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    (Original post by GHOSH-5)
    Did you get this for Q4 off that 2000 paper?
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    k = 2001, 667, 87, 23, 3
    Wait, back to Q3, how do we prove that on the triangle ABC:

    , CB + d > AB + AC? (where d is the perpendicular from point A to CB)
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    (Original post by refref)
    Wait, back to Q3, how do we prove that on the triangle ABC:

    , CB + d > AB + AC? (where d is the perpendicular from point A to CB)
    Consider if the legs AB and AC are lengths x and y respectively. Find CB + d in terms of x and y, and then prove that the inequality is true for all pairs of positive reals x, y.
    Spoiler:
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    You should find that CB + d =  \sqrt{x^2+y^2} + \dfrac{xy}{\sqrt{x^2+y^2}} = \dfrac{x^2+y^2 + xy}{\sqrt{x^2+y^2}}
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    So we need to prove that this is greater than x + y; try multiplying the inequality by  \sqrt{x^2+y^2} and then squaring the whole thing.
    Also did you get 1260 and 5376 for Q5 of the same paper?
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    (Original post by GHOSH-5)
    Consider if the legs AB and AC are lengths x and y respectively. Find CB + d in terms of x and y, and then prove that the inequality is true for all pairs of positive reals x, y.
    Spoiler:
    Show
    You should find that CB + d =  \sqrt{x^2+y^2} + \dfrac{xy}{\sqrt{x^2+y^2}} = \dfrac{x^2+y^2 + xy}{\sqrt{x^2+y^2}}
    Spoiler:
    Show
    So we need to prove that this is greater than x + y; try multiplying the inequality by  \sqrt{x^2+y^2} and then squaring the whole thing.
    Also did you get 1260 and 5376 for Q5 of the same paper?
    I am stuck on the most simple step is seems...how do we get d in terms of x and y?
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    (Original post by refref)
    I am stuck on the most simple step is seems...how do we get d in terms of x and y?
    There are various ways. You could consider doing it with coordinate geometry, treating the legs as appropriate axes. Or perhaps, suppose it is the point D on the hypotenuse with AD perp. to BC. Suppose DB = m; hence CD = root(x^2+y^2) - m. Then use pythagoras on triangles ADB and ADC and equate them to get AD. I think the neatest way is to consider the area of the triangle in two different ways: it is base x perp. height/2 = product of legs/2 = hypotenuse x d/ 2.
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    (Original post by GHOSH-5)
    Did you get this for Q4 off that 2000 paper?
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    k = 2001, 667, 87, 23, 3
    Yes. But I dont think I have a real proof

    How did you do it?

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    also I may have used wolfram to find out the prime factorisation of 2001...but we are allowed computers in the exam aren't we?
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    (Original post by refref)
    Yes. But I dont think I have a real proof

    How did you do it?

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    also I may have used wolfram to find out the prime factorisation of 2001...but we are allowed computers in the exam aren't we?
    No calculators, or indeed computers, in the exam.
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    I proved inductively that:

    For all  t \geq 0 , the following are true:

     a_{4t} = 4tk + 1

     a_{4t+1} = k-1

     a_{4t+2} = (4t+3)k - 1

     a_{4t+3} = 1

    Hence we need to find if either (4t+3)k-1, k-1 or 4tk+1 are ever equal 2000. The last case clearly is not true, as the LHS would be congruent to 1 mod 4, but the RHS would be congruent to 0 mod 4.

    For k-1 = 2000, clearly k = 2001 fits the bill.

    For (4t+3)k -1 = 2000 => (4t+3)k = 2001 = 3 x 23 x 29. So clearly both k and 4t+3 must be congruent to 3 mod 4 and must be positive integer divisors of 2001, which are namely 3, 23, 3 x 29 and 23 x 29. And that should be proof enough.
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    (Original post by GHOSH-5)
    No calculators, or indeed computers, in the exam.
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    I proved inductively that:

    For all  t \geq 0 , the following are true:

     a_{4t} = 4tk + 1

     a_{4t+1} = k-1

     a_{4t+2} = (4t+3)k - 1

     a_{4t+3} = 1

    Hence we need to find if either (4t+3)k-1, k-1 or 4tk+1 are ever equal 2000. The last case clearly is not true, as the LHS would be congruent to 1 mod 4, but the RHS would be congruent to 0 mod 4.

