Oxford MAT 2013/2014

I'm struggling to make sense of this diagrammatically. Can anyone help?

Also, have I gone mad, or is this forgetting the little bit at the corner?

The second attachment is just wrong! You're correct, it DOES neglect the corner. As for the first, take an equation Ax+By=C. Think about how it'll look as you vary the signs of A and B between positive and negative. You're looking for the case where there will be a FINITE number of positive integer solutions, that is, a finite portion of the first quadrant will satisfy Ax+By<C. This immediately rules out positive slope of x axis. Negative slope can be achieved by keeping A and B both positive or by keeping them both negative. But in the second case, the region satisfying Ax+By<C is the upper region, so there are infinite solutions. In the first case, it is the lower region, so it has finitely many solutions. Now replace A, B by the suitable constants and set them positive. Does this make sense? (I don't remember the question properly, so I might have got something wrong. Please tell me if I messed up or didn't explain properly.)

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Makes perfect sense, thank you so much!

You're welcome!
One question, if you don't mind. Err, where is that second attachment from? Is that on any website?

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Also, have I gone mad, or is this forgetting the little bit at the corner?

Does anyone know where/if the mark scheme for the 2006 paper can be found? I can't seem to find it anywhere...

you are indeed correct
I calculated r=3-2root2

there isn't one but we discussed our answers on this thread yesterday

around here: http://www.thestudentroom.co.uk/showthread.php?p=45034414#post45034414

me too

i actually got (root2 - 1)/ (root2 +1), which when I put into my calculator gives me what you got.

yay, i just simplified it myself by multiplying and dividing by (root2 - 1)

how did you get that?

Hey guy's can someone help me on section A question H.

The answers say it's pi / 9 - 1/6 but I keep getting pi / 9 - root 3 / 12.

Can someone check where the error is in my method?

What I've done is found the sector, with radius 1 angle 2pi/3 (using inverse trig). Then I took away the triangle from it which was side length 1 and 1 and angle 2pi/3.

I then doubled this to find the shaded area, giving me 2pi /3 - root3 / 2.
I then divided by 6, the area of the rectangle.

Thanks

Does it matter if we do the exam in pencil or pen?

The square root of 2 (from the centre to the top right corner) is equal to 1 + 2r + the little bit.

Find the little bit in terms of r and then solve the above equation for r. I'm sure you can do it