HSC 2013 MX2 Marathon (archive) (2 Viewers)

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Sy123

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Re: HSC 2013 4U Marathon

So it is obvious how you WANT students to do this question so I won't write that solution up. But just a neat observation, the notion of a sequences "derivative" from http://community.boredofstudies.org/showthread.php?t=292775 can be used to solve this question kind of nicely. It also allows you to generalise the result to an arbitrary number of nested summations!

(The differentiation operator removes sums from the LHS in the same way that differentiation is inverse to integration.)
Oh no not this again haha, I still only half understand this concept.
 

seanieg89

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Re: HSC 2013 4U Marathon

Oh no not this again haha, I still only half understand this concept.
I think you probably use something equivalent to it whenever you prove something like this by induction, without really thinking about it in that way.
 

cutemouse

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Re: HSC 2013 4U Marathon

Find the Cartesian equation of the locus for all points z such that |z+i|+|z-i| = 2.

Hint: Think before performing "mindless algebra".
 

GoldyOrNugget

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Re: HSC 2013 4U Marathon

Each state in the country Bostralia has a set number of electoral votes (one can think of this as being the number of 'points' this state is worth. it is weighted by population.) The candidate for presidency who wins the most votes in any given state obtains all the points for that given state. The winner of the presidential election is the candidate who obtains the most points in total.

Assuming there are two candidates for presidency, what is the largest proportion of national votes that you can obtain whilst still losing?
We win some of the states and lose some of the states. Each state can be considered its own independent unit so the specific breakdown of voters inside each state determines only the winner for that state. So to maximise total voters, for each state we lost, we can assume we lost it by as small a margin as possible, and for each state we won, we can assume we got 100% of voters in that state.

Also, observe that 'losing' and 'winning' are in a sense distributive over states: losing the two states A and B is accomplished with the same amount of voters as losing the hypothetical combined state A+B. e.g. if you have a state of 10 and a state of 6, we lose if we have <5 voters in the first and <3 in the second. If we have a state of 10+6=16, we lose if we have <(5+3) voters in it. Furthermore, because people are integral quantities, the combined state will always allow for more people in a loss than its constituent states (in the above example, we lose with a maximum of 7 in the combined state, but a maximum of 4+2 in the constituent states). With a more formal argument, I believe you can treat any set of state wins and losses as two states: a state we won and a state we lost. The maximum proportion of voters would occur when for these two states, B has one more person than A, and we win A with 100% of voters and lose B with 49.99..% of voters. For sufficiently large populations, 75% of the population can vote for us and we can still lose.
 
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Find the Cartesian equation of the locus for all points z such that |z+i|+|z-i| = 2.

Hint: Think before performing "mindless algebra".
is z greater than/equal to -i and less than/equal to i? Or x=0, -1<y<1?

inb4igetthiswrongagain
 
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Sy123

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Re: HSC 2013 4U Marathon

is z greater than/equal to -i and less than/equal to i? Or x=0, -1<y<1?

inb4igetthiswrongagain
I havent done the question yet, but do you mean Im(z) >= -1 Im(z) <= 1

Because it is impossible to have complex inequalities.
 
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Re: HSC 2013 4U Marathon

I havent done the question yet, but do you mean Im(z) >= -1 Im(z) <= 1

Because it is impossible to have complex inequalities.
well I interpreted the question as the sum of the distances of z from i and -i is equal to 2, so Re(z)=0 and -1<Im(z)<1?
is it some sort of a weird ellipse? Forgive me if I'm wrong since I haven't started conics yet lol
 
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seanieg89

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Re: HSC 2013 4U Marathon

We win some of the states and lose some of the states. Each state can be considered its own independent unit so the specific breakdown of voters inside each state determines only the winner for that state. So to maximise total voters, for each state we lost, we can assume we lost it by as small a margin as possible, and for each state we won, we can assume we got 100% of voters in that state.

Also, observe that 'losing' and 'winning' are in a sense distributive over states: losing the two states A and B is accomplished with the same amount of voters as losing the hypothetical combined state A+B. e.g. if you have a state of 10 and a state of 6, we lose if we have <5 voters in the first and <3 in the second. If we have a state of 10+6=16, we lose if we have <(5+3) voters in it. Furthermore, because people are integral quantities, the combined state will always allow for more people in a loss than its constituent states (in the above example, we lose with a maximum of 7 in the combined state, but a maximum of 4+2 in the constituent states). With a more formal argument, I believe you can treat any set of state wins and losses as two states: a state we won and a state we lost. The maximum proportion of voters would occur when for these two states, B has one more person than A, and we win A with 100% of voters and lose B with 49.99..% of voters. For sufficiently large populations, 75% of the population can vote for us and we can still lose.
Correct answer, although I lost track of your logic halfway through. Will read more carefully later, it sounds like you knew what you were doing.
 

SpiralFlex

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Re: HSC 2013 4U Marathon

well I interpreted the question as the sum of the distances of z from i and -i is equal to 2, so Re(z)=0 and -1<Im(z)<1?
is it some sort of a weird ellipse? Forgive me if I'm wrong since I haven't started conics yet lol
It is an ellipse. The sum of the distances from the foci is constant namely twice that of the major axis of the ellipse.
 

cutemouse

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Re: HSC 2013 4U Marathon

If I'm not mistaken, it is not an elllipse, but a line...
 

deswa

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Re: HSC 2013 4U Marathon

Nah its defs an ellipse- you are taking the sum of the distances from two points and that equals a constant -> one definition of ellipse.

A line would be something like: |z-i|=|z+i|
 

seanieg89

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Re: HSC 2013 4U Marathon

Nah its defs an ellipse- you are taking the sum of the distances from two points and that equals a constant -> one definition of ellipse.

A line would be something like: |z-i|=|z+i|
Well if the constant is small enough the locus will just be a point, or even nonexistent.
 

SpiralFlex

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Re: HSC 2013 4U Marathon

I havent done the question yet, but do you mean Im(z) >= -1 Im(z) <= 1

Because it is impossible to have complex inequalities.
You are right in saying complex numbers have no order but Im(z) is a function to find the real number of the imaginary component.
 

SpiralFlex

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Re: HSC 2013 4U Marathon

I havent done the question yet, but do you mean Im(z) >= -1 Im(z) <= 1

Because it is impossible to have complex inequalities.
You are right in saying complex numbers have no order but Im(z) is a function to find the real number of the imaginary component.
 

cutemouse

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Re: HSC 2013 4U Marathon

Nah its defs an ellipse- you are taking the sum of the distances from two points and that equals a constant -> one definition of ellipse.

A line would be something like: |z-i|=|z+i|
In most cases yes.

But for my example, geometrically speaking, |z-i| + |z+i| is the sum of the distances from any point z to the points i and -i. So the minimum value its minimum value will be 2.

So, if I'm correct, then the locus of |z-i| + |z+i| = 2 will be the line x=0 for -1 <= y <= 1...
 
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Carrotsticks

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Re: HSC 2013 4U Marathon

In most cases yes.

But for my example, geometrically speaking, |z-i| + |z+i| is the sum of the distances from any point z to the points i and -i. So the minimum value its minimum value will be 2.

So, if I'm correct, then the locus of |z-i| + |z+i| = 2 will be the line x=0 for -1<y<1...
You are correct. Your locus there is a degenerate ellipse with foci at plus/minus i.

But your inequalities should not be strict. y can most certainly be equal to 1 and -1.

PS: Part of what you typed did not come out, and it is not coming out when I quote it either. It can be seen by clicking 'quote' on your post.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Suppose three complex numbers make an equilateral triangle on an argand diagram. Show that:

 
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