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  1. seanieg89

    Fun problem!

    For a regular n-gon I am unsure of your definition of this region, as we we can draw several circular arcs containing two vertices centred at any given vertex... Do you simply mean the region that consists of all points with distance at most one from every vertex?
  2. seanieg89

    Fun problem!

    I got: (1-\sqrt{3})+\pi/3 for the square.
  3. seanieg89

    Irrationality question.

    Yep, for finite repeating strings a GP approach will work or the old (10^k)x-1 trick to obtain a number with finitely many nonzero digits to the right of it's decimal point. I think the harder/more tedious thing is showing that this is a NECESSARY condition for rationality. I can think of a way...
  4. seanieg89

    I guess we would both score above the average that year then! :).

    I guess we would both score above the average that year then! :).
  5. seanieg89

    cheers for the rep :).

    cheers for the rep :).
  6. seanieg89

    Ah cool, they are essentially the same :). Nice question, where's it from?

    Ah cool, they are essentially the same :). Nice question, where's it from?
  7. seanieg89

    thanks for the rep :). is that the way the question is meant to be done?

    thanks for the rep :). is that the way the question is meant to be done?
  8. seanieg89

    Difficult Complex Numbers Question

    $By the fundamental theorem of algebra, we may factorise $P$ into real monic irreducible quadratic and linear factors.$\\$As $|P(i)|<1$, we have that at least one of these factors has modulus less than one at $z=i$. From the triangle inequality, this is impossible for monic linear factors so we...
  9. seanieg89

    Difficult Complex Numbers Question

    Cool, will have a crack at it :).
  10. seanieg89

    Difficult Complex Numbers Question

    Or worse, a constant polynomial P(z) = c, with 0 < |c| < 1.
  11. seanieg89

    Difficult Complex Numbers Question

    Are you sure that is true? Consider the polynomial P(z)=z/2...
  12. seanieg89

    Inequality.

    No, AM-GM is the only essential step really. Spotting a neat factorisation isn't entirely necessary, althought it makes the solution look nicer. One could also do the question like this, just using AM-GM and grouping suitable terms together: (a+b+c)\geq 3/2, abc\leq 1/8$ by AM-GM$.\\$So...
  13. seanieg89

    Inequality.

    Asianese, your solution doesn't quite work. There are a few inequality signs the wrong way around which ruin your idea for proving the inequality. It can be fixed but the entire second half of your argument will look different. Good solution carrotsticks, essentially the same as mine.
  14. seanieg89

    Inequality.

    a,b,c\in\mathbb{R}^+$ satisfy: $(a+b)(a+c)(b+c)=1. \\$Prove that: $ab+ac+bc\leq 3/4.
  15. seanieg89

    Irrationality question.

    Related: Prove or disprove that a real number is rational if and only if its decimal expansion eventually consists of a repeating string.
  16. seanieg89

    Irrationality question.

    "You can find your phone number in any irrational number" Prove this statement or provide a counter-example.
  17. seanieg89

    Integration question

    I would think thats WHY abs value signs would be necessary. Say you were integrating over an interval where tan is negative...I remember Cambridge not doing this though so I don't know what the BOS expect.
  18. seanieg89

    Question on Prime Numbers

    My logic: -If either n or 11n+32 has a factor d with |d|>1, then 1<|d| is a positive divisor of n(11n+32). -n(11n+32) is prime => |d|=n(11n+32) => n(11n+32) divides either n or 11n+32. This means that one of the two factors must be +-1. -Check the few values of n which make one of the...
  19. seanieg89

    interesting dilemma

    Have to be more explicit about what the infinite power tower means. I think the standard meaning of such a notation is as follows: Let S(n) be the sequence defined by: S(1)=x, S(n+1)=x^S(n). x^x^x^x^... is lim (n->inf) S(n) provided this limit exists. Under this interpretation, known as...
  20. seanieg89

    interesting dilemma

    No, same issue.
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