MAT Prep Thread 2018

Assuming you mean (-1, 1):

Clearly y = 1, x = -1 satisfies the equation so whatever the value of $\theta$, the line must pass through (-1, 1). The gradient is $\sin \theta$ which takes values between 0 and $2\pi$ so the gradient follows the sine curve and will result in the line rotating by up to $2\pi$ as $\theta$ varies.

If you're struggling to visualise this then try this:

https://www.desmos.com/calculator/r9jmooqxlz

Shouldn't the gradient be $tan(\theta)$?

Yes you’re right.

@Notnek , shouldn't the answer be 1? I don't get the symmetry argument. I tried to solve it , and ended up with only 1 answer.

Since theta ranges from 0(inclusive) to 2pi, for every value of tan theta, there are 2 corresponding values of theta. As the area is the function of tan theta, it follows that there are 2 solutions.

@Notnek , shouldn't the answer be 1? I don't get the symmetry argument. I tried to solve it , and ended up with only 1 answer.

@Notnek , shouldn't the answer be 1? I don't get the symmetry argument. I tried to solve it , and ended up with only 1 answer.

https://www.desmos.com/calculator/loqezcbzzr

Try moving the slider from 0 to 2pi. You get an identical change between 0 and pi and between pi and 2pi. So A achieves it's maximum once between 0 and pi and by symmetry it must achieve it again a second time between pi and 2pi.

But how do you come up with this in the exam?
When playing with desmos and visualizing the situation it makes sense.
But I had no idea how to visualize the question in the exam.

Just realised that some of my comments might seem a little rude. I am sorry if this is the case. I am just very confused on this problem for a very long time.
But really appreciate the help @Notnek

But how do you come up with this in the exam?
When playing with desmos and visualizing the situation it makes sense.
But I had no idea how to visualize the question in the exam.

Just realised that some of my comments might seem a little rude. I am sorry if this is the case. I am just very confused on this problem for a very long time.
But really appreciate the help @Notnek

But how do you come up with this in the exam?
When playing with desmos and visualizing the situation it makes sense.
But I had no idea how to visualize the question in the exam.

Just realised that some of my comments might seem a little rude. I am sorry if this is the case. I am just very confused on this problem for a very long time.
But really appreciate the help @Notnek

Your post wasn't rude - don't worry.

The key part of this method is to realise that the line is doing a full rotation about (-1,1) as theta varies. Then if you make a sketch it's not hard to see that the maximum will occur as shown in the diagram in the solution. Then this picture has to occur twice since the line is doing a full rotation, so the answer is 2.

Because I'm not a MAT/STEP expert it's hard for me to give more advice on how best to approach this question. @DFranklin or @RichE may be able to help? It's question G in the 2017 paper.

I'll post a solution.

I am terrible at explaining, but i hope this could help somewhat.
for the equation of the line you can notice that when y=1 and x=-1, LHS=RHS=0 regardless the value of theta, meaning the line will cross (-1,1) at any given value of theta.
Also we know the other equation is a circle and that (-1,1) definitely lies within the circle (you can prove by plugging the number in)
Lastly, by that explanation, we can tell that if there is only one value of tan theta for the idea of 'minimum area'
By tan graph, we know from 0 to 2pi, there are two solutions for one value. ( hope that makes sense)
(My guess for the idea of symmetry is that the area is minimised when the shape is symmetrical, maybe)

Thank you



Thanks mate



Thanks. Oh and btw hope your exams went well .

Your post wasn't rude - don't worry.

The key part of this method is to realise that the line is doing a full rotation about (-1,1) as theta varies. Then if you make a sketch it's not hard to see that the maximum will occur as shown in the diagram in the solution. Then this picture has to occur twice since the line is doing a full rotation, so the answer is 2.

Because I'm not a MAT/STEP expert it's hard for me to give more advice on how best to approach this question. @DFranklin or @RichE may be able to help? It's question G in the 2017 paper.

But how do you come up with this in the exam?
When playing with desmos and visualizing the situation it makes sense.
But I had no idea how to visualize the question in the exam.

