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# Official TSR Mathematical Society Tweet

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1. Re: Official TSR Mathematical Society
guys you all possibly know this. but i have a problem on game theory. the thing is that it assumes none of the people involved in the game will work collaboratively with the other people involved. what if they do. are we sure that they will get the highest profit ( none goes to prison)?
What's the game?
2. Re: Official TSR Mathematical Society
not all the calculators have that ability. but out of the interest remember to use 3.1416 as pi. it helps accuracy.
thanks, it's just that i need to give some of my answers in terms of pi (eg 3[pi]/2)

so do you think i will have to use another calculator for the exam? (not something i really want to do since i am more used to the way my calculator works)
3. Re: Official TSR Mathematical Society
(Original post by poyyo)
thanks, it's just that i need to give some of my answers in terms of pi (eg 3[pi]/2)

so do you think i will have to use another calculator for the exam? (not something i really want to do since i am more used to the way my calculator works)
ok. if you want my opinion dont go for a new calculator. find your answer in radians and divide them by 3.1416( eg if you get 1.5708 radians divide it by 3.1416 and what youll get is 1/2) then times it by pi (e.g. 1(pi)/2) but if you want the method of working with a fancy calculator ill get one of my friends to post it tonight on this thread. hope it helps
by the way tell me what exam you have. itll be better.
Last edited by mashmammad; 29-05-2012 at 15:58.
4. Re: Official TSR Mathematical Society
(Original post by sputum)
What's the game?
ok this is the game theory:
two people have stolen something, police can not arrest any of them if they dont confess.if one of them confesses the other one is arrested and jailed for one year, and the one that has confessed is freed. its the same argument for the other person. but if they both confess each will be prisoned for 6 moths. what do you think they will do?
5. Re: Official TSR Mathematical Society
ok this is the game theory:
two people have stolen something, police can not arrest any of them if they dont confess.if one of them confesses the other one is arrested and jailed for one year, and the one that has confessed is freed. its the same argument for the other person. but if they both confess each will be prisoned for 6 moths. what do you think they will do?
Cooperatively they will shut the something up
6. Re: Official TSR Mathematical Society
(Original post by sputum)
Cooperatively they will shut the something up
thats exactly what they dont do. look the first one will think:
if i shut up he'll confess and ill be prisoned, if he doesnt however, if i confess and he'll be prisoned and ill be free, how ever if he has already confessed Ill confess so that my sentence is halved. so for these reasons the first one confesses. the second guy will think the same way, so he'll confess as well, so they'll both end up in prison, for a period of 6 months. its called john nash's game theory.
7. Re: Official TSR Mathematical Society

guys you all possibly know this. but i have a problem on game theory. the thing is that it assumes none of the people involved in the game will work collaboratively with the other people involved. what if they do. are we sure that they will get the highest profit ( none goes to prison)?
if they work collaboratively they both 'win'
unless I'm not understanding something about the question
8. Re: Official TSR Mathematical Society
(Original post by sputum)

if they work collaboratively they both 'win'
unless I'm not understanding something about the question
yes if they do work collaboratively. but the whole point is that people are selfish and they go for personal profit.
and end up with something less than what they can get
9. Re: Official TSR Mathematical Society
yes if they do work collaboratively. but the whole point is that people are selfish and they go for personal profit.
and end up with something less than what they can get
well we can all take something good from it then
10. Re: Official TSR Mathematical Society
ok. if you want my opinion dont go for a new calculator. find your answer in radians and divide them by 3.1416( eg if you get 1.5708 radians divide it by 3.1416 and what youll get is 1/2) then times it by pi (e.g. 1(pi)/2) but if you want the method of working with a fancy calculator ill get one of my friends to post it tonight on this thread. hope it helps
by the way tell me what exam you have. itll be better.
oh wow i just tried it and that's much better, thank you so much!!!!!!!!!!!!!!!!!!!!!!!! i have aqa maths core 3 and core 4 exams by the way

honestly thank you really i am so grateful i was freaking out before about this but really thanks!!!!
11. (Original post by poyyo)
oh wow i just tried it and that's much better, thank you so much!!!!!!!!!!!!!!!!!!!!!!!! i have aqa maths core 3 and core 4 exams by the way

honestly thank you really i am so grateful i was freaking out before about this but really thanks!!!!
No worries

This was posted from The Student Room's Android App on my HTC Wildfire
12. Re: Official TSR Mathematical Society
BMO1 1966

A regular polygon ABCD.... is such that .

How many sides does the polygon have?

I'm hoping to see a better solution than mine. I ended up using De Moivre's theorem which seems inappropriate for BMO1.
13. Re: Official TSR Mathematical Society
(Original post by BabyMaths)
BMO1 1966

A regular polygon ABCD.... is such that .

How many sides does the polygon have?

I'm hoping to see a better solution than mine. I ended up using De Moivre's theorem which seems inappropriate for BMO1.
There is no need for anything other than basic geometry here...you should be able to express AC and AD in terms of AB using trig. If that seems a bit like a slog, you might make use of the fact that if this is true in a regular polygon, in particular, a certain triangle, consisting or sides with length AB,AC,AD has special properties that you can analyse to get the answer. (this latter method might not work...my brain isn't good enough to run through the whole solution in my head, but I've got it down to something that looks easy to solve)
14. Re: Official TSR Mathematical Society
Hmmm,

With I arrived at the equation .

That much was easy. I then disposed of the possibilities n=3, n=4 and n=6.

It turns out that there are 7 sides.
15. Re: Official TSR Mathematical Society
Anyone interested?

I gambled on looking at n=7 rather than n=5.

Letting De Moivre's theorem gives

Spot the factor c+1 and get

and factor a bit more

etc.

But is there an easier way?
16. Re: Official TSR Mathematical Society
(Original post by TheMagicMan)
There is no need for anything other than basic geometry here...
Any chance you could post this basic geometry solution. I confess I'm not too good at geometry.
17. Re: Official TSR Mathematical Society
(Original post by BabyMaths)
Any chance you could post this basic geometry solution. I confess I'm not too good at geometry.
neither am I...it was always my great olympiad weakness

I'll have a go later at it. There is definitely a simple solution as I've done the question before.
18. Re: Official TSR Mathematical Society
Hey guys, i've got a very simple percentage I need to work out associated with my A2s - it isn't anywhere near as complex as A2 mathematics, but I still can't do it haha (arts student here)

BASICALLY in one of my A levels:
I achieved 83% on the section of the A level worth 50%
I achieved 87% on the section of the A level worth 20%

How much must I get in the final section, woth 30%, to achieve 80% overall?

19. Re: Official TSR Mathematical Society
(Original post by swbp)
...
Are you interested how to work it out, or just in the actual number?
If these are rough percentages, anything more than should do.
20. Re: Official TSR Mathematical Society
Are you interested how to work it out, or just in the actual number?
If these are rough percentages, anything more than should do.
No, I just needed a figure. Thanks !