HSC 2013 MX2 Marathon (archive) (1 Viewer)

InfinityBlade · Nov 9, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
It is true for all real a and b (assuming a =/= b since there is a strict inequality there), since (a-b)^2 > 0 for all real a and b where a=/=b.

Not all real values. Equality holds when a,b=0, a,b=1 both of which include a=b. There should be an equal sign too.

Carrotsticks · Nov 9, 2012

Re: HSC 2013 4U Marathon

InfinityBlade said:
Not all real values. Equality holds when a,b=0, a,b=1 both of which include a=b. There should be an equal sign too.

Yes, that is exactly why I said in my quote "assuming a=/=b" twice.

Carrotsticks said:
It is true for all real a and b (assuming a =/= b since there is a strict inequality there), since (a-b)^2 > 0 for all real a and b where a=/=b.

seanieg89 · Nov 9, 2012

Re: HSC 2013 4U Marathon

Fun little question.

a) A whole grade consisting of several (>1) classes participates in a mathematics competition. In each class the average female score is higher than the average male score. Is it possible for the average male score over the entire grade to be higher than the average female score over the entire grade?

b) What if all classes are the same size?

c) What if each class has an equal gender ratio?

Justify all answers with either a proof or a counterexample.

(Nothing too difficult, perhaps more apt for the 2U or 3U forum but I think its cool.)

RealiseNothing · Nov 9, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Fun little question.

a) A whole grade consisting of several (>1) classes participates in a mathematics competition. In each class the average female score is higher than the average male score. Is it possible for the average male score over the entire grade to be higher than the average female score over the entire grade?

b) What if all classes are the same size?

c) What if each class has an equal gender ratio?

Justify all answers with either a proof or a counterexample.

(Nothing too difficult, perhaps more apt for the 2U or 3U forum but I think its cool.)

I am going to answer the parts of this question in seperate posts as I solve them:

a) Yes, it is possible for the average male score to be higher than the average female score over the entire grade.

This is quite easy to see actually. The average male/female score of each class has a certain "weighting". The more students of one gender there is, the stronger the weighting of the class average. For example, a class consisting of 2 boys and 2 girls will have a far less affect on the entire cohort compared to a class of 20 boys and 20 girls, and hence the larger class has a higher "weighting".

Therefore it should be easy to see why this situation is possible. All we have to do is let the male "weighting" of one class absolutely dominate the female "weighting", and let the rest of the remaining classes have a very small "weighting". As a general counterexample:

Let the first class have $n$ amount of boys, and $k$ girls. Give the restriction $1 \leq k \leq n$

Suggest now that there are $j$ other classes, all containing $p$ amount of boys and girls.

All we need to do is make the variables $p~k~j$ as small as we like in realistic terms, and make $n$ as big as we like, and the "weighting" of $n$ will reach a certain point where it dominates the combined weighting of the other variables, and hence the average male score over the entire grade will be higher than the average female score.

RealiseNothing · Nov 9, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Fun little question.

a) A whole grade consisting of several (>1) classes participates in a mathematics competition. In each class the average female score is higher than the average male score. Is it possible for the average male score over the entire grade to be higher than the average female score over the entire grade?

b) What if all classes are the same size?

c) What if each class has an equal gender ratio?

Justify all answers with either a proof or a counterexample.

(Nothing too difficult, perhaps more apt for the 2U or 3U forum but I think its cool.)

b) Very similar to part (a). Consider there are $n$ classes with $x$ amount of people in each class. If we let the male "weighting" of one class dominate so much over the females with a minute difference in averages, then go on to let the remaining classes be rubbish, there comes a point where the male will once again outscore the females.

For example, let $(i)B$ denote the amount of boys in the class, and $(j)G$ denote the amount of girls in the class.

If we have $n$ classes, with $x$ amount of people, we can produce the following scenario:

For $n-1$ classes, with $x=20$ (20 people in each class), we let the gender ratio be $(19)B ~to~ (1)G$ . We give all the boys a score of 99%, and the one girl a score of 100%. The female>male class average still holds here.

Now for the remaining class, swap the ratio around and have $(1)B ~to~ (19)G$ . Give all the girls 1%, and the one boy 0%. The female>male class average also holds true here.

However the total cohort's average is:

Boys; $\frac{99\times19(n-1)}{19(n-1)+1}$

Girls; $\frac{100(n-1)+19}{19+(n-1)}$

Hence we can see, that for small values of $n$ , the overall male average will be higher than the overall female average.

RealiseNothing · Nov 9, 2012

Re: HSC 2013 4U Marathon

I'll do part (c) later, but my intuition is telling me it's not possible.

seanieg89 · Nov 9, 2012

Re: HSC 2013 4U Marathon

Good. You will have no trouble with c). The part of this problem I consider 'interesting' is that people's intuitions may fool them for a and b.

In a vaguely similar style I made up the following question based on something a friend asked me during the US election:

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?