Just realised that some of my comments might seem a little rude. I am sorry if this is the case. I am just very confused on this problem for a very long time.
But really appreciate the help @Notnek

So here's my thought process as I go through this:

Rewrite $(y-1) \cos \theta = (x+1) \sin \theta$ as $\dfrac{y-1}{x+1} = \dfrac{\sin \theta}{\cos \theta}$. Now I immediately recognized this as a line through (-1, 1) with angle theta (*) (note that I don't even need to know whether this is measured from horizontal, vertical etc at this stage).

Quick sketch persuades us there's only one position for the line that maximizes the area, and of course it's the same line whether we go "backwards" or "forwards", so there are two choices for theta.

(*) Supposing I hadn't realised this: If I write Y = y-1, X = x+1 and rearrange I get $\dfrac{Y}{\sin \theta} = \dfrac{X}{\cos \theta}$. Writing t for the quotient, I get $Y = t \sin \theta, X = t \cos \theta$, which hopefully makes it obvious.

As you would be applying post-qualification, you will have had an extra year's education than most candidates. For that reason, you would be expected to do a bit better in the MAT than others.

Gavin

Hello.
Sorry for asking this after such a long time. But I got back my IB grades and exceeded the Oxford offer so I am going to be reapplying this year. I am fully aware that I am expected to perform better this time around in comparison to other applicants in the MAT. But I just wanted to know whether there is any upper band score that guarantees an interview?

I know that the Physics department has a cutoff score for the PAT and so was wondering whether there is anything similar on the MAT.
As far as the graph goes for last year I noticed that scores above 80 guaranteed an interview, but it looked weird that some applicants with scores between 75 - 80 were not shortlisted. Could this be because of gap years?

Another part where I was looking for a bit of clarification was: when Professor Gavin Lowe mentioned that gap year applicants are expected to perform a bit better is it possible to quantify what this would mean for the year 2017 for instance. The average score amongst applicants shortlisted was around 67 so would a gap year applicant be expected to score something around 75? 80? to be shortlisted.

If Its of any help, my IB grades are below if that helps in any way.

Any help is appreciated .

No need for apology. There certainly is no hard cut-off upper band score, for a variety of reasons:

MAT scores vary a little from year to year (depending on the difficult of the test and quality of the applicant field) and consequently an absolute score would never be used in any such way as you describe. At best this would need to be a score relating to the general distribution of scores or the MAT scores attained by the top fraction.



Consequently I would be surprised if Physics are rigidly tied to a PAT score.

Also, with the MAT there is considerable discretion with applications in light of other information. Often applicants' information paints a coherent picture, but there are more than a few occasions where contextual information is important.

A score of 75-80 on the MAT would in general have earned you an interview that you, but would you have felt the same if one of those applicants was already in the second year of a degree elsewhere, or the UCAS personal statement did not mention maths at all and went continually on about actuarial studies? Surely these are reasons for tutors to gauge the context of the application. I don't think, though, that gap years would raise the same concerns.



As Gavin wrote the original comment then I will wait for him to reply, but given my comments about the holistic nature of our treatment of applications I trust you won't be expecting a specific numerical value in any sense.

Hello.
I was wondering if Professors @gavinlowe and @RichE could help me clarify a couple of questions .

Sorry for asking this after such a long time. But I got back my IB grades and exceeded the Oxford offer so I am going to be reapplying this year. I am fully aware that I am expected to perform better this time around in comparison to other applicants in the MAT. But I just wanted to know whether there is any upper band score that guarantees an interview?
I know that the Physics department has a cutoff score for the PAT and so was wondering whether there is anything similar on the MAT.
As far as the graph goes for last year I noticed that scores above 80 guaranteed an interview, but it looked weird that some applicants with scores between 75 - 80 were not shortlisted. Could this be because of gap years?

Another part where I was looking for a bit of clarification was: when Professor Gavin Lowe mentioned that gap year applicants are expected to perform a bit better is it possible to quantify what this would mean for the year 2017 for instance. The average score amongst applicants shortlisted was around 67 so would a gap year applicant be expected to score something around 75? 80? to be shortlisted.

If Its of any help, my IB grades are below if that helps in any way.

Any help is appreciated .