HSC 2013 MX2 Marathon (archive) (5 Viewers)

hayabusaboston · May 23, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
no, its not.

Naturally.

braintic · May 23, 2013

Re: HSC 2013 4U Marathon

braintic said:
Since sin^2(0) = cos^2 (90), sin^2 (3) = cos^2 (87), sin^2 (6) = cos^2 (84), sin^2 (93) = cos^2 (177), ...
Then the sum of sin^2 is equal to the sum of cos^2.

But the sum of the (sum of sin^2 and cos^2) is the sum of (1) which is 60.
So the answer is 30.

Actually we are summing 61 terms so my answer looks wrong - it should be 30.5
Yet I took the time to add it all up on a calculator and got 30. What's going on here?

seanieg89 · May 23, 2013

Re: HSC 2013 4U Marathon

Think about which cos^2(theta) you are pairing up with sin^2(90). You are essentially just double-counting something.

braintic · May 24, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Think about which cos^2(theta) you are pairing up with sin^2(90). You are essentially just double-counting something.

Ahh, got it. Thanks.

Sy123 · May 26, 2013

Re: HSC 2013 4U Marathon

$Runners who run with a constant speed, Fred, Alistair, Maggie, and Bob run around a 500 metre track. In the time it takes Bob to finish the race, Alistair has only run 300 metres. When Maggie has finished the race, Fred has run 450 metres. When Alistair finishes the race, Maggie would have been halfway through her second lap if she kept on running$

$When Bob finishes the race, how far was he ahead of Fred?$

seanieg89 · May 28, 2013

Re: HSC 2013 4U Marathon

$Let $f$ be a continuous function from the interval $[0,1]$ to itself. Prove that there is some $\alpha\in[0,1]$ with $f(\alpha)=\alpha$. \\ \\By way of example, show that the claim in this question is no longer true if we drop the assumption of continuity.$

seanieg89 · May 28, 2013

Re: HSC 2013 4U Marathon

An infinitesimally small billiard ball is at the edge of a circular table of unit radius and is hit in a direction that makes an acute angle of theta with the radius joining the starting position to the centre of the circle.

Find the minimum, maximum, and average distance of the ball from the centre of the circle. We assume that there is no friction so our ball never slows down.

Describe the set of points that the ball will eventually pass through and how this set depends on theta.

seanieg89 · May 28, 2013

Re: HSC 2013 4U Marathon

$Let $f$ be a continuous function on $\mathbb{R}$ and define the sequence of functions $(F_n)_{n=1}^\infty$ recursively by the conditions:\\ \\ $F_0(x)=f(x)\\ F_{n+1}'(x)=F_n(x)\\ F_n(0)=0$\\ \\ for all non-negative integers $n$. \\ \\ Prove that we cannot choose $f$ and $\alpha>0$ such that $F_k(\alpha)=1$ for all non-negative integers $k$.\\ \\ You may assume without proof the identity:\\ \\ $\lim_{n\rightarrow\infty}\left(\sum_{j=0}^n \frac{x^j}{j!}\right)=e^x$.$

Sy123 · May 28, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
An infinitesimally small billiard ball is at the edge of a circular table of unit radius and is hit in a direction that makes an acute angle of theta with the radius joining the starting position to the centre of the circle.

Find the minimum, maximum, and average distance of the ball from the centre of the circle. We assume that there is no friction so our ball never slows down.

Describe the set of points that the ball will eventually pass through and how this set depends on theta.

I'll leave the previous question to someone else.

Here are my answers to this one:

$$Max$ = 1$

$$Min$ = \sin \theta$

$$Average$ = \sin \theta \cdot \ln \left |\frac{1+\cos \theta}{1-\cos \theta} \right |$

$$If the ball undergoes a 'period', whereby it hits the same intersection again, i.e.$

$360n = m(180-2\theta)$

$$For some integers$ \ \ n, m$

$m \geq 2n$

$\theta = 90 - 180 \cdot \frac{n}{m}$

$Therefore if theta is rational, then there is only a finite number of points whereby the ball hits the edge of the table, however if theta is irrational, then the ball continues on to hit the table at an infinite number of points$

Part 1: Congruent Triangle Proof, max and min distance, h is the height of the triangle

Part 2: Average distance

Is the logic correct?

EDIT: Actually...... For the average distance do I have to divide by area of the triangle?
Because integration is just summing the distances, I'm guessing I have to divide by the area....right?

So the new average:

$$Average$ \ \ = \sec \theta \cdot \ln \left |\frac{1+\cos \theta}{1-\cos \theta} \right |$

seanieg89 · May 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
I'll leave the previous question to someone else.

