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Re: 2012 HSC MX2 Marathon Analysis ftw indeed, but symmetry is of universal importance in mathematics. Side note: I didn't like the diffgeom course so much because the course was nearly completely embedded in R^3. It does make the study of differential geometry on manifolds easier to...

The solutions cluster near the line 1km south of the north pole and near the south pole. I intended the question to mean that P could not have been within 1km from the north pole, for then the man would not have been able to walk 1km north.

(True, by spherical coordinates I more meant in terms of circles of latitude, didnt want to give away the form of the solution though.) Re you response, the circle of latitude that is 0.5km north of the equator wont be a solution, only the one 0.5km south. But the main point of the question is...

Re: 2012 HSC MX2 Marathon Umm no, largarithmic is correct...the integral does not converge. The hypotheses of L'Hospitals rule are not even satisfied here.

(A variant with exact solutions can be found by removing the final westward leg of his journey and posing the same question I think.)

(And the equation describing the solutions can be found using only high school geometry, its just a little tricky to visualise naturally. I don't think it is possibe to find a solution to this equation of a nice form...unless I am mistaken and wolfram alpha is also lying to me. Although it is...

Essentially carrotsticks, as long as you are aware that these lines correspond to the lines of constant latitude/longitude. (Don't entirely get your notation but its not terribly important.)

(His journey is not a 'square' technically, as the E-W segments are not geodesics on the sphere, they are lines of constant latitude. So in this case the angles at the 'vertices' ARE pi/2. But yes, the fact that we are working on a sphere is why the question is nontrivial.) I am fairly...

Thought up this question: A man stands at some point P on the surface of a spherical planet of radius 1000km. He walks 1km north, 1km east, 1km south, and 1km west and ends up exactly where he started. What are the possible locations of P? (Use spherical coordinates to state your answer.) Have...

Re: 2012 HSC MX2 Marathon A simpler solution to the integration question: https://docs.google.com/open?id=0B5x34d2OjtrQWlgwblItMGFUc0MzYUw3eG4zLWNodw

Re: 2012 HSC MX1 Marathon Wouldn't necessarily call any of the school courses "fun" but at least 4U had the potential for some of the questions to be interesting. Higher mathematics is a lot more "fun" because there is room for creativity and originality.

That one wasn't particularly nice. I meant more things like basic applications of Fermat's little theorem, basic results about primes etc.

Minimal polys are not needed to prove a "conjugate root theorem" of the sort used here are they? Define \mathbb{Q}[\sqrt{5}]:=\{a+b\sqrt{5}:a,b\in\mathbb{Q}\}. This subset of the reals is closed under addition and multiplication. (It is in fact a subfield of \mathbb{R}). Define the...

Some elementary number theory, it is one of the most attractive and accessible subjects for a high school student.

Maybe. As far as I know its pretty hard to do these days, and does not give you that much of an edge in return. Poker can be very profitable though, as casinos/online rooms don't care how much money you win off OTHER people.

For me, pure > applied. Statistics doesn't really feature on the scale since it too has applied AND theoretical components. Of the major branches of pure mathematics my favourite is analysis, especially complex analysis and harmonic analysis. Abstract algebra, number theory and certain aspects...

Re: 2012 HSC MX2 Marathon $Assume that $P(z)$ is not the zero polynomial.\\Suppose $\alpha\neq 0$ is a root of $P(z)$ with minimal argument in the range $0<\arg(\alpha)<2\pi$.\\ Then taking the principal square root of $\alpha$, we have $P(\alpha^{1/2})^2=P(\alpha)=0$ and $P(\alpha^{1/2})=0$...

pure ftw. though some aspects of applied are reasonably cool...like game theory.

Re: 2012 HSC MX2 Marathon An easy polynomials question: $Let $x_1,\ldots,x_{n+1}$ be $n+1$ distinct real numbers. Let $P(x)$ and $Q(x)$ be real polynomials of degree $n$ such that $P(x_j)=Q(x_j)$ for $j=1,\ldots,n+1.$ Prove that $P(x)=Q(x)$ for all real $x.

For Galois theory it is probably best to first take the 2nd year algebra course, but Metric Spaces is quite manageable without prereqs I think...