HSC 2014 MX2 Marathon (archive) (1 Viewer)

Sy123 · Jun 10, 2014

Re: HSC 2014 4U Marathon

$\\ $The quadratic equation$ \ x^2+ax+b+1 = 0 \ $has integers$ \ a,b \ $and has integer roots$ \\ \\ $Show that$ \ a^2+b^2 \ $cannot be prime$$

aDimitri · Jun 11, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $The quadratic equation$ \ x^2+ax+b+1 = 0 \ $has integers$ \ a,b \ $and has integer roots$ \\ \\ $Show that$ \ a^2+b^2 \ $cannot be prime$$

$\text {In order to avoid confusion (because I used alpha and beta as roots), I'm going to call a and b, p and q respectively} \\ \text {Let the roots of the quadratic be given by } \alpha \text { and }\beta \\ \alpha + \beta = -p \\ p^{2} = (\alpha + \beta)^{2} \\ q + 1 = \alpha\beta \\ q^{2} = (\alpha\beta - 1)^{2} \\ p^{2} + q^{2} = (\alpha + \beta)^{2} + (\alpha\beta - 1)^{2} \\ = \alpha^{2}\beta^{2} + \alpha^{2} + \beta^{2} + 1 \\ (\alpha^{2} + 1)(\beta^{2} + 1) \\ \text {Since }\alpha\text { is an integer, } \alpha^{2} + 1 \text { is an integer greater than one, and a factor of } p^{2}+q^{2} \\ \text {Hence, }p^{2} + q^{2}\text { is not a prime.}$

Sy123 · Jun 20, 2014

Re: HSC 2014 4U Marathon

$\\ $Let$ \ \triangle ABC \ $be an equilateral triangle. Let$ \ Q \ $be any point on$ \ BC \ $and let$ \ P \ $be the meeting point of$ \ AQ \ $and the circumscribed circle of$ \ \triangle ABC \\ \\ $Prove that$ \ \frac{1}{PQ} = \frac{1}{PB} + \frac{1}{PC}$

Ikki · Jun 22, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Let$ \ \triangle ABC \ $be an equilateral triangle. Let$ \ Q \ $be any point on$ \ BC \ $and let$ \ P \ $be the meeting point of$ \ AQ \ $and the circumscribed circle of$ \ \triangle ABC \\ \\ $Prove that$ \ \frac{1}{PQ} = \frac{1}{PB} + \frac{1}{PC}$

We know from diagram (triangle inside the circle):
APxPQ=BPxPC

Working backwards from what to prove:
BPxPC=PQx(PB+PC)
BPxPC=PQxBC

If we equate what to prove and what we know:
BC=AP? But BC=AB
so, AB=AP which makes no sense. (Does not necessarily have to be true and shouldn't)

Sy123 · Jun 22, 2014

Re: HSC 2014 4U Marathon

Ikki said:
We know from diagram (triangle inside the circle):
APxPQ=BPxPC

Working backwards from what to prove:
BPxPC=PQx(PB+PC)
BPxPC=PQxBC

If we equate what to prove and what we know:
BC=AP? But BC=AB
so, AB=AP which makes no sense. (Does not necessarily have to be true and shouldn't)

You have made a mistake in your first line, Q is the intersection between the 2 chords, not P.

So from your proof it now is:

AQ x QP = BQ x QC

--

I can reveal my proof if you wish, or I can let you have another go

Davo_01 · Jun 24, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
You have made a mistake in your first line, Q is the intersection between the 2 chords, not P.

So from your proof it now is:

AQ x QP = BQ x QC

--

I can reveal my proof if you wish, or I can let you have another go

This is what i did:
$$ By considering similar triangles , we can prove$ \dfrac{BQ}{AB}=\dfrac{PQ}{PC}\; $and$\; \dfrac{QC}{AC}=\dfrac{PQ}{PB}$

$Adding these two equations:$

$\dfrac{BQ}{AB}+\dfrac{QC}{AC}=1=PQ\left(\dfrac{1}{PB}+\dfrac{1}{PC}\right)$

$\therefore \dfrac{1}{PQ}=\dfrac{1}{PB}+\dfrac{1}{PC}$

Davo_01 · Jun 24, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Let$ \ a,b \ $is real, and such that$ \ \ \frac{a+b}{a-b} + \frac{a-b}{a+b} = 6 \ $and$ \ |a| \neq |b| \\ \\ $Find the value of$ \\ \\ \frac{a^3+b^3}{a^3-b^3} + \frac{a^3-b^3}{a^3+b^3}$

I got 18/7?