    For k-1 = 2000, clearly k = 2001 fits the bill.

    For (4t+3)k -1 = 2000 => (4t+3)k = 2001 = 3 x 23 x 29. So clearly both k and 4t+3 must be congruent to 3 mod 4 and must be positive integer divisors of 2001, which are namely 3, 23, 3 x 29 and 23 x 29. And that should be proof enough.
    I'm not too familiar with using induction on things that are not simple series...simple divisibility and simple matricies (fp1...) so I didnt prove that. But basically did it the same way, I just need to prove more things.

    Could you show me how we use induction to prove the first bit? Thanks (but only if you can be bothered, you dont have to )
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    (Original post by refref)
    I'm not too familiar with using induction on things that are not simple series...simple divisibility and simple matricies (fp1...) so I didnt prove that. But basically did it the same way, I just need to prove more things.

    Could you show me how we use induction to prove the first bit? Thanks (but only if you can be bothered, you dont have to )
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    Check it's true for our base case, where t = 0:
     a_{4(0)} = 1 = 4 (0) \cdot k +1
     a_{4(0)+1} = k - 1
     a_{4(0)+2} = 3k - 1 = (4 (0) +3)k -1
     a_{4(0)+3} = 1

    So the base case is true.

    Suppose it is true for some t = p; i.e.  a_{4p} = 4pk +1, a_{4p+1} = k-1, a_{4p+2} = (4p+3)k- 1, a_{4p+3} = 1

    Then:

     a_{4p+3+1} = a_{4(p+1)} = 4(p+1)k + 1

     a_{4(p+1)+1} = (4(p+1)+1)k - (4(p+1)k + 1) = k - 1

     a_{4(p+1)+2} = (4(p+1)+2)k + (k-1) = (4(p+1)+3)k - 1

     a_{4(p+1)+3} = (4(p+1)+3)k - ((4(p+1)+3)k-1) = 1

    So it is true for  t = p+1

    And hence by induction, it is true for all integers  t \geq 0
    There's another example of this half way down page two, where I prove something about Fibonacci numbers.
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    (Original post by GHOSH-5)
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    Check it's true for our base case, where t = 0:
     a_{4(0)} = 1 = 4 (0) \cdot k +1
     a_{4(0)+1} = k - 1
     a_{4(0)+2} = 3k - 1 = (4 (0) +3)k -1
     a_{4(0)+3} = 1

    So the base case is true.

    Suppose it is true for some t = p; i.e.  a_{4p} = 4pk +1, a_{4p+1} = k-1, a_{4p+2} = (4p+3)k- 1, a_{4p+3} = 1

    Then:

