HSC 2017 MX2 Marathon (archive) (1 Viewer)

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davidgoes4wce

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Re: HSC 2017 MX2 Marathon

Not sure if its me or the book error but I had a complex number question

(6+i)(a+bi)=2

My solution for



My book has a different solution to that.
 

Green Yoda

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Re: HSC 2017 MX2 Marathon

Not sure if its me or the book error but I had a complex number question

(6+i)(a+bi)=2

My solution for



My book has a different solution to that.
I also got the same answer as you.
 

He-Mann

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Re: HSC 2017 MX2 Marathon

The book's answer is right, you can check it by squaring it to get 5 + i.
 

pikachu975

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Re: HSC 2017 MX2 Marathon

I spent 5 minutes on this and couldn't solve it.





x^2 - y^2 + 2xyi = 5+i

Comparing real and imaginary
2xy = 1
y = 1/2x

x^2 - y^2 = 5
x^2 - 1/4x^2 = 5
4x^4 - 20x^2 - 1 = 0
x^2 = (20 +- sqrt(20^2 - 4(4)(-1))/2(4)
x^2 = (20+- sqrt(416))/8
x^2 = (20 +- 4sqrt(26))/8
x^2 = (5 +- sqrt(26))/2

But x is real so take the positive case only
x = +- sqrt[(5+sqrt26)/2]
 

jathu123

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Re: HSC 2017 MX2 Marathon



apologies if the question isn't clear (hence my two examples)
 
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si2136

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Re: HSC 2017 MX2 Marathon

Is there a formula for finding the sum of all points (angles where it's spiky) on an n pointed star? For example, for a 5 pointed star, it is 180 degrees
 

1729

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Re: HSC 2017 MX2 Marathon

Is there a formula for finding the sum of all points (angles where it's spiky) on an n pointed star? For example, for a 5 pointed star, it is 180 degrees
For higher n, there becomes multiple ways to construct the star, each with their own angle sum. ie. the points of the star may create an n-gon, or another star, as shown for n = 7, 11. Note in each case, the first star creates an n-gon by joining every second vertex, while the others create other stars.



 
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Sy123

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Re: HSC 2017 MX2 Marathon

This may not be in the spirit of the question, but if you had something different in mind with a more geometric proof, I'd like to see it. Let be the centre of circle . Consider the kite , and apply the cosine rule to the side from both and . Letting, . We obtain that, . A similar expression is obtained for . From this we obtain,

. As we have that is monotonic increasing, and so is maximised if and only if is maximised. Using the appropriate formulae we obtain,

, through calculus or the AM-GM inequality we obtain a maxima when that is unique and so on. However this sends suggesting that .
 

Sy123

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Re: HSC 2017 MX2 Marathon







 
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pikachu975

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Re: HSC 2017 MX2 Marathon

In how many ways can seven identical cats be put into three identical pens so that all of the pens are occupied? You must state reasoning. (2 marks)

Why can we not use stars and bars? I.e. 9C2
 

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seanieg89

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Re: HSC 2017 MX2 Marathon

This may not be in the spirit of the question, but if you had something different in mind with a more geometric proof, I'd like to see it. Let be the centre of circle . Consider the kite , and apply the cosine rule to the side from both and . Letting, . We obtain that, . A similar expression is obtained for . From this we obtain,

. As we have that is monotonic increasing, and so is maximised if and only if is maximised. Using the appropriate formulae we obtain,

, through calculus or the AM-GM inequality we obtain a maxima when that is unique and so on. However this sends suggesting that .
This is what I had in mind.

On the topic of a more geometric solution, I suspect there is a way to kill it quicker using nonsyllabus techniques (inversive geometry perhaps), but it is not immediately apparent to me, I might have a crack later this week.
 
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