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Did anyone get anywhere with the simplification?

However, I think I have my own solution to the problem. I will post it very soon, once I have checked it

Wel it's optional.....to a degree. In the way that, you don't get to choose whether or not you do calculus, but you can choose whether or not you do maths as a whole after what will be...year 10 (i believe)(when you're about 15/16).

Those who choose to continue with the subject get taught the same course regardless of ability, ie calculus.

The way it work is that everyone in the year sits what is called a Standard Grade in maths, then the next level - Higher Maths - is optional. If you pass that exam then you can either choose to stop or continue on and do Advanced Higher Maths in your last year. It is Higher and Advanced Higher which include the calculus etc.. but obviously Advanced Higher is much harder than Higher and Higher than Standard Grade.

Do you understand my ramblings?

I am not sure, I can only assume that it is from when he passed the jogger.

Yes, that's what we are doing this year and have done last year...

basic and further integration (inverse trig functions, partial fractions and integration, integration by parts....),

the same with differentiation (implicit differentiation, parametric equations, logarithmic differentiation....)

complex numbers,

sequences and series,

proofs,

etc.

Yes, we have done area under a curve in the first integration unit.

Sorry if i rambled a bit, but just thought i'd explain some of the topics we have done.

For learning everything you know from books some of your solutions are excellent. Even if you had gone to school i'd be impressed.

As for the problems, the sphere one was one my teacher showed me. However, the rather difficult logic/story problem about the walkers came from a small local maths paper copied for me by my teacher. So unfortunately they didn't come from a book as such. Apologies

Nope sorry, I'm from sunny (?) UK, Scotland to be exact. (98.6% of them are maths idiots also)!

The UK's maths system is also too slow, I've taught myself most of the complex stuff I know that is above the level at which i'm working, from various internet sites and textbooks.

"So your in trig then? Or did you get AP and get calculus?". I don't really understand that bit since i'm not familiar with how the courses in America work. We just tend to do one broad course covering every topic at some point.

There you go mikau, I finally posted an introduction . At the moment I am in "year 12" (is that what its called in America?) - the last year of high school. I am only 17, but may do maths at university next year. Therefore I will be studying high school maths at the moment, hence the reason why I was pleased to see irspower's alternive proof for my spheres question - becuase it involves maths I don't currently know.

Hope that helped

**Graeme**- Replies: 8

Hi, i have recently joined the forum.

I am Graeme, as you can see from the name . I am 17, and in last year of high school, or the twelfth grade (correct?) as the Americans call it.

I enjoy maths a lot and possibly hope to do it at university for a career in maths.

Thats about all.

Hope that helped, mikau

**Graeme**- Replies: 1

Here's a new one for you maths brains to puzzle over Though I'm sure it's not TOO difficult.

In the diagram linked, the square has two of its vertices on the circle and the other two lie on a tangent to the circle. What is the ratio of the area of the square to the area of the circle?

irspow wrote:

I just worked the now easy problem in reverse.

if R = radius of bowl and r = radius of ball and C = number of balls

C = 1; R = 2r

C = 2; R = r(1 + √2) ≈ 2.414r

C = 3; R = r(1 + √(7/3)) ≈ 2.527r

C = 4; R = r(1 + √3) :asmp 2.732r

C = 5; R = r(1 + √[(4sin²54°)/(sin²72°) + 1]) ≈ 2.973r

C = 6; R = r(1 + √5) ≈ 3.236r

C = 7; R ≈ 3.512r

C = 8; R ≈ 3.798r

In general for this equation;

R = r(1 + √((4sin²a / sin²b) + 1))

where a = [180 - (360/C)] / 2 and b = 360/C and C = number of balls.

Could you please explain how you worked out this formula. Apologies for the trouble but I can't really work it out.

Maybe it's too early and i'm tired

irspow wrote:

Once again, I show my stupidity....I couldn't figure out how to do it in two dimensions, much less using a single sphere, all the while making it much more difficult.

Thanks for the proof and the visualization. Now I can see exactly how Graeme was looking at the problem.

Graeme, sorry for any inconvenience that I may have caused you.

No problem at all, I appreciate all sorts of approaches to problems. Since I am only young, I haven't even touched on the more complex maths, so alternative approaches are always interesting.

mikau wrote:

Come to think of it, my not noticing I could use graeme's method was so idiotic its unreal. My equation and graph skill is much higher then my geometry. In my oppinion the simplest solution is the best and graeme's was the simplest. Hats off to you graeme!

Great problem!

Thank you very much mikau.

mikau wrote:

Maybe. I'm not sure where you got that formula.

I think the reason I didn't see the obvious solution was I did not assume the line drawn from the small circle to the edge of the large, had the angle as the line from the center of the large circle to the point of tangency.

Lets see if I can prove it.

(edit) dang. It seems obvious enough but I can't seem to prove it. I'll keep working on it but maybe you guys can get it before me.

Prove this!

http://i21.photobucket.com/albums/b299/ … raeme2.jpg

If a small circle B is inside a large circle A, and circle B is tangent to circle A at point C, prove the line drawn through point A to point B passes through the point C. (the point of tangency). Point A and point B are the centerpoints of circles A and B.

I have an explanation/proof for the line and the points for spheres, however I presume it works in principle for circles also.

Whenever two spheres touch each other (one inside the other or outsides touching), their point of contact will be on the line through their centers. You can see this intuitively if you simplify the problem and picture one ball at the bottom of the bowl--clearly the contact is at the very bottom of the bowl. (You can prove this rigorously by rotating the bowl and sphere about the line connecting their centers. The bowl and

sphere will remain in place because of their symmetry, so their point of contact must as well, meaning it is on the line of symmetry, which only touches the bowl at the bottom). Back to this more complex case, if you froze the spheres in place in the bowl and rolled the bowl so that you look straight down through its center and through the center of one of the balls, it looks exactly like the one-ball case (with two extra balls in the way).

Thank you very much irspow for your kind words and help with this.

I will post pictures of my possible solution as soon as possible (within 24 hours hopefully) and hopefully it can clear things up about my idea.

Thanks again

Oh I can assure you that I didn't make this one up off the top of my head. This came from a small maths paper and I will be waiting in anticipation for the "solution".

I know exactly how you feel, I have spent a long time on this also to no avail. However, I am only 17, so I thought that I would open it up to the more qualified mathematicians on here who may have been able to offer an approach I wasn't familiar with.

this doesn't sound like a particularly great question

I finally became a member lol. Now irspow can find out who I am

Anyway, thanks for any help with my solution i posted

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