HSC 2014 MX2 Marathon (archive) (2 Viewers)

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Ikki

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Re: HSC 2014 4U Marathon

We know from diagram (triangle inside the circle):
APxPQ=BPxPC

Working backwards from what to prove:
BPxPC=PQx(PB+PC)
BPxPC=PQxBC

If we equate what to prove and what we know:
BC=AP? But BC=AB
so, AB=AP which makes no sense. (Does not necessarily have to be true and shouldn't)
 

Sy123

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Re: HSC 2014 4U Marathon

We know from diagram (triangle inside the circle):
APxPQ=BPxPC

Working backwards from what to prove:
BPxPC=PQx(PB+PC)
BPxPC=PQxBC

If we equate what to prove and what we know:
BC=AP? But BC=AB
so, AB=AP which makes no sense. (Does not necessarily have to be true and shouldn't)
You have made a mistake in your first line, Q is the intersection between the 2 chords, not P.

So from your proof it now is:

AQ x QP = BQ x QC

--

I can reveal my proof if you wish, or I can let you have another go
 

Davo_01

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Re: HSC 2014 4U Marathon

You have made a mistake in your first line, Q is the intersection between the 2 chords, not P.

So from your proof it now is:

AQ x QP = BQ x QC

--

I can reveal my proof if you wish, or I can let you have another go
This is what i did:






 
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RealiseNothing

what is that?It is Cowpea
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Re: HSC 2014 4U Marathon

Hey dunjaa and Sy do one of you think you could explain part i of this answer to me? I understand the rest of it but can't quite follow the first step. Thanks.
His solution to part (i) is wrong.
 

kev-

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Re: HSC 2014 4U Marathon

Hello, I need help
3 letters, A B C. How many words can be made with n letters where A occurs zero or an even number of times. (n is an even positive integer)
 

dunjaaa

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Re: HSC 2014 4U Marathon

Woops, I kind of screwed up the transitivity lol. I typed the solution straight out and didn't realise that I couldn't do it via that method.
 

dunjaaa

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Re: HSC 2014 4U Marathon

My solution was ugly but I got OB^2=26cm. Yeah, I went brute force with angles, I'm trying to think of a more elegant geometrical solution.
 

Sy123

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Re: HSC 2014 4U Marathon

My solution was ugly but I got OB^2=26cm. Yeah, I went brute force with angles, I'm trying to think of a more elegant geometrical solution.
Yea I did it via angles as well
 
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