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# Mathematics Applicants 2012 watch

1. (Original post by dnumberwang)
I've got an email from Imperial too, just an hour ago
imperial do exist afterall! imperial aint as welcoming as the other unis
2. (Original post by anyone_can_fly)
not exactly.

I'd say I am 40% troll, 60% human. So you could argue that I'm actually a hybrid.
3. (Original post by ilyking)
not exactly.

I'd say I am 40% troll, 60% human. So you could argue that I'm actually a hybrid.
In some ways similar to your associate, the dreaded man-bear-pig? Although remembering he is 50% man, 50% bear and 50% pig.
4. (Original post by Aristotle's' Disciple)
In some ways similar to your associate, the dreaded man-bear-pig? Although remembering he is 50% man, 50% bear and 50% pig.
Manbearpig is greater than 1! he is 1.5!
5. (Original post by Reminisce)
Clue:

Spoiler:
Show
You should be aware of the angles of an equilateral triangle... Now, what do you know of the properties of the tangent trigonometric function?

Spoiler:
Show
One side has a gradient of one. Another has a gradient of tan(60) since the tangent is the opposite side divided by the adjacent - hence the gradient and the other is the negative of that (think about shape and how you would use tan to find the gradient). Then multiply.
The answer is, 2(1 - sqrt 3) ? I've tried using your method, by letting one side of gradient to be 1, so the other one should be -sqrt 3, and one more is 0.

I don't know whether it works, but I add all gradients (0, 1, -sqrt3) by 1 (since I couldn't multiply everything by 0 and get 0), and multiply all resultant gradients which would yield 2(1 - sqrt 3).

tan (m1 - m2) = (m1 - m2)/(1 + m1m2),
since m1 - m2 = 60, tan 60 = sqrt 3 = (m1 - m2)/(1 + m1m2).

Do the same for m2 and m3, m1 and m3, and by using a little bit of algebra, you can find m1m2m3?

I'm trying to use method above but... seem to be stuck. Yesterday I was too tired so I couldn't think of the logic (it was 3am).
6. Did anyone get offers for Warwick G100 or acknowledgement from Imperial G100?
7. (Original post by Nicholasng925)
The answer is, 2(1 - sqrt 3) ? I've tried using your method, by letting one side of gradient to be 1, so the other one should be -sqrt 3, and one more is 0.

I don't know whether it works, but I add all gradients (0, 1, -sqrt3) by 1 (since I couldn't multiply everything by 0 and get 0), and multiply all resultant gradients which would yield 2(1 - sqrt 3).

tan (m1 - m2) = (m1 - m2)/(1 + m1m2),
since m1 - m2 = 60, tan 60 = sqrt 3 = (m1 - m2)/(1 + m1m2).

Do the same for m2 and m3, m1 and m3, and by using a little bit of algebra, you can find m1m2m3?

I'm trying to use method above but... seem to be stuck. Yesterday I was too tired so I couldn't think of the logic (it was 3am).
Yeah that's right (the second part), I amended my assumption in the first post to the more general one in the post after if you saw.
8. (Original post by mpc1)
Did anyone get offers for Warwick G100 or acknowledgement from Imperial G100?
nope....but i emailed warwick to change my course from g100 to g103
9. (Original post by Reminisce)
Yeah that's right, I amended my assumption in the first post to the more general one in the post after if you saw.
Alright, thank you so much!
10. (Original post by Nicholasng925)
The answer is, 2(1 - sqrt 3) ? I've tried using your method, by letting one side of gradient to be 1, so the other one should be -sqrt 3, and one more is 0.
This must be wrong. If one side has a gradient of 1. the another side does not have a gradient of 0 as this implies a 45 degree angle

And even so, if one gradient is zero, the the product is zero
11. (Original post by Nicholasng925)
Interview in 2 more days, with Magdalene College's Doctoral of Studies Stuart Martin, and written test in 4 days!!! *Panicked*

On some random note, anyone minds to solve this question:

Find the product of gradients (m1m2m3) of an equilateral triangle. Hint: Equilateral triangle is the clue. If possible, please show the working cause I'm lost now!
This question is impossible. If one side runs up the x axis the product of the gradients is obviously zero. However it is easily possible to fix the triangle such that the product of the gradients is not 0
12. (Original post by TheMagicMan)
This question is impossible. If one side runs up the x axis the product of the gradients is obviously zero. However it is easily possible to fix the triangle such that the product of the gradients is not 0
Yeah between the rotation of 90 and 0 (where 90 would have a vertical side which creates a problem and 0 where one side is horizontal), the gradient should be non-zero.

Actually, a rotation of 30 degrees would cause one side to be vertical then a rotation of 60 would cause another side to be horizontal...

EDIT: so yeah, the product should be non-zero at angles other than those between 90 and 0.
13. You'll have to do the question in terms of variables, I suspect.
14. (Original post by Reminisce)
Yeah between the rotation of pi/2 and 0 (where pi/2 would have a vertical side which creates a problem and 0 where one side is horizontal), the gradient should be non-zero.
Exactly. If you introduce a new variable theta which is the angle of the 'base' above the x axis, then it should be realtively elementary using trig and addition formulae to get an explicit formula in terms of theta
15. (Original post by mpc1)
Did anyone get offers for Warwick G100 or acknowledgement from Imperial G100?
I got an acknowledgement for G104
16. With regards to the question about gradients of lines, perhaps vectors could simplify matters...
No offers
17. (Original post by TheMagicMan)
Exactly. If you introduce a new variable theta which is the angle of the 'base' above the x axis, then it should be realtively elementary using trig and addition formulae to get an explicit formula in terms of theta
Explicit formula, yes. But not sure what form they are expecting the final answer to be, and don't think simplifying the double angles, etc. will give a "nice " result.
18. (Original post by twig)
Explicit formula, yes. But not sure what form they are expecting the final answer to be, and don't think simplifying the double angles, etc. will give a "nice " result.
I would be expecting it to be in the form or something of that ilk

I get which is not too complicated as a formula
19. (Original post by mpc1)
Did anyone get offers for Warwick G100 or acknowledgement from Imperial G100?
I got an acknowledgement for Imperial G100 yesterday. I haven't heard from Warwick for G100 yet, but my friend, also for G100, got an offer last Thursday
20. (Original post by TheMagicMan)
I would be expecting it to be in the form or something of that ilk

I get which is not too complicated as a formula
Yep, I got that, not that complicated but not very neat either...

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