Oxford MAT- Nov 4th 2015

Which bit of the solution are you not following?

I wish I could but this is beyond my expertise.

Wait for a purist/problem solver to arrive.

Pn+1 (Xn - Yn, Yn +Xn)

I don't follow how we arrive at the new coordinates; It's beyond vague (to me) on the Mark Sch.

I see you've answered this before and talk of a 90 deg. anticlockwise rotation. I understand that this gives (-y, x); but why are we doing this? why does it answer the question?

Because we've found out that (in terms of the final change in position), P_n is the same as a move of (x, y).

$P_{n+1}$ is P_n, followed by a 90 degree anticlockwise rotation, followed by doing P_n again (followed by a turn the other way to restore orientation, but that's not so important here).

A 90 degree rotation sends the vector (x, y) to the vector (-y, x) and so
doing P_n after a 90 degree rotation will be equivalent to a translation (-y, x).

Adding the two, we get

$(x_{n+1}, y_{n+1}) = (x_n - y_n, x_n + y_n)$

Hello TeeEm

https://www.maths.ox.ac.uk/system/files/attachments/test12.pdf - Question

https://www.maths.ox.ac.uk/system/files/attachments/websolutions12.pdf - Answer

Could you please explain question 5(vi)?

Why is there no partial credit for multiple choice? It is incredibly frustrating

Too ...much ...maths

Which bit of the solution are you not following?

Sorry I was late!
I see dfranklin has cleared the problem as always

Posted from TSR Mobile

Hey, I'm having some trouble formulating a solution to parts 3 and 4 for Question two on the 2010 MAT paper.

https://www.maths.ox.ac.uk/system/files/attachments/test10.pdf

I have had a look at the solutions however I'd quite like to find a different method which would be more understandable?

Hey, I'm having some trouble formulating a solution to parts 3 and 4 for Question two on the 2010 MAT paper.

https://www.maths.ox.ac.uk/system/files/attachments/test10.pdf

I have had a look at the solutions however I'd quite like to find a different method which would be more understandable?

When I did this paper my solution for part 3 was about (0.5,0.5) would be equidistant from 4 points, hence as r gets bigger the circle "eats" 4 points at once periodically, hence its a multiple of 4.
And for part 4 I talked about how the centre defined there cannot be equidistant to any points, hence as r increases the circle "eats" one point at a time, hence every value of k is achieved. My solution was obviously a lot more indepth but I don't have it to hand. Hope you get the idea and can make your own solution.

When I did this paper my solution for part 3 was about (0.5,0.5) would be equidistant from 4 points, hence as r gets bigger the circle "eats" 4 points at once periodically, hence its a multiple of 4.
And for part 4 I talked about how the centre defined there cannot be equidistant to any points, hence as r increases the circle "eats" one point at a time, hence every value of k is achieved. My solution was obviously a lot more indepth but I don't have it to hand. Hope you get the idea and can make your own solution.

I had the exact same idea but cannot find my solution. This is one of the hardest mat questions made i reckon, a beautiful one albeit.


Posted from TSR Mobile

Thankyou. I did indeed have a solution along those lines but was wondering if there was a more formal approach to expressing it? Not too sure how formal a solution they want for this question since some parts they seem to want more rigorously answered and other parts not.

If the radius is less than the radius required to reach the first 4 points around it, N would equal 0 - this still counts as a multiple of 4 right? as 0 x 4 = 0

Yes it would.


Posted from TSR Mobile

Thankyou. I did indeed have a solution along those lines but was wondering if there was a more formal approach to expressing it? Not too sure how formal a solution they want for this question since some parts they seem to want more rigorously answered and other parts not.

If the radius is less than the radius required to reach the first 4 points around it, N would equal 0 - this still counts as a multiple of 4 right? as 0 x 4 = 0

Yes but be careful when saying this, r cannot be 0 as defined in the question.

Oh no, I was stating that if r was 0.25, then surely it wouldn't meet the first 4 points. Would having a radius of 0.25 be viable?