HSC 2012 MX1 Marathon #1 (archive) (1 Viewer)

[bleakarcher]

Re: 2012 HSC MX1 Marathon

^ 4/3

 

[SpiralFlex]

Re: 2012 HSC MX1 Marathon

New question,

Let

New limits: [1,-1]



LaTeX errored.

Use substitution.



My turn! Let's turn things up!

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

lol nobody has yet tried my question ='(

 

[nightweaver066]

Re: 2012 HSC MX1 Marathon

Let

New limits: [1,-1]



LaTeX errored.

Use substitution.



My turn! Let's turn things up!

Or,

 

[hayabusaboston]

Re: 2012 HSC MX1 Marathon

lol nobody has yet tried my question ='(

what is your question Mr Carrotsticks?

 

[nightweaver066]

Re: 2012 HSC MX1 Marathon

Some trigonometry.

 

[SpiralFlex]

Re: 2012 HSC MX1 Marathon

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

lol nobody has yet tried my question ='(

I posted it earlier but I don't think I made it look like a question.

It is known that all triangles can be circumscribed within a circle.

Prove that the smallest possible ratio Circle (the one circumscribing the triangle) : Triangle occurs when the triangle is an equilateral triangle.

 

[bleakarcher]

Re: 2012 HSC MX1 Marathon

Carrotsticks, I am still unsure on how you determined the nature of the roots i.e. which ones were real and which ones were complex?

 

[hayabusaboston]

Re: 2012 HSC MX1 Marathon

I posted it earlier but I don't think I made it look like a question.

ah man this is bulldust, im bloody latex incompetent. Someone teach me please?

 

[nightweaver066]

Re: 2012 HSC MX1 Marathon

The case where







The case where









I like this question. haha

 

[SpiralFlex]

Re: 2012 HSC MX1 Marathon

I posted it earlier but I don't think I made it look like a question.

Max/min?

 

[SpiralFlex]

Re: 2012 HSC MX1 Marathon

The case where







The case where









I like this question. haha

Correct. It is meant to be semi-annoying.

 

[Examine]

Re: 2012 HSC MX1 Marathon

My knowledge of Maths is only Year 10 but I took a try and ended with + or - 9, and + or - 3.

EDIT: Realised I forgot the inequalities :L

 

[nightweaver066]

Re: 2012 HSC MX1 Marathon

Correct. It is meant to be semi-annoying.

Perhaps if it was triple absolute values, then maybe.

 

[Examine]

Re: 2012 HSC MX1 Marathon

I've got two questions:
1. How do you format the working out on the computer like that?
2. So if you take the absolute out of x when it is |x|< 9 it turns to -9 < x < 9?

 

[deswa1]

Re: 2012 HSC MX1 Marathon

Disclaimer: This is from a 4U paper but 3U students can do it.

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Carrotsticks, I am still unsure on how you determined the nature of the roots i.e. which ones were real and which ones were complex?

A cubic always has 3 roots. The question is, how many of them are real?

Case #1: All 3 roots are real and distinct (meaning no two roots are the same)

Case #2: All 3 roots are real, but not all are distinct (This is the case when there's a double root and a single root)

Case #3: 1 root is real, but the others are complex (We MUST have at least 2 real roots because all the coefficients are real, meaning that the conjugate root theorem kicks in)

Now, we found that the sum of roots (squared) is equal to 1.

Suppose we have Case #1:

All 3 roots must be within the interval 0 < x < 1 in order for the sum of them (squared) to be equal to 1. The roots could be positive, but when squared it will be positive. They could ALL be negative. However when squared, they will still be positive! The moment one of the roots exceeds 1, the sum of the squares of the other 2 roots must be negative to 'balance' it. However, it's impossible for the other 2 roots to be real, since the only numbers that give negative values when squared... are complex numbers!

Suppose we have Case #2:

Again, both roots must be within the interval 0 < x < 1 (which in short is denoted by (0,1). If I use a square bracket [0,1], it means it is inclusive).

The trivial case would be to have a double root at 0, and the single root at 1 (or visa versa). However by simply substituting x=0 and 1 into the equation, we can see that roots occur at neither.

This means that the double root AND the single root has to FIT within the interval (0,1), which is very unlikely. Reason being, (0,1) is a small interval, and the coefficients of the polynomial look too 'normal' for anything weird like that to occur.

This leaves Case #3, which is how I made my deduction (and it is the most probabilistic solution).

 

[nightweaver066]

Re: 2012 HSC MX1 Marathon

I've got two questions:
1. How do you format the working out on the computer like that?
2. So if you take the absolute out of x when it is |x|< 9 it turns to -9 < x < 9?

1. Using Latex. You use this to write up equations:
www.codecogs.com/latex/eqneditor.php

Then you input them in to this forum by inserting whatever you typed in to these brackets

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Max/min?

I don't know =)

In the HSC, they do not always tell you what method to you use, do they?

Remember that the HSC is changing. Learn to apply your knowledge now.

Use whatever techniques are at your arsenal to try to prove it.