HSC 2013 MX2 Marathon (archive) (4 Viewers)

[RealiseNothing]

Re: HSC 2013 4U Marathon

If Alice is allowed to toss a coin 10 times and Bob 9 times, then what is the probability that Alice gets more heads than Bob?

Anyway I get

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

1/2 ?

Yep I agree.

 

[Sy123]

Re: HSC 2013 4U Marathon

Exactly, well done. Are you a student or a teacher?

*please be a teacher*

*please be a teacher*

 

[gustavo28]

Re: HSC 2013 4U Marathon

Exactly, well done. Are you a student or a teacher?

I'm a student.

The answer is 1/2. What methods did people use?

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

I'm a student.

The answer is 1/2. What methods did people use?

Find the probability that after 9 throws each they are equal, alice is ahead, bob is ahead. Then if alice is ahead now she already wins, if bob is ahead then alice cant win, and if it is equal it comes down to alice's last toss.

 

[twinklegal19]

Re: HSC 2013 4U Marathon

*please be a teacher*

*please be a teacher*

ahahaha

sy vs realise vs gustavo

kgo

 

[polertics]

Re: HSC 2013 4U Marathon

For what it's worth, here's how I did it.
Alice has n+1 tosses. Bob has n tosses. Define winning in the following way. Alice wins if she gets more heads than Bob. Bob wins if she does not.
Bob wins if and only if Alice throws more tails than Bob
Alice wins if and only if Alice throws more heads than Bob
So, the likelihood of Alice throwing more heads than Bob is the same as that of her throwing more tails than Bob. Hence the likelihood of her winning is ½.

 

[gustavo28]

Re: HSC 2013 4U Marathon

Consider a triangle ABC with points D, E, F on BC, CA and AB respectively. Suppose that AD, BE and CF are concurrent. Prove that (BD/DC)*(CE/EA)*(AF/FB)=1

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

Consider a triangle ABC with points D, E, F on BC, CA and AB respectively. Suppose that AD, BE and CF are concurrent. Prove that (BD/DC)*(CE/EA)*(AF/FB)=1

Let the point they meet at be P. Now consider triangles ABC and PBC. Since the base BC is common we can deduce that:

where XYZ = area of triangle XYZ.

Similarly for the other 2 sides:







Which gives you your result.

 

[polertics]

Re: HSC 2013 4U Marathon

Let the point of concurrency be G. We can use the following theorem:
Thm: If you have a triangle ABC with D on the interior of BC. Then ABD/ACD=BD/CD.
(This is because if you let the height from A be h. Then ABD=h(BD) and ACD=h(CD).)
So, BD/CD=ABD/ACD=BGD/CGD. if a/b=c/d then we are able to prove that a/b=(a-c)/(b-d). This is because ad=bc implies that ab-ad=ab-bc which in turn implies that a/b=(a-c)/(b-d).
Now, BD/CD=(ABD-BGD)/(ACD-CGD)=ABG/ACG.
Similarly, CE/EA=(BCG)/(ABG)
And AF/FB=(ACG)/(BCG).

 

[RayRay]

Re: HSC 2013 4U Marathon

Let the point they meet at be P. Now consider triangles ABC and PBC. Since the base BC is common we can deduce that:

where XYZ = area of triangle XYZ.

Similarly for the other 2 sides:







Which gives you your result.

I actually haven't seen that before, using the area, interesting...

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

Wait I just realised I didn't answer the question properly lol. woops.

Same sort of idea though I'm fairly sure.

 

[polertics]

Re: HSC 2013 4U Marathon

Let P(x)=x^5-5cx+1 where c is real
1) If c<0, prove that P has exactly one real root and that it is negative
2) Prove that if P has exactly 3 distinct real roots, then c>(1/4)^(4/5)

 

[gustavo28]

Re: HSC 2013 4U Marathon

For what it's worth, here's how I did it.
Alice has n+1 tosses. Bob has n tosses. Define winning in the following way. Alice wins if she gets more heads than Bob. Bob wins if she does not.
Bob wins if and only if Alice throws more tails than Bob
Alice wins if and only if Alice throws more heads than Bob
So, the likelihood of Alice throwing more heads than Bob is the same as that of her throwing more tails than Bob. Hence the likelihood of her winning is ½.

Yep, thats probably the best way to do it. You can also an induction sort of thing.

 

[gustavo28]

Re: HSC 2013 4U Marathon

Let the point of concurrency be G. We can use the following theorem:
Thm: If you have a triangle ABC with D on the interior of BC. Then ABD/ACD=BD/CD.
(This is because if you let the height from A be h. Then ABD=h(BD) and ACD=h(CD).)
So, BD/CD=ABD/ACD=BGD/CGD. if a/b=c/d then we are able to prove that a/b=(a-c)/(b-d). This is because ad=bc implies that ab-ad=ab-bc which in turn implies that a/b=(a-c)/(b-d).
Now, BD/CD=(ABD-BGD)/(ACD-CGD)=ABG/ACG.
Similarly, CE/EA=(BCG)/(ABG)
And AF/FB=(ACG)/(BCG).

There's a slight error in that the area of ABD is equal to h/2(BD). But that doesn't affect the proof at all, which is really good apart from that.

 

[Sy123]

Re: HSC 2013 4U Marathon

Let P(x)=x^5-5cx+1 where c is real
1) If c<0, prove that P has exactly one real root and that it is negative
2) Prove that if P has exactly 3 distinct real roots, then c>(1/4)^(4/5)

----

 

[Sy123]

Re: HSC 2013 4U Marathon







edit: my bad

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

Let

 

[superSAIyan2]

Re: HSC 2013 4U Marathon

let k be a real number; k>=4

show that for every positive real number b, there exists a positive real number a such that 1/a + 1/b = k/(a+b)

Its from HSC 2010

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

let k be a real number; k>=4

show that for every positive real number b, there exists a positive real number a such that 1/a + 1/b = k/(a+b)

Its from HSC 2010

Cross multiply etc, form a quadratic, discriminant.