Here are my answers to this one:

$$Max$ = 1$

$$Min$ = \sin \theta$

$$Average$ = \sin \theta \cdot \ln \left |\frac{1+\cos \theta}{1-\cos \theta} \right |$

$$If the ball undergoes a 'period', whereby it hits the same intersection again, i.e.$

$360n = m(180-2\theta)$

$$For some integers$ \ \ n, m$

$m \geq 2n$

$\theta = 90 - 180 \cdot \frac{n}{m}$

$Therefore if theta is rational, then there is only a finite number of points whereby the ball hits the edge of the table, however if theta is irrational, then the ball continues on to hit the table at an infinite number of points$

Part 1: Congruent Triangle Proof, max and min distance, h is the height of the triangle

Part 2: Average distance

Is the logic correct?

EDIT: Actually...... For the average distance do I have to divide by area of the triangle?
Because integration is just summing the distances, I'm guessing I have to divide by the area....right?

So the new average:

$$Average$ \ \ = \sec \theta \cdot \ln \left |\frac{1+\cos \theta}{1-\cos \theta} \right |$

Close, but with the "average" part of the question you don't want to be integrating with respect to the angle alpha, you want to integrate w.r.t something that changes at a constant rate with respect to time.

(And you will have to divide by how much this thing changes in one straight-line segment of the trajectory.)

Sy123 · May 28, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Close, but with the "average" part of the question you don't want to be integrating with respect to the angle alpha, you want to integrate w.r.t something that changes at a constant rate with respect to time.

(And you will have to divide by how much this thing changes in one straight-line segment of the trajectory.)

Why do we need to integrate with respect to something that changes at a constant rate?

seanieg89 · May 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Why do we need to integrate with respect to something that changes at a constant rate?

Because "average" means average with respect to time.

This is a continuous analogue of rolling a dice 5 times, adding the totals and dividing by 5. Each "roll" is given an equal weighting in our average for rolling dice, and similarly, each infinitesimal period of time must be given an equal weighting when taking our continuous average of the distance function. Another way of thinking about it is that alpha will vary faster during the parts of the balls trajectory when it is close to its minimum distance from the origin, and hence the distance at these values of alpha should be "weighted less".

If we integrate with respect to something that changes at a constant rate this is just a linear change of variables away from integrating with respect to time...which is the correct thing to do when we want to take a time average.

Sy123 · May 28, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Because "average" means average with respect to time.

This is a continuous analogue of rolling a dice 5 times, adding the totals and dividing by 5. Each "roll" is given an equal weighting in our average for rolling dice, and similarly, each infinitesimal period of time must be given an equal weighting when taking our continuous average of the distance function. Another way of thinking about it is that alpha will vary faster during the parts of the balls trajectory when it is close to its minimum distance from the origin, and hence the distance at these values of alpha should be "weighted less".

If we integrate with respect to something that changes at a constant rate this is just a linear change of variables away from integrating with respect to time...which is the correct thing to do when we want to take a time average.

Yep I get it. I'll see what I can do with this though.

seanieg89 · May 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yep I get it. I'll see what I can do with this though.

The end expression isn't that pretty but it is certainly an MX2 integral.

Sy123 · May 28, 2013

Re: HSC 2013 4U Marathon

Allowing x to be the distance to the point on the circular table to the actual ball, then we have a triangle of sides, d (the distance), x and 1, with angle theta. (I am assuming constant speed for the ball)

So by the cosine rule:

$d= \sqrt{x^2-2x\cos \theta + 1}$ Theta is constant.

$$Average$ = \frac{1}{2\cos \theta} \int_0^{2\cos \theta} \sqrt{x^2-2x\cos \theta+1} \ dx$

$2\cos \theta \ \ $is the base of the triangle$$

Then through a trig substitution after completing the square, you get (if I haven't made any mistakes)

$\frac{1}{2} + \frac{\sin^2 \theta}{4\cos \theta} \ln \left |\frac{1+\cos \theta}{1-\cos \theta} \right |$

Great question.

GoldyOrNugget · Jun 1, 2013

Re: HSC 2013 4U Marathon

How many 6-letter words are there whose letters are sorted in alphabetical order, and no letter is repeated? (using a loose definition of 'word')

braintic · Jun 1, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
How many 6-letter words are there whose letters are sorted in alphabetical order, and no letter is repeated? (using a loose definition of 'word')

26C6

Sy123 · Jun 2, 2013

Re: HSC 2013 4U Marathon

$Someone swings a rock of mass$ \ \ m \ \ $attached to a string, in a vertical circle in uniform circular motion$

$By considering Tension in the string, show that IF the rock makes a full revolution, then the velocity$ \ \ v \ \ $must be$

$v > \sqrt{gr} \ \ \ $where$ \ \ g \ \ $is the gravitational acceleration constant, and$ \ \ r \ \ $is the length of the string$$

seanieg89 · Jun 3, 2013

Re: HSC 2013 4U Marathon

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.

braintic · Jun 3, 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.

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

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