Sy123 · Jun 26, 2014

Re: HSC 2014 4U Marathon

Davo_01 said:
This is what i did:
$$ By considering similar triangles , we can prove$ \dfrac{BQ}{AB}=\dfrac{PQ}{PC}\; $and$\; \dfrac{QC}{AC}=\dfrac{PQ}{PB}$

$Adding these two equations:$

$\dfrac{BQ}{AB}+\dfrac{QC}{AC}=1=PQ\left(\dfrac{1}{PB}+\dfrac{1}{PC}\right)$

$\therefore \dfrac{1}{PQ}=\dfrac{1}{PB}+\dfrac{1}{PC}$

Yep well done, this was my proof:

$\angle BPQ = \angle BCA$

$\angle CBA = \angle PCQ$

(By angle in same segment)

(Using $A_{ABC}$ as the area of triangle ABC)

$A_{PBC} = A_{PBQ} + A_{PQC}$

$\\ \frac{1}{2} PB \cdot PC \sin \frac{2\pi}{3} = \frac{1}{2} PB \cdot PQ \sin \frac{\pi}{3} + \frac{1}{2} PQ \cdot PC \sin \frac{\pi}{3}\\ \\ \Rightarrow \ PB \cdot PC = PB \cdot PQ + PQ \cdot PC \\ \\ \Rightarrow \ \frac{1}{PQ} = \frac{1}{PB} + \frac{1}{PC}$

Davo_01 said:
I got 18/7?

Yes I got this as well

Sy123 · Jun 26, 2014

Re: HSC 2014 4U Marathon

Here is a proof I devised for the AM-GM inequality of the third degree

$\\ $i) Show that$ \ a^3 + b^3 \geq ab(a+b) \ $for positive$ \ a,b \\ \\ $ii) Show that$ \ a^3 + b^3 + c^3 \geq \frac{1}{2} (a(b^2+c^2) + b(a^2+c^2) + c(a^2+b^2)) \\ \\ $iii) Hence deduce that$ \\ \\ \frac{x+y+z}{3} \geq \sqrt[3]{xyz} \ $for positive real$ x,y,z$

dunjaaa · Jul 13, 2014

Re: HSC 2014 4U Marathon

Screen shot 2014-07-13 at 9.10.08 PM.png

Nice proof!

Axio · Jul 21, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
View attachment 30662 Nice proof!

Hey dunjaa and Sy do one of you think you could explain part i of this answer to me? I understand the rest of it but can't quite follow the first step. Thanks.

RealiseNothing · Jul 21, 2014

Re: HSC 2014 4U Marathon

Axio said:
Hey dunjaa and Sy do one of you think you could explain part i of this answer to me? I understand the rest of it but can't quite follow the first step. Thanks.

His solution to part (i) is wrong.

Kurosaki · Jul 21, 2014

Re: HSC 2014 4U Marathon

For part I, equivalent proof is
$a^3+b^3-ab(a+b) \geq 0$
$LHS=(a+b)(a^2-ab+b^2)-ab(a+b) \\ =(a+b)(a^2-2ab+b^2) \\ =(a+b)(a-b)^2 \geq 0$ since a and b are positive.
We can then rearrange, to get
$a^3 + b^3 \geq ab(a+b) \\ \square$

kev- · Jul 21, 2014

Re: HSC 2014 4U Marathon

Hello, I need help
3 letters, A B C. How many words can be made with n letters where A occurs zero or an even number of times. (n is an even positive integer)

dunjaaa · Jul 21, 2014

Re: HSC 2014 4U Marathon

Woops, I kind of screwed up the transitivity lol. I typed the solution straight out and didn't realise that I couldn't do it via that method.

Sy123 · Jul 25, 2014

Re: HSC 2014 4U Marathon

dunjaaa · Jul 25, 2014

Re: HSC 2014 4U Marathon

My solution was ugly but I got OB^2=26cm. Yeah, I went brute force with angles, I'm trying to think of a more elegant geometrical solution.

Sy123 · Jul 25, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
My solution was ugly but I got OB^2=26cm. Yeah, I went brute force with angles, I'm trying to think of a more elegant geometrical solution.

Yea I did it via angles as well

Fizzy_Cyst · Jul 26, 2014

Re: HSC 2014 4U Marathon

aDimitri · Jul 26, 2014

Re: HSC 2014 4U Marathon

Fizzy_Cyst said:

$I = \frac{2\ln(\sec\alpha + \tan\alpha)}{3} + C \\ \text{Where } \sin\alpha = \frac{2}{3}\left(\cos\theta + \frac{1}{2}\right)$

cbb trying to simplify it LOL

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