HSC 2013-14 MX1 Marathon (archive) (1 Viewer)

Drongoski · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

SpiralFlex said:
Sure?

Thanks for pointing out Spiral - was going to put the 2 but in my huwwy, forgot. Ha ha.

Edit:
Turns out I had more than that wrong.

Sy123 · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

$Using the Binomial Theorem and/or Combinatorics$

$$Prove that$ \ \ \ ^{n^2} C_2 = ^n C_1 \cdot ^nC_2 \left (1+^nC_1 \right )$

This is a little harder as per request

You may not expand the whole thing hoping for it to fall into place, be creative

HeroicPandas · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

Drongoski said:
$\int \frac {1+cos x}{1-cos x} dx = \int \frac {2cos^2 {\frax {x}{2}}}{2sin^2 \frac{x}{2}} dx = \int cosec^2 \frac {x}{2} dx = - 2cot \frac {x}{2} + C$

Someone please post another question.

sorry, u are wrong

isn't cos/sin = cot, not cosec?

Drongoski · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
sorry, u are wrong

isn't cos/sin = cot, not cosec?

Sir, you are right! Corrected!

Sy123 · Jan 28, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Using the Binomial Theorem and/or Combinatorics$

$$Prove that$ \ \ \ \binom{n^2}{2} = \binom{n}{1} \binom{n}{2} \left (1+\binom{n}{1} \right )$

This is a little harder as per request

You may not expand the whole thing hoping for it to fall into place, be creative

Making it easier:

$Using the identity$

$(1+x)^{n^2} = (1+x)^n (1+x)^n (1+x)^n \dots (1+x)^n$

vbzxwgy · Jan 29, 2013

Re: HSC 2013 3U Marathon Thread

you have three dice:

die a has sides: 2,2,4,4,9,9.

die b has sides: 1,1,6,6,8,8.

die c has sides: 3,3,5,5,7,7.

1. what is the average score of die a,b,c respectively.

2. what are the probabilities of:

i. a single roll of a being higher than a single roll of b

ii. a single roll of b being higher than a single roll of c

iii. a single roll of c being higher than a single roll of a.

Sy123 · Feb 2, 2013

Re: HSC 2013 3U Marathon Thread

$On the graph of$ \ \ \ x^2+y^2=r^2 \ \ \ $a line is drawn from the origin$ \ O \ $ to point$ \ P \ $which is a variable point that lies on the graph$

$PO \ \ $makes an angle of$ \ \ \theta \ \ $with the$ \ \ x$-axis$$

$$A tangent to the graph at$ \ \ P \ \ $is constructed, it intersects the$ \ \ x-$axis, at$ \ \ R \ \ $and intersects the$ \ \ y$-axis, at$ \ \ S$

$Find$ \ \ \theta \ \ $such that the grey shaded area in the diagram above is a minimum$

asianese · Feb 2, 2013

Re: HSC 2013 3U Marathon Thread

Sy I just did the above question roughly and I got a minimum? Probs did something wrong.

Sy123 · Feb 2, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
Sy I just did the above question roughly and I got a minimum? Probs did something wrong.

Yes its a typo haha, it is supposed to be a minimum, if you notice the maximum area is actually infinite so yeah there is no real max. Fixing it now

Sy123 · Feb 2, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$On the graph of$ \ \ \ x^2+y^2=r^2 \ \ \ $a line is drawn from the origin$ \ O \ $ to point$ \ P \ $which is a variable point that lies on the graph$

$PO \ \ $makes an angle of$ \ \ \theta \ \ $with the$ \ \ x$-axis$$

$$A tangent to the graph at$ \ \ P \ \ $is constructed, it intersects the$ \ \ x-$axis, at$ \ \ R \ \ $and intersects the$ \ \ y$-axis, at$ \ \ S$

$Find$ \ \ \theta \ \ $such that the grey shaded area in the diagram above is a minimum$

Adding another part, maybe some 2012ers will remember this part if they saw the marathon last year, but here is another way to prove it using the above result:

$$Construct a normal to$ \ \ SR \ \ $ at the point$ \ \ S$

$Using the above result find$ \ \ \theta \ \ $such that the normal at$ \ \ S \ \ $is a tangent to the circle$

braintic · Feb 2, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$On the graph of$ \ \ \ x^2+y^2=r^2 \ \ \ $a line is drawn from the origin$ \ O \ $ to point$ \ P \ $which is a variable point that lies on the graph$

$PO \ \ $makes an angle of$ \ \ \theta \ \ $with the$ \ \ x$-axis$$

$$A tangent to the graph at$ \ \ P \ \ $is constructed, it intersects the$ \ \ x-$axis, at$ \ \ R \ \ $and intersects the$ \ \ y$-axis, at$ \ \ S$

$Find$ \ \ \theta \ \ $such that the grey shaded area in the diagram above is a minimum$

Since the area of the circular quadrant is a constant, surely the problem boils down to simply minimizing the area of the triangle ORS.

asianese · Feb 2, 2013

Re: HSC 2013 3U Marathon Thread

The answer is as one would expect, tau/8.

Sy123 · Feb 2, 2013

Re: HSC 2013 3U Marathon Thread

braintic said:
Since the area of the circular quadrant is a constant, surely the problem boils down to simply minimizing the area of the triangle ORS.

