ATAR Notes: Forum
VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: brightsky on August 10, 2010, 09:42:28 pm
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Does anyone have any AIMO past papers (preferably with solutions) that they are willing to post up (excluding the practice paper on the AMT website)? Thanks. :)
Also, good luck to anyone doing the AIMO this Thursday. :D
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is a perfect square where
is an integer. Find the largest possible value of
. Any help appreciated.
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Are you able to give me a scanned copy of just 2007-2008? I'm working on them now. ::)
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is a perfect square where
is an integer. Find the largest possible value of
. Any help appreciated.
AIMO 2007 Q7.
Suppose x^2-19x+94 = m^2 where m is a non-neg integer. Then x^2-19x+94-m^2 = 0
Go from there.
Ah, solved. Thanks Azure!
Next:
5. Find
where
and
are non-zerp solutions of the system of equations:


9. Find a prime, p, with the property that for some larger prime number, q, both 2q - p and 2q + p are prime numbers. Prove that there is only one such prime p.
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Probably not the best way, but the equations given are ugly anyway :P:
Subtract from the first equation the second times 5. You get:


=5x^2(-17+y))
(y-17))
So either x=0. y=5x, or y=17, lolz. Now you can plug these in to the original equation to see what u get in each case.
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Probably not the best way, but the equations given are ugly anyway :P:
Subtract from the first equation the second times 5. You get:


=5x^2(-17+y))
(y-17))
So either x=0. y=5x, or y=17, lolz. Now you can plug these in to the original equation to see what u get in each case.
Ooh, thanks kamil!
I got a related question:
Real numbers
are linked by the two equations:


Determine the largest value for
.
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The Arithmetic-Quadratic mean inequality states that:

Using this, 
So
 \leq 40-2\sqrt{a^2+b^2+c^2+d^2} = 40-2\sqrt{400-e^2})
Hmm well if this happens to be right so far then I guess you could solve for e.
EDIT: Oh wait, AM-QM only works for positive numbers
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9. Find a prime, p, with the property that for some larger prime number, q, both 2q - p and 2q + p are prime numbers. Prove that there is only one such prime p.
p=3, q=5 works.
Now suppose that there exists some other prime with this property, call it p. Thus
(it is easy to see p cannot be 2)
Thus 

Case 1: 
Then one of p or -p is -1 mod 3. Thus one of 2q-p or 2q+p is a multiple of 3. Note also that 3<q<2q-p<2q+p Since q>q.
It follows that one of these numbers must be a multiple of 3 greater than 3, hence not prime.
The other case is basically the same.
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Real numbers
are linked by the two equations:


Determine the largest value for
.
First I tried to reason by analogy, if instead of a, b, c, d, e we only had c, d, e then we can think about it geometrically in R^3. We can have e as the vertical axis, then we have a plane (first equation) and a sphere (second equation) intersecting in some sort of rotated circle in R^3 and we're trying to find the highest point on the circle, which we visually see occurs when c=d. By analogy we may expect that equality occurs when a=b=c=d, and generally when dealing with symmetric inequalities this is a good idea.
If a=b=c=d then the first equation gives us that a = b = c = d = 10 - e/2. We try to analyse the effect of deviation of the values of a, b, c, d away from 10 - e/2 (our suspected best point). So we let a = 10 - e/2 + a', and similarly for b,c,d, and we think of a' as representing a small deviation. Now we know that a' + b' + c' + d' = 0 from the first equation, this is very convenient. Plugging these into the second equation results in massive cancellation, and we're left with
(assuming I did this right), which tells us our maximum occurs when 5e - 80 = 0, or e = 16.
It's interesting to think about why symmetry causes the massive cancellation, and why performing a change of coordinates (a,b,c,d) -> (a',b',c',d') to localize our variables at our suspected maximum is a good idea.
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https://drive.google.com/open?id=1E3l2RGQvOg4C0V-LtzacIIKc--psVw2z
These has all the past AIMO papers with SOLUTIONS from 1998 to 2017.
Enjoy.