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STEP Maths I,II,III 1987 Solutions watch

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    i think i have the very first part of question 2 for step I

    The bit about the relationship between R, r and n.

    My idea was:

     \frac{Area of larger circle - Area of smaller circle}{Area of Little Circles}

    which should be:

     \frac{(R+r)^2 - (R-r)^2}{r^2} = n

    after expanding and subtraction it cancels to:

     \frac{4R}{r} = n

    This is probably wrong but i thought id at least give it ago lol
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    (Original post by thedemon13666)
    i think i have the very first part of question 2 for step I

    The bit about the relationship between R, r and n.

    My idea was:

     \frac{Area of larger circle - Area of smaller circle}{Area of Little Circles}

    which should be:

     \frac{(R+r)^2 - (R-r)^2}{r^2} = n

    after expanding and subtraction it cancels to:

     \frac{4R}{r} = n

    This is probably wrong but i thought id at least give it ago lol
    I don't know, but what you have might not work because of the area of the gaps between the circles if you see what I mean.

    I'm not entirely sure what it's on about.

    I can think of quite a few relationships between R, r and n but I don't know what it's looking for exactly. Probably has something to do with the later parts.

    eg:
    The circumferences of the circles: (is the plural of circumference circumferences? :/)
    2\pi(R-r)<2n\pi(r)<2\pi(R+r)

    Or (this one im not sure about), if you consider the third circle of radius R you would get:

    2nr\approx 2 \pi R

    nr\approx \pi R

    I doubt this one is much use as what you are asked to prove seems to be very precise :s.

    This third circle of radius R that would cut all of the smaller circles in half for the second part but didn't have anything near what you're supposed to prove.

    Maybe one of the STEP veterans could give me a little hint?

    Or perhaps if you were to consider a third circle of radius R that would cut all of the smaller circles in half. Thats what I tried to do for the second part but didn't have anything near what you're supposed to prove.
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    well the later parts are onabout area's so i guessed to follow relationships to do with areas.
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    they all cancelled out

    i tried to do a similar thing for the second part and ended up with 1/2 like in the answer

    but i cant for the life of me see where the roots are coming from
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    Okay, what I did was this:

    Considered a third circle of radius R which would cut all the smaller circles in half, only giving us the section of the band that we are interested in.

    Calculated the area of this half of the band to be: \pi R^2-\pi(R-r)^2

    From that you would have to take away the area of n half circles of radius r. Doesn't quite come out as what is there though :/.
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    thats exactly what i did, doesnt look similar, least the half is in there
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    (Original post by thedemon13666)
    they all cancelled out

    i tried to do a similar thing for the second part and ended up with 1/2 like in the answer

    but i cant for the life of me see where the roots are coming from
    1/2 of the answer is alot closer than I have gotten
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    no i mean at least the fraction 1/2 is in there :P

    do you think maybe they have used their relationship and subbed in to make their equation look different?
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    (Original post by thedemon13666)
    no i mean at least the fraction 1/2 is in there :P

    do you think maybe they have used their relationship and subbed in to make their equation look different?
    It's possible, it's abit vague through, write down a relationship that must hold for all of these. You could write down any number of things. The only thing I haven't tried is trying to work out the area of a portion of the object and then multiply it by n.
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    Thats the problem with step.

    I don't know how i managed to pass the AEA if i can't even do Step I.
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    (Original post by Square)
    I can think of quite a few relationships between R, r and n but I don't know what it's looking for exactly. Probably has something to do with the later parts.
    There's an exact formula relating R,r and n. You will need to do some trig: think of n as determining an angle (e.g. if n = 12 then the angle between circle centers is 30o) and try to find a right angled triangle...
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    Square, your first post, STEP 1 section, is a bit messed up. I did question 4 and you've put it as question 3 (also, there isn't even a question 4 option there).
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    Ooops, sorry.

    Cheers DFranklink.
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    are you on the right track now?
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    Blargh, still not entirely sure.

    I got:

    \displaystyle n=\frac{\pi}{\arctan(\frac{r}{R}  )}

    From considering a right angled triangle from the centre of a smaller circle, to the edge of a smaller circle, to the centre of the circle with radius R.

    \displaystyle\theta=\arctan(\fra  c{r}{R})

    \displaystyle\frac{2\pi}{\arctan  (\frac{r}{R})}=2n - If you think about it there will be 2 thetas per smaller circle. (badly explained)

    \displaystyle n=\frac{\pi}{\arctan(\frac{r}{R}  )}
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    and wherd that come from?
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    (Original post by Square)
    Blargh, still not entirely sure.

    I got:

    \displaystyle n=\frac{\pi}{\arctan(\frac{r}{R}  )}

    From considering a right angled triangle from the centre of a smaller circle, to the edge of a smaller circle, to the centre of the circle with radius R.

    \displaystyle\theta=\arctan(\fra  c{r}{R})

    \displaystyle\frac{2\pi}{\arctan  (\frac{r}{R})}=2n - If you think about it there will be 2 thetas per smaller circle. (badly explained)

    \displaystyle n=\frac{\pi}{\arctan(\frac{r}{R}  )}
    I think you've drawn/labelled your triangle wrong. Note that when r=R, you should have n = 6. (Get 7 pennies and try it...)
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    im already lost

    long day

    dont think ill be much help tonight
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    (Original post by DFranklin)
    I think you've drawn/labelled your triangle wrong. Note that when r=R, you should have n = 6. (Get 7 pennies and try it...)
    Okies, the pennies work, but its still not working on paper.

    Not resting until I get this one!
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    (Original post by generalebriety)
    Depends. Are you talking about 1987? No.
    To be honest, the difficulty of the STEP I 1987 questions is all over the shop. At a glance, Q1, Q4 are very easy questions (much easier than you'd expect to see on a current paper), while some of the others look fairly tough. I think the examiners were still feeling their way at this point, and so hedged their bets by having a couple of easy questions and a few tough ones.
 
 
 
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