HSC 2013 MX2 Marathon (archive) (1 Viewer)

seanieg89 · Jun 3, 2013

Re: HSC 2013 4U Marathon

braintic said:
How can one person have an average? Average number of friends per ... what?

The situation described is a probabilistic one, not a definite one. Friendship is not a certain relationship.

So average = average number of friends over all equally likely friendship arrangements. (there are 2^{nC2} of these)

braintic · Jun 3, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
The situation described is a probabilistic one, not a definite one. Friendship is not a certain relationship.

So average = average number of friends over all equally likely friendship arrangements. (there are 2^{nC2} of these)

So do you mean, if Bob was successively placed into a large number of such towns, what is is the average numbers of friends Bob would make per town?

seanieg89 · Jun 3, 2013

Re: HSC 2013 4U Marathon

braintic said:
So do you mean, if Bob was successively placed into a large number of such towns, what is is the average numbers of friends Bob would make per town?

Yep. Just like flipping a number of coins 10 times and asking what the average number of heads would be: the problem is imprecision in the English language. When formal probability/statistics definitions (such as expectation) are used these ambiguities disappear but yeah, that obviously is not high school material.

Hopefully people who want to have a crack at it understand what it is asking now.

Edit: And in the many-towns formulation of the question, b) can be phrased as:

"For each town we can calculate the friendcount of each of Bob's friends. We can then average these friendcounts to end up with a number X. Find the average value of X over all possible towns."

GoldyOrNugget · Jun 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
There are n people who live in a small town and the probability of any given pair of them being friends is 0.5 (friendship is of course mutual). One of these people is Bob.

a) What is the average number of friends Bob will have?

b) What is the average number of friends that friends of Bob will have?

c) Come up with some sort of heuristic way of explaining what a) and b) tell you.

I'll start this off.

a)

let A be the expected (average, whatever) number of friends: A = 0*P(no friends) + 1*P(1 friend) + 2*P(2 friends) + ... + (n-1)*P(n-1 friends)

P(no friends) = 0.5^(n-1)
P(1 friend) = (n-1)C1 * 0.5^1 * 0.5^(n-2) = (n-1)C1 * 0.5^(n-1)
P(2 friends) = (n-1)C2 * 0.5^2 * 0.5^(n-3) = (n-1)C2 * 0.5^(n-1)
...
P(k friends, 0 ≤ k < n) = (n-1)Ck * 0.5^(n-1)

$kP(k \text{ friends}) = k\binom{n-1}{k}\left(\frac{1}{2}\right)^{n-1}$

$\mathbb{E}[A] = \sum_{k=1}^{n-1} kP(k \text{ friends}) = \frac{1}{2^{n-1}} \left(\sum_{k=1}^{n-1} k\binom{n-1}{k}} \right) = \frac{1}{2^{n-1}} \left(\sum_{k=1}^{n-1} \binom{n-1}{k}} + \sum_{k=2}^{n-1} (k-1)\binom{n-1}{k}} \right) = 1 + \sum_{k=2}^{n-1} (k-1)\binom{n-1}{k}} \right) = \frac{n-1}{2}$

The expansion/simplification proved to be tricky and I almost drowned in a sea of latex. Maple came to the rescue. Someone else should do it properly, but I'm pretty sure it involves repeatedly using the fact that $\sum_{k=0}^n \binom{n}{k}=2^n$

c) I've seen this before, presented as a paradox. it's pretty cool. i won't spoil it further.

seanieg89 · Jun 4, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
I'll start this off.

a)

let A be the expected (average, whatever) number of friends: A = 0*P(no friends) + 1*P(1 friend) + 2*P(2 friends) + ... + (n-1)*P(n-1 friends)

P(no friends) = 0.5^(n-1)
P(1 friend) = (n-1)C1 * 0.5^1 * 0.5^(n-2) = (n-1)C1 * 0.5^(n-1)
P(2 friends) = (n-1)C2 * 0.5^2 * 0.5^(n-3) = (n-1)C2 * 0.5^(n-1)
...
P(k friends, 0 ≤ k < n) = (n-1)Ck * 0.5^(n-1)

$kP(k \text{ friends}) = k\binom{n-1}{k}\left(\frac{1}{2}\right)^{n-1}$

$\mathbb{E}[A] = \sum_{k=1}^{n-1} kP(k \text{ friends}) = \frac{1}{2^{n-1}} \left(\sum_{k=1}^{n-1} k\binom{n-1}{k}} \right) = \frac{1}{2^{n-1}} \left(\sum_{k=1}^{n-1} \binom{n-1}{k}} + \sum_{k=2}^{n-1} (k-1)\binom{n-1}{k}} \right) = 1 + \sum_{k=2}^{n-1} (k-1)\binom{n-1}{k}} \right) = \frac{n-1}{2}$

The expansion/simplification proved to be tricky and I almost drowned in a sea of latex. Maple came to the rescue. Someone else should do it properly, but I'm pretty sure it involves repeatedly using the fact that $\sum_{k=0}^n \binom{n}{k}=2^n$

c) I've seen this before, presented as a paradox. it's pretty cool. i won't spoil it further.

