People interested in mathematics???

Tatendrang said:
I suggest you pick up a college textbook. It's never too early to start.
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thanks for advice, but in that moment I would like to keep going in competitive olympiad style math. In the next year(11 grade) I'll start learning calculus and higher algebra
 
I read my Masters in mathematics at university. Now I tutor students from GCSE to university-level. Still have a passion for the subject but I do occasionally get bored by modern syllabuses. Here's a good one accessible for all levels:

Q1. Two circles of radius 1 go through each other's centres. What is the exact area of overlap?

Post answers in a spoiler so as not to ruin it for others!
 
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2pi/3-sqrt(3)/2
High school geometry, I hope I didn't mess up lol. The general case would be lot harder, probably. Might need polar coordinates.
 
Caligula said:
2pi/3-sqrt(3)/2
High school geometry, I hope I didn't mess up lol. The general case would be lot harder, probably. Might need polar coordinates.
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You got it! There is also the (still 2D) case of three circles of radius 1 going through each other's centres. It's not much harder than the problem above. But then it's interesting to ask why it's impossible to draw any more than 3 circles of equal radius that fulfil this criterion.

I actually never spent much time thinking about the general n-dimensional case. You mean like 3 spheres going through each other's centres? I'm sure there's some elegant answer... Polar coordinates seem like a good bet. 3Blue1Brown has a video on something similar, YouTube "The hardest problem on the hardest test".
 
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Another question a student asked me about the other day. From the Oxford entrance test. Harder than above but very satisfying.
qn2.png
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Sputnik I said:
You got it! There is also the (still 2D) case of three circles of radius 1 going through each other's centres. It's not much harder than the problem above. But then it's interesting to ask why it's impossible to draw any more than 3 circles of equal radius that fulfil this criterion.
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(pi-sqrt(3))/2. In order to have 4 circles we would need 4 points all the same distance apart and that's impossible. At least on a plane.
Maybe on a surface of a sphere it could be done?

Sputnik I said:
I actually never spent much time thinking about the general n-dimensional case. You mean like 3 spheres going through each other's centres? I'm sure there's some elegant answer... Polar coordinates seem like a good bet. 3Blue1Brown has a video on something similar, YouTube "The hardest problem on the hardest test".
Click to expand...
I actually meant arbitrarily large circles arbitrarily distant from each other. But going up in dimensions might be more interesting. I have had that 3b1b video on my Watch Later list for ages lol.
 
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i) a^(n+1)-b^(n+1)
ii) no because n^2-1 = (n-1)(n+1) and a product is a prime only if one of the factors is 1
iii) 2 and nothing else because n^3+1 = (n+1)(n^2-n+1) and you know the rest
I'll leave the rest for tomorrow. Thanks for these :)
 
Let n>=2. Consider the number (n! +k) where 2 <= k <= n. This is always divisible by k. But this is a sequence of (n-1) consecutive numbers, so we're short by one. Instead, for any fixed n, we could just change the sequence to ((n+1)! + k), where 2 <= k <= n+1. Each of these is again divisible by k giving a sequence of n composite numbers.

Examples:
n = 2: (2+1)! = 6. Sequence is 8,9
n = 3: (3+1)! = 24. Sequence is 26, 27, 28
n = 4: (4+1)! = 120. Sequence is 122, 123, 124, 125

Very enjoyable question! Was stuck in a boring online meeting when I saw it and couldn't stop scribbling :D
 
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Finishing Sputnik's problems:
iv) not prime. Using the equality in i) and the fact that 2015 is divisible by 5 i.e. 2015 = 5k where k is some integer we get 3^2015-2^2015 = (3^k-2^k)(3^(4k)+3^(3k)*2^k+3^(2k)*2^(2k)+3^k*2^(3k)+2^(4k))
v) no. The expression is greater than k^3 but not large enough for the next cube which is (k+1)^3=k^3+3k^2+3k+1.
deeroo said:
Show that there exists for every number n>=2 a sequence of n consecutive numbers such that all n numbers are composite(not prime)
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(n+1)!+2, (n+1)!+3, ..., (n+1)!+n+1
 
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