HSC 2014 MX2 Marathon ADVANCED (archive) (3 Viewers)

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FrankXie

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Re: HSC 2014 4U Marathon - Advanced Level

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

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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
 

Chlee1998

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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

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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
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 coz opposite angles are supplementary. Thus CH is parallel to BG.

ABGC is cyclic and imply . Thus BH is parallel to CG.

This proves BGCH is a parallelogram. Done!
 

Sy123

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Re: HSC 2014 4U Marathon - Advanced Level

 

FrankXie

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Re: HSC 2014 4U Marathon - Advanced Level

A method beyond HSC scope is:

Let
.
Find , then integrate. Finally the desired sum is .

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

Chlee1998

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Re: HSC 2014 4U Marathon - Advanced Level

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 coz opposite angles are supplementary. Thus CH is parallel to BG.

ABGC is cyclic and imply . 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

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Re: HSC 2014 4U Marathon - Advanced Level

Show that for any real numbers , where is a positive integer,

 

Carrotsticks

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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

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



where .
 

glittergal96

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Re: HSC 2014 4U Marathon - Advanced Level

Show that for any real numbers , where is a positive integer,

You can do it using a similar trick to my last inequality proof (last page depending on your forum formatting), we can scale so (By replacing each with , both sides of the inequality are scaled by the same factor of . Similarly with b.)

Then

Exactly the same trick works for the integral version.
 
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Sy123

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Re: HSC 2014 4U Marathon - Advanced Level

Show that for any real numbers , where is a positive integer,

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)















 
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FrankXie

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Re: HSC 2014 4U Marathon - Advanced Level

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















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

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Re: HSC 2014 4U Marathon - Advanced Level

View attachment 31287
hmm can someone check the "legitness" while you do it...
so confusing a question. I believe 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?
 

RealiseNothing

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Re: HSC 2014 4U Marathon - Advanced Level

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 for some integer if and only if is prime.
lol this was the last question of my MATH2988 exam today
 
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