Sy123 · Nov 9, 2012

Re: HSC 2013 4U Marathon

$The roots of the equation $ z^n=r $ are plotted on the Argand diagram. The points are then connected by straight lines$

$i) What polygon is formed with the connection of all the roots?$

$$ii) By considering the arguments of each of the complex numbers (or otherwise). Show that the area of this polygon is$ \\ \\ A=\frac{nr^2}{2}\sin \frac{2\pi}{n}$

$iii) Explain what happens as n approaches infinity$

$$iv) By employing a clever substitution. Using parts ii and iii, prove the area of the circle is $ \pi r^2$

deswa · Nov 9, 2012

Re: HSC 2013 4U Marathon

@Sy- actually you can- I'll give a quick counter example. Consider 2 classes.

Class 1: Female scores are (10). Male scores are (9,9,9). Average female score is 10, average male score is 9.
Class 2: Female scores are (2,2,2,2). Male scores are (1). Average female score is 2, average male score is 1.

All together, average female score is (10+2+2+2+2)/5=3.6
Average male score is (9+9+9+1)/4 = 7

Sy123 · Nov 9, 2012

Re: HSC 2013 4U Marathon

deswa said:
@Sy- actually you can- I'll give a quick counter example. Consider 2 classes.

Class 1: Female scores are (10). Male scores are (9,9,9). Average female score is 10, average male score is 9.
Class 2: Female scores are (2,2,2,2). Male scores are (1). Average female score is 2, average male score is 1.

All together, average female score is (10+2+2+2+2)/5=3.6
Average male score is (9+9+9+1)/4 = 7

The stupid flaw with my idea is that I counted the averages of the classes as a whole one unit when calculating total average. So for example average of class 1 for females is 90 males 80, average for class 2 females 95 males 85

Then total average for females is (90+95)/2 and (80+95)/2 for males. Which of course is absolutely wrong and you decided to post before I quickly deleted.

twinklegal19 · Nov 9, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
An easier one:

$\text{If } z $ is a complex number on the Argand diagram$ \\ \\ \text{Describe the locus } \frac{1}{z}+\frac{1}{\bar{z}}=1$

(remember to post a question after you answer)

$\\\frac{1}{z}+\frac{1}{\overline{z}}=1\\\\\overline{z}+z=z\overline{z}\\\\$let $z=x+iy\\\\x+iy+x-iy=x^2+y^2\\\\2x=x^2+y^2\\\\x^2-2x+1+y^2=1\\\\(x-1)^2+y^2=1\\\\$therefore the locus of z is a circle with unit radius and centre (1,0)$

RealiseNothing · Nov 9, 2012

Re: HSC 2013 4U Marathon

$\sum_{k=1}^{2012} i^k$

Sy123 · Nov 9, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
$\\\frac{1}{z}+\frac{1}{\overline{z}}=1\\\\\overline{z}+z=z\overline{z}\\\\$let $z=x+iy\\\\x+iy+x-iy=x^2+y^2\\\\2x=x^2+y^2\\\\x^2-2x+1+y^2=1\\\\(x-1)^2)+y^2=0\\\\$therefore the locus of z is a circle with unit radius and centre (1,0)$

Not quite

iBibah · Nov 9, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
$\sum_{k=1}^{2012} i^k$

0

Basically just doing the sum of i^1 to i^4 it equals 0. It then repeats, so the sum to 2012 would also be zero.

EDIT:

$i+i^{2}+i^{3}+i^{4}=i-1-i+1=0\\$Therefore every four terms the sum is zero. Hence at the 2012th term, the sum will be zero.$$

Sy123 · Nov 9, 2012

Re: HSC 2013 4U Marathon

I am hoping people have learnt Binomial Theorem from Extension 1:

$$Expand $ (1+i)^{4n} \\ \\ $Show that $ \binom{4n}{0}-\binom{4n}{2}+\binom{4n}{4}-\binom{4n}{6}+...+\binom{4n}{4n} = (-4)^n$

twinklegal19 · Nov 11, 2012

Sy123 said:
Not quite

I still haven't figured it out yet

is it the "mindless algebra"? lol

bleakarcher · Nov 11, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
I still haven't figured it out yet is it the "mindless algebra"? lol

Notice the point (0,0) representing z=0 lies on the circle lz-1l=1, initially we were given (1/z)+(1/conjugate(z))=1 and so z=/=0.

Sy123 · Nov 11, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
I still haven't figured it out yet is it the "mindless algebra"? lol

Yes it is the 'mindless algebra', think carefully about what z can be

SpiralFlex · Nov 11, 2012

Re: HSC 2013 4U Marathon

Guys are you up to graphs/conics yet?

bleakarcher · Nov 11, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
Guys are you up to graphs/conics yet?

Well yeah pretty much. My class has already finished Complex Numbers and almost finished Graphs.

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