HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

SpiralFlex

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Re: HSC 2012 Marathon :)

I'll pop by to make some questions later.
 

GoldyOrNugget

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Re: HSC 2012 Marathon :)

the derivative cos(x) + 1 is always >= 0 so f(x) is always increasing, so it has at most one root. f(0.4) < 0, f(0.6) > 0, so f(x) = 0 for some 0.4<x<0.6. Then just halve the interval.

How do you do math notation?
 

Sy123

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Re: HSC 2012 Marathon :)

the derivative cos(x) + 1 is always >= 0 so f(x) is always increasing, so it has at most one root. f(0.4) < 0, f(0.6) > 0, so f(x) = 0 for some 0.4<x<0.6. Then just halve the interval.

How do you do math notation?
To do math notation you type normal latex code within the tags [.tex] and [./tex] (without the dots)

Also for the hexagon question I havent quite got the answer yet but Im quite close, I feel like I need to somehow prove that BP=BC since everything falls nicely from there. I havent done it yet but you could consider constructing a circle around the regular hexagon (its justified since it is regular), and then applying circle properties or something.
 

Sy123

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Re: HSC 2012 Marathon :)

Another question I made up:

 

kenkap

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Re: HSC 2012 Marathon :)

CAN somebody please check my aproach for Sy123's question part (v)

i subbed the equation of normal in equation of the semicircle (function) making a quadratic equation in x...since it only touches the curve, i made the discriminant=0 but it gave me answers of theta=0,90,180 degrees which seems weird??

where am i going wrong?? is my method correct??
 

Sy123

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Re: HSC 2012 Marathon :)

CAN somebody please check my aproach for Sy123's question part (v)

i subbed the equation of normal in equation of the semicircle (function) making a quadratic equation in x...since it only touches the curve, i made the discriminant=0 but it gave me answers of theta=0,90,180 degrees which seems weird??

where am i going wrong?? is my method correct??
Nope, your method is correct, I did it and it yielded my results (45, 135)

Did you end up with:

 

GoldyOrNugget

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Re: HSC 2012 Marathon :)

For Sy's projectile problem (this is quite a convoluted method):

 
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Sy123

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Re: HSC 2012 Marathon :)



Yes I cant think of any other good questions from other topics, Id rather not have easy questions but sharing questions that are nice and moderate to hard.
 
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GoldyOrNugget

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Re: HSC 2012 Marathon :)

I think that question is missing information. Right now it's just a scenario with two unrelated projectiles. Do they hit each other or something?
 

Sy123

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Re: HSC 2012 Marathon :)

I think that question is missing information. Right now it's just a scenario with two unrelated projectiles. Do they hit each other or something?
Oh of course, sorry for that, yes the particles collide at maximum height. Sorry for the inconvinience
 

GoldyOrNugget

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Re: HSC 2012 Marathon :)

I found that one much easier. Now back to this stupid hexagon >.<



EDIT: derp, got the terms the wrong way around. fixed now.
 

Sy123

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Re: HSC 2012 Marathon :)

I found that one much easier. Now back to this stupid hexagon >.<



EDIT: derp, got the terms the wrong way around. fixed now.
Very nice way of looking at it, I did it by treating (t+T) as a single parameter and found equation of projectile B then solved from there.

And I too must get back to the hexagon, post a solution if you find it out :)
 

GoldyOrNugget

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Re: HSC 2012 Marathon :)

If you represent the hexagon's vertices as points on a plane and do lots of coordinate geometry, you obtain the following expression for the angle:



And if you type that into a calculator, it does indeed give 80o. I don't know how to simplify it manually though.
 

Carrotsticks

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Re: HSC 2012 Marathon :)

Okay worked out the hexagon Q.

HINT: Construct AC and it follows that angle BAC = angle BCA = 30 degrees (isosceles triangle). Work from there and focus purely on ABCP.
 

seanieg89

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Re: HSC 2012 Marathon :)

Yep, exactly. The question itself is a pretty standard induction...but it raises other interesting questions about this construction.
 

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