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HSC 2013 MX2 Marathon (archive) (1 Viewer)

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Sy123

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Re: HSC 2013 4U Marathon

It's just backwards multiplication of two numbers. The original number ends in 6, and when multiplied by 4 gets the same number but with a 6 out the front instead of the end:



Multiply by the 4 and carry the 2 (since you multiply by the units and carry the tens)



Multiply by the 8 and carry the 1:



Multiply by the 3 and carry the 3:



Multiply by the 5 and carry the 1:



Multiply by the 1 and carry the 2:



Since we have just 6, we stop here, and so our number is all the numbers we multiplied - 648351 - but we reverse it as we did the multiplcation in the reverse order to get 153846.
Wow......
Very intelligent solution.

Mine was simplifying the number where it must satisfy:

13n = 2 x 10^{m} - 2 for some integer m, and then I just computed a couple of values until I arrived at what I needed.
Though your method completely trumps mine haha
 

Sy123

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Re: HSC 2013 4U Marathon

(Another early IMO question).



 

RealiseNothing

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Re: HSC 2013 4U Marathon

I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.

Consider the following diagram, where PQ is a tangent to the circle:







An alternative proof for the tangents from an external point theorem.
 
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Sy123

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Re: HSC 2013 4U Marathon

I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.

Consider the following diagram, where PQ is a tangent to the circle:







An alternative proof for the tangents from an external point theorem.
I'll give someone else a go for this one, good question =)
 

HeroicPandas

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Re: HSC 2013 4U Marathon

Do you even know how much u confused me with ur polynomials question, where u made it a^2 not a, that had me going for hours and hours trying to figure out how u got a^2 and it turns out it was a all along lol. Btw If U have a cool conics question can u plz give to me.
LOL! do the other polynomial question i posted on the 3U marathon thread
 

Sy123

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Re: HSC 2013 4U Marathon

I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.

Consider the following diagram, where PQ is a tangent to the circle:







An alternative proof for the tangents from an external point theorem.
From the tangent-secant theorem:














ii)

 

Capt Rifle

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Re: HSC 2013 4U Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{1-x}{1-\sqrt{x}}\: dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{1-x}{1-\sqrt{x}}\: dx" title="\int \frac{1-x}{1-\sqrt{x}}\: dx" /></a>
 

RealiseNothing

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Re: HSC 2013 4U Marathon

I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.

Consider the following diagram, where PQ is a tangent to the circle:







An alternative proof for the tangents from an external point theorem.
From tangent-secant theorem:





Also since M is the midpoint of A and B:



Note the AM-GM inequality:



Let and





Also for the second part, equality holds in the AM-GM inequality when and so . This only occurs when M coincides with Q, and so . Now construct the second tangent and let it touch the circle at point J. Equality holds again when . Hence as both are arbitrary tangents and give the minimum possible value of .
 

Sy123

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Re: HSC 2013 4U Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{1-x}{1-\sqrt{x}}\: dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{1-x}{1-\sqrt{x}}\: dx" title="\int \frac{1-x}{1-\sqrt{x}}\: dx" /></a>


==========

 

Trebla

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Re: HSC 2013 4U Marathon

Hmm perhaps I should give hints for the second part of my question?
 

jyu

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Re: HSC 2013 4U Marathon

I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.

Consider the following diagram, where PQ is a tangent to the circle:







An alternative proof for the tangents from an external point theorem.
PQ is a tangent, so PQM is a right angle. Without stating the obvious, PM > PQ. PM = PQ when the radius = 0.
 

Sy123

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Re: HSC 2013 4U Marathon

Moving on from part (i)



Equate real parts of both sides, also make the substitution:















Makes the LHS side zero, on the RHS there is a polynomial of degree n in cot^2 theta, and the theta that makes cot^2 theta a root is above.
Now taking the sum of all roots.



Now, cotangent is periodic by pi, therefore the following applies:



 
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Sy123

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Re: HSC 2013 4U Marathon

PQ is a tangent, so PQM is a right angle. Without stating the obvious, PM > PQ. PM = PQ when the radius = 0.
We do not know that M is the centre.
 
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