Nice.

As an alternate solution to a) note that the problem is invariant under the {friendship}<--->{non-friendship} symmetry, so if x is the avg number of friends of Bob we must have x = (n - 1) - x.

GoldyOrNugget · Jun 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Nice.

As an alternate solution to a) note that the problem is invariant under the {friendship}<--->{non-friendship} symmetry, so if x is the avg number of friends of Bob we must have x = (n - 1) - x.

so much more elegant than mine >.<

RealiseNothing · Jun 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
There are n people who live in a small town and the probability of any given pair of them being friends is 0.5 (friendship is of course mutual). One of these people is Bob.

a) What is the average number of friends Bob will have?

b) What is the average number of friends that friends of Bob will have?

c) Come up with some sort of heuristic way of explaining what a) and b) tell you.

a) $\frac{n-1}{2}$ as the probability of some one being his friend is 50/50, and so the average should sort itself out to be half the people in the town (besides himself)?

b) Bob's friends are already friends with bob, and there is a 50/50 chance they will be friends with the other $n-2$ people in the town, and so the average amount of friends they have is $1+\frac{n-2}{2} = \frac{n}{2}$

c) I guess the thing the problem is trying to point out is that Bob is an arbitrary person, so why are the two probabilities different? If so, this is because in part a) Bob was not guaranteed any friends, whilst in part b) they were guaranteed atleast one friend by already being friends with Bob, which will alter their average amount of friends to be a bit higher.

Sy123 · Jun 6, 2013

Re: HSC 2013 4U Marathon

$$A projectile is fired in a medium of no air resistance, and lands on an inclined plane of angle$ \ \ \alpha$

$It is fired with some velocity, at$ \ \ \theta \ \ $to the horizontal, the projectile is fired from the start of the inclined plane$

$The projectile lands on the inclined plane, of distance$ \ \ l \ \ $to the origin$

$Find$ \ \ \theta \ \ $in terms of alpha, such that$ \ \ l \ \ $is maximised$

Sy123 · Jun 7, 2013

Re: HSC 2013 4U Marathon

$Prove that$

$1+ \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{n^2} < 2 - \frac{1}{n}$

study1234 · Jun 8, 2013

Re: HSC 2013 4U Marathon

How do you do this projectile motion question? It's from Patel Ex 8B question 7.

Sy123 · Jun 8, 2013

Re: HSC 2013 4U Marathon

study1234 said:
View attachment 28202

How do you do this projectile motion question? It's from Patel Ex 8B question 7.

Deriving the general equations of motion

$x= Ut \cos \theta$

$y = -\frac{1}{2}gt^2 + Ut \sin \theta + h$

Their equation reminds me of the cartesian equation of motion due to the tan theta and secant squared, so finding that:

$t = \frac{x}{U \cos \theta}$

$\Rightarrow \ y = \frac{-gx^2}{2U^2} \sec^2 \theta + x\tan \theta + h$

$$When$ \ \ x=R \ \ y=0$

$\therefore \ \ 0 = -\frac{gR^2}{2U^2} \sec^2 \theta + R \tan \theta + h$

$\therefore gR^2 \sec^2 \theta - 2U^2R \tan \theta - 2hU^2 = 0$

Next part:
I did it in a weird way, where I got tan theta before I got R_1
Now, since you are posting in the 4U forum, I am going to assume that I am allowed to do this:

Implicitly differentiating:

$gR^2 \sec^2 \theta - 2U^2R \tan \theta - 2hU^2 = 0 \ \ | \frac{dR}{d\theta}$

$\frac{g}{2U^2} (2R R' \sec^2 \theta + R^2 2\sec^2 \theta \tan \theta) = R' \tan \theta + \sec^2 \theta R$

$$When$ \ \ R'=0 \ \ R=R_1$

$\therefore \ \frac{g}{2U^2} R_1^2 \sec^2 \times 2 \theta \tan \theta = R_1\sec^2 \theta$

$\therefore \ \ \tan \theta = \frac{U^2}{gR_1}$

=====

$gR^2(1+\tan^2 \theta) - 2U^2 R \tan \theta - 2hU^2 = 0$

$\tan \theta = \frac{U^2}{gR_1} \ \rightarrow R=R_1$

$gR_1^2 (1+ \frac{U^4}{gR_1^2}) - 2U^2 R_1 \frac{U^2}{gR_1} - 2hU = 0$

.
.
.

