechanics Question.pdf

RealiseNothing · Nov 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
I'm not sure if there is a HSC method for this question, but feel free to try.

Just in case there isn't, try this:

Prove that $(p-1)!+1 = kp$ for some integer $k$ if and only if $p$ is prime.

lol this was the last question of my MATH2988 exam today

Squar3root · Nov 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
lol this was the last question of my MATH2988 exam today

it was also the 5th question in the tutorial 1 problems

HSC 2014 MX2 Marathon ADVANCED (archive) (2 Viewers)

Active Member

Active Member

This too shall pass

Active Member

Member

Member

Active Member

This too shall pass

Active Member

Member

Active Member

Retired

Active Member

Active Member

This too shall pass

Active Member

Active Member

what

what is that?It is Cowpea

realest nigga

Users Who Are Viewing This Thread (Users: 0, Guests: 2)