I never implied otherwise.

asianese said:
The answer is as one would expect, tau/8.

?

Shadowdude · Feb 2, 2013

Re: HSC 2013 3U Marathon Thread

ugh tau

$\tau = 2\pi$

because apparently tau is a lot more prevalent in formulae so they feel that tau should be the "correct" mathematical constant over pi

Sy123 · Feb 3, 2013

Re: HSC 2013 3U Marathon Thread

Shadowdude said:
ugh tau

$\tau = 2\pi$

because apparently tau is a lot more prevalent in formulae so they feel that tau should be the "correct" mathematical constant over pi

I'm not a fan of it

============================

$Prove using the Binomial Theorem/Combinatorics that$

$\sum_{j=m}^{n} \binom{j}{m} \equiv \binom{m}{m} + \binom{m+1}{m} + \binom{m+2}{m}+ \dots + \binom{n}{m} = \binom{n+1}{m+1}$

HeroicPandas · Feb 3, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
I'm not a fan of it

============================

$Prove using the Binomial Theorem/Combinatorics that$

$\sum_{j=m}^{n} \binom{j}{m} \equiv \binom{m}{m} + \binom{m+1}{m} + \binom{m+2}{m}+ \dots + \binom{n}{m} = \binom{n+1}{m+1}$

Here is one attempt

$\binom{m}{m} + \binom{m+1}{m} + \binom{m+2}{m}+ \dots + \binom{n}{m} = \binom{n+1}{m+1} \\ \\$Using, $\binom{n}{r} = \binom{n}{n-r} $to change every term on the LHS \\ \\ $\therefore 1 + \binom{m+1}{1} + \binom{m+2}{2}+ \dots + \binom{n}{n-m} = \binom{n+1}{n-m} \\ \\ 1 + \binom{m+1}{1} + \binom{m+2}{2}+ \dots + \binom{n}{n-m-1} = \binom{n+1}{n-m} -\binom{n}{n-m} \\ \\ $Using $\binom{n}{r-1} + \binom{n}{r} = \binom{n+1}{r} $to change the RHS \\ \\ $\therefore 1 + \binom{m+1}{1} + \binom{m+2}{2}+ \dots + \binom{n}{n-m-1} = \binom{n}{n-m-1} \\ \\ $Keep doing this and you will end up with 0 = 0, hence LHS = RHS$

Sy123 · Feb 3, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
Here is one attempt

$\binom{m}{m} + \binom{m+1}{m} + \binom{m+2}{m}+ \dots + \binom{n}{m} = \binom{n+1}{m+1} \\ \\$Using, $\binom{n}{r} = \binom{n}{n-r} $to change every term on the LHS \\ \\ $\therefore 1 + \binom{m+1}{1} + \binom{m+2}{2}+ \dots + \binom{n}{n-m} = \binom{n+1}{n-m} \\ \\ 1 + \binom{m+1}{1} + \binom{m+2}{2}+ \dots + \binom{n}{n-m-1} = \binom{n+1}{n-m} -\binom{n}{n-m} \\ \\ $Using $\binom{n}{r-1} + \binom{n}{r} = \binom{n+1}{r} $to change the RHS \\ \\ $\therefore 1 + \binom{m+1}{1} + \binom{m+2}{2}+ \dots + \binom{n-1}{n-m-1} = \binom{n}{n-m-1} \\ \\ $Keep doing this and you will end up with 0 = 0, hence LHS = RHS$

I'm assuming in your last line its a typo for (as fixed above) the last term on the LHS of your last term used to be $\binom{n}{n-m-1}$
So I'm assuming its:

$\binom{n-1}{n-m-1}$

Its an interesting method, but I will leave it to someone else to say whether its valid or not. I'm not sure whether its entirely valid or not.

HeroicPandas · Feb 3, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
I'm assuming in your last line its a typo for (as fixed above) the last term on the LHS of your last term used to be $\binom{n}{n-m-1}$
So I'm assuming its:

$\binom{n-1}{n-m-1}$

Its an interesting method, but I will leave it to someone else to say whether its valid or not. I'm not sure whether its entirely valid or not.

ok

HeroicPandas · Feb 3, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Making it easier:

$Using the identity$

$(1+x)^{n^2} = (1+x)^n (1+x)^n (1+x)^n \dots (1+x)^n$

can u please reveal the solution to this question?

Sy123 · Feb 3, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
can u please reveal the solution to this question?

For that one:

Using that identity, equate the co-efficient of x^2 on both sides:

Now to select 2 x's from the RHS, we need to either pick x from 1 (1+x)^n and 1 x from (1+x)^n, this makes x^2
And the ways we can do this is:

First pick 2 brackets from the n number of brackets:

$\binom{n}{2}$

In each of these brackets, we need to pick the x term, so that is

$\binom{n}{1} \cdot \binom{n}{1}$

So we combine both of these to get:

$\binom{n}{2} \binom{n}{1}^2$

Now we can also get x^2 via picking 1 x^2 from 1 bracket, to do this pick 1 bracket out of the n, and in that bracket pick the x^2 co-efficient:

$\binom{n}{1} \binom{n}{2}$

Add both of these together, and equate co-efficent on LHS

$\binom{n^2}{2} = \binom{n}{1} \binom{n}{2} \left (1+\binom{n}{1}\right )$

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