$R_1 = \frac{U}{g} \sqrt{U^2 + 2gh}$

===

From the angle of projection:

$R_1 = \frac{U^2}{g} \cot \theta$

Equating this with the other form:

$\frac{U^2}{g} \cot \theta = \frac{U}{g} \sqrt{U^2+ 2gh}$

$h = \frac{U^2(\cot^2 \theta - 1}{2g}$

Next just use that:

$\cot 2\theta = \frac{2\cos^2 \theta - 1}{2\sin \theta \cos \theta}$

Using:

$\tan \theta = \frac{U^2}{gR_1}$

Use Pythagoras's Theorem to find sin theta and cosine theta, substitute into cotangent 2 theta, you should find an expression which resolves into the formula for h above.

HeroicPandas · Jun 8, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$A projectile is fired in a medium of no air resistance, and lands on an inclined plane of angle$ \ \ \alpha$

$It is fired with a velocity of$ \ \ \theta \ \ $to the horizontal, the projectile is fired from the start of the inclined plane$

$The projectile lands on the inclined plane, of distance$ \ \ l \ \ $to the origin$

$Find$ \ \ \theta \ \ $in terms of alpha, such that$ \ \ l \ \ $is maximised$

"Fired with a velocity of theta to the horizontal"

anything wrong with this? or ntohing wrong?

AnimeX · Jun 8, 2013

Re: HSC 2013 4U Marathon

How would you do this question?

P is a point on the ellipse x^2 /a^2 + y^2 / b^2 =1 with a centre O. A line is drawn from O, parallel to the tangent to the ellipse at P and meets the ellipse at Q. Prove that the area of triangle OPQ is independent of the position P.

Sy123 · Jun 8, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
"Fired with a velocity of theta to the horizontal"

anything wrong with this? or ntohing wrong?

Yep, its supposed to read fired with a velocity, at theta to the horizontal, in your final answer theta will be independent of V.

iBibah · Jun 9, 2013

Sy123 said:
$Prove that$

$1+ \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{n^2} < 2 - \frac{1}{n}$

*for n>1

asianese · Jun 9, 2013

Re: HSC 2013 4U Marathon

Consider the curve $y=\dfrac{1}{x^2}$ and consider areas.

asianese · Jun 9, 2013

Re: HSC 2013 4U Marathon

AnimeX said:
How would you do this question?

P is a point on the ellipse x^2 /a^2 + y^2 / b^2 =1 with a centre O. A line is drawn from O, parallel to the tangent to the ellipse at P and meets the ellipse at Q. Prove that the area of triangle OPQ is independent of the position P.

Find gradient of tangent at P. Find eqn of tangent at P. Since tangent meets ellipse again, solve the ellipse and the tangent. Find the coordinates of O P and Q. Find their area.

Makematics · Jun 9, 2013

Re: HSC 2013 4U Marathon

AnimeX said:
How would you do this question?

P is a point on the ellipse x^2 /a^2 + y^2 / b^2 =1 with a centre O. A line is drawn from O, parallel to the tangent to the ellipse at P and meets the ellipse at Q. Prove that the area of triangle OPQ is independent of the position P.

oh god, pls :'( not this question... lost my only mark on it in my half-yearly fml...

Makematics · Jun 9, 2013

Re: HSC 2013 4U Marathon

asianese said:
Find gradient of tangent at P. Find eqn of tangent at P. Since tangent meets ellipse again, solve the ellipse and the tangent. Find the coordinates of O P and Q. Find their area.

Just to add a bit, use (acostheta,bsintheta) as P. Also i dont think the tangent meets the ellipse again? Should be
1. Find the gradient of the tangent at P
2.Use the point-gradient formula to find the equation of the line through the origin and parallel to the tangent.
3. Solve the equations of the line and the ellipse simultaneously to find the coordinates of Q.
4. Finding the area is the challenging bit. Use the perpendicular distance formula from the origin to PQ and then use A=1/2 (PQ) (perp. distance). After heavy algebra, you should get 1/2 ab units^2 as the area iirc. Or you could use the determinant method, which is not allowed according to my teachers -.-

Sy123 · Jun 9, 2013

Re: HSC 2013 4U Marathon

$Without repeated iterations of Newton's Method of Approximation$

$$Find a rational approximation of$ \ \ \pi \ \ $of an error less than$ \ \ 10^{-3}$

$Justify your answer$

EDIT: On hindsight its actually very simple, 3141/1000 will do =(

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