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1. A circle is not a polygon, it doesn't make sense to talk about an infinitely sided polygon. 2. How can you "logically deduce" that the circle of circumference 1 is the unique curve that maximises A ? This is in fact a harder problem than the one you are trying to prove!

Had another look at this problem. There are some things that are a little tricky to rigorously prove, but here is the outline. 0. Use a compactness argument to show that there must exist a maxima of A/L^2 over the space of n-gons. 1. Show that the max of A/L^2 over all n-gons increases as n...

I own about 0.1BC atm but I used to have a fair bit more. I don't know enough about finance to speculate much but as a consumer I do like the idea of a decentralised anonymous currency and I hope it catches on more than it already has.

Cool, I thought the questions were a little too well-crafted to be asked by someone who didn't have a pretty good idea of their answers.

It's mostly just that notation for the situation is a little cumbersome I guess. Just keeping the idea "the derivative of the inverse is the derivative of the inverse" is good enough to write down any formula you like for it. I will leave these questions for a little while to give others an...

Well the inverse function theorem is more the statement that if a function is continuously differentiable in a region and has nonzero derivative at a point (or nonsingular Jacobian if we are dealing with higher dimensions), then the function with a small domain containing this point has an...

This fact is true whenever a,b,c,d are rational numbers and b,d have irrational square roots. (So b,d > 0. If b or d were negative, then this fact is still true but for different reasons.) Assume a is not equal to c, which implies that b must also be unequal to d. Then...

Well I don't mean doing illegal things, I just mean that if h,g can be zero in your problem, then you will get several solutions. (Or something like that.)

1+\pi = 2+(\pi-1) are two ways of writing the same number that have different "rational and irrational parts". How would you define the rational part of a real number unambiguously?

Well, that's cool if people state that these things are nonzero in the hypotheses of the working but it completely changes the situation. One solution can become infinitely many.

Ignoring potential division by zero problems of course.

Re: HSC 2014 4U Marathon Ah okay, I thought it might be this. This is actually a kind of similar idea to how one proves that the definition of a Riemann integral as a limit of Riemann sums makes sense and is unambiguous. But yeah, any argument you construct using integrals is implicitly using...

Re: HSC 2014 4U Marathon Huh, what do you mean by this? My solution is also based on Riemann sums. S=\sum_{k=1}^n \frac{1}{\sqrt{2nk-k^2}}=\sum_{k=1}^n \frac{1}{n}\frac{1}{\sqrt{2k/n-(k/n)^2}}\rightarrow \int_0^1 \frac{dx}{\sqrt{2x-x^2}}=\frac{\pi}{2}. By the way, using Riemann sums...

One of my first ideas was similar to your most recent post Realise, the logic isn't quite right as is though. Think about your last line. If the idea is to get an arbitrary polygon and show that we can make its area bigger by step by step making adjacent sides equal then you are missing the fact...

Re: HSC 2014 4U Marathon $Let $y_k=na_k$ and let $S_k$ be the $k$-th elementary symmetric sum of the set $\{y_k^{-1}\}_{k=1}^n.$\\ \\ Our constraint on the $a_j$ translates to $\sum_k y_k=n$, and so by AM-GM we have $\prod_k y_k\leq 1.$\\ \\ Applying AM-GM to the sum $S_k$ and using the...

We cannot assume that the polygon is regular or cyclic unfortunately. "Cutting" into triangles is a promising approach though.

$Prove that the area of a polygon with perimeter $1$ is alway smaller than $\frac{1}{4\pi}$, and that we can find polygons with perimeter $1$ that have area arbitrarily close to $\frac{1}{4\pi}$. $ NB: This is a polygonal version of the isoperimetric inequality. The general isoperimetric...

Re: HSC 2014 4U Marathon It's fine, all steps are obviously biconditional so it doesn't matter what end you start at. And you need this biconditionality to get the "iff" statement.

Re: HSC 2014 4U Marathon I must admit, the AM-GM solution would be a bit harder to find if I didn't know Lagrange multipliers. (As the appropriate definition of y_k could potentially take a little time to spot.) LM told me the equality conditions (each y_k as defined above must be equal.)...

Re: HSC 2014 4U Marathon $Realise's inequality screams AM-GM. \\ \\ Indeed if we define $y_k:=\frac{1}{k}(a_k/a_{k+1})^{1/k}$, we can directly apply AM-GM to the sum $\sum_{k=1}^n\sum_{j=1}^k y_k$ containing $\frac{n(n+1)}{2}$ terms to obtain the result as well as the equality conditions.\\ \\...