     a_{4p+3+1} = a_{4(p+1)} = 4(p+1)k + 1

     a_{4(p+1)+1} = (4(p+1)+1)k - (4(p+1)k + 1) = k - 1

     a_{4(p+1)+2} = (4(p+1)+2)k + (k-1) = (4(p+1)+3)k - 1

     a_{4(p+1)+3} = (4(p+1)+3)k - ((4(p+1)+3)k-1) = 1

    So it is true for  t = p+1

    And hence by induction, it is true for all integers  t \geq 0
    There's another example of this half way down page two, where I prove something about Fibonacci numbers.
    Thankyou
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    (Original post by GHOSH-5)
    Consider if the legs AB and AC are lengths x and y respectively. Find CB + d in terms of x and y, and then prove that the inequality is true for all pairs of positive reals x, y.
    Spoiler:
    Show
    You should find that CB + d =  \sqrt{x^2+y^2} + \dfrac{xy}{\sqrt{x^2+y^2}} = \dfrac{x^2+y^2 + xy}{\sqrt{x^2+y^2}}
    Spoiler:
    Show
    So we need to prove that this is greater than x + y; try multiplying the inequality by  \sqrt{x^2+y^2} and then squaring the whole thing.
    Also did you get 1260 and 5376 for Q5 of the same paper?
    I have never really 'proved' anything in an inequality before. I did this one today though, multiplying the inequ by sqrt (x^2 + y^2) and expanding/rearranging things got to x^2y^2 > 0. This is obviously true, but does it prove the original inequality?

    And I cant do the dwarfs question. I'll try again tonight

    (Did you know that six out of seven dwarfs are not happy?)
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    (Original post by refref)
    I'll let you know

    (there was no option for delivery, meh, I'll just email them right now to find how long it will take)
    have u got a reply from UKMT yet ? thx
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    (Original post by refref)
    I have never really 'proved' anything in an inequality before. I did this one today though, multiplying the inequ by sqrt (x^2 + y^2) and expanding/rearranging things got to x^2y^2 > 0. This is obviously true, but does it prove the original inequality?
    Yes it does. It should be clear that you can put a '\iff' sign between each step, hence each statement is true if and only if another statement is true.
    And I cant do the dwarfs question. I'll try again tonight
    I realise I've made a large mistake anyway, so don't take the answers I had above as correct.
    (Did you know that six out of seven dwarfs are not happy?)
    :sigh:
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    (Original post by lovmaths)
    have u got a reply from UKMT yet ? thx
    Nope :mad:

    (Original post by GHOSH-5)
    Yes it does. It should be clear that you can put a '\iff' sign between each step, hence each statement is true if and only if another statement is true.

    I realise I've made a large mistake anyway, so don't take the answers I had above as correct.

    :sigh:
    Ok

    I want my book! :mad:
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    (Original post by refref)
    Nope :mad:



    Ok

    I want my book! :mad:
    Well, i'm going to order it today Olympiad Primer
    i can't wait no more- hehe only 1month left till BMO!!!
    and i don't wanna get 0/60
    :P
    anyway, thanks
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    I'm pushing my other subjects aside and actually doing work in the weekdays

    I'll be happy with anything above 20 I suppose...aiming for 30 haha.
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    (Original post by refref)
    I'm pushing my other subjects aside and actually doing work in the weekdays

    I'll be happy with anything above 20 I suppose...aiming for 30 haha.
    i'll be happy with just getting 1 question right and i'm aming for 20+ hehe
    the first 2 questions seem to be easier than the rest :P
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    2001/1. Find all two-digit integers N for which the sum of the digits
    of 10^N − N is divisible by 170.
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    (Original post by Hughh)
    2001/1. Find all two-digit integers N for which the sum of the digits
    of 10^N − N is divisible by 170.
    I like this one, it's number theoryish. (obviously)
    Are you asking for help or as a question to try?
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    Easiest way to start is to let 10^N-N=(10^N-1)-(N-1)
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    (Original post by lovmaths)
    i'll be happy with just getting 1 question right and i'm aming for 20+ hehe
    the first 2 questions seem to be easier than the rest :P
    They sent it yesterday by second class post. I suppose it should come this week.
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    (Original post by refref)
    They sent it yesterday by second class post. I suppose it should come this week.
    *edit*

    Actually it came tody....I really dont know how thats possible.

    But awesome.

    (Original post by GHOSH-5)
    Yes it does. It should be clear that you can put a '\iff' sign between each step, hence each statement is true if and only if another statement is true.

    I realise I've made a large mistake anyway, so don't take the answers I had above as correct.

    :sigh:
    I have all the answers now (I was only expecting answers for a few years...but it seems it has them all. We were right (well, obviously) on Q's 3 and 4. Do you have this book? If not, if you want any answers nudge me.
 
 
 
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