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Re: HSC 2013 4U Marathon Yep, nice work.

hahaha Pathetic.

Re: HSC 2013 4U Marathon $An n-sided regular polygon has its sides such that none are vertical$ $Let$ \ \ m_1, \ m_2, \ m_3, \dots , \ m_n $Be the gradients of all the sides of the polygon$ $Prove that$ m_1 m_2 + m_2 m_3 + m_3 m_4 + \dots + m_{n-1} m_n + m_n m_1 = -n

Re: HSC 2013 4U Marathon Use gradient formula, equate gradients, then turn the sum into products using those formulae that are easily derivable, we end up with \frac{2\sin((q-p)/2) \cos((q+p)/2)}{-2\sin((q+p)/2) \sin((q-p)/2)} = \frac{2\sin((s-r)/2) \cos((s+r)/2}{-2\sin((s+r)/2)...

Re: HSC 2013 4U Marathon $Find$ I=\int \frac{dx}{1+\sqrt{x}}

Re: HSC 2013 4U Marathon Ah yes your way is a bit faster than my one.

Re: HSC 2013 4U Marathon Yeah I noticed, your new one isn't much harder its just a bit long. Is there any cool trick we can do to make it much faster (except for trinomial expansion :P)?

Re: HSC 2013 4U Marathon \int \frac{dx}{1+e^x} = \int \left(1-\frac{e^x}{1+e^x} \right ) \ dx = x-\ln(1+e^x) + c

Re: HSC 2013 4U Marathon Eh, not sure what word to put there, Compute, Find?

Re: HSC 2013 4U Marathon $Evaluate$ I=\int\frac{1}{x^2\sqrt{1+x^2}} \ dx

Those who dream by night in the dusty recesses of their minds wake in the day to find that all was vanity; but the dreamers of the day are dangerous men, for they may act their dream with open eyes, and make it possible - Lawrence of Arabia

Re: HSC 2013 3U Marathon Thread As Q approaches P it is clear that the length QO is decreasing, and that the minimum distance is when Q is at P. So therefore the shortest distance from O to Q is when Q coincides with P. Now within HSC geometry, the shortest distance is always perpendicular...

Re: HSC 2013 3U Marathon Thread $15 Balls of equal unit radius are bounded by a large triangle like a Billiard rack, as shown below$ $Find the area of empty space$ $Consider all calculations within 2 dimensions$

4U too hard for you?

Re: HSC 2013 4U Marathon First lets establish what we know about the tan inverse function, due to: \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta} It then follows that: \tan^{-1} A + \tan^{-1}B = \tan^{-1} \frac{A+B}{1-AB} So lets try and split up the tan...

Re: HSC 2013 4U Marathon $Prove that the polynomial has no real roots$ p(x) = \frac{x^6}{6} + \frac{x^5}{5} + \frac{x^4}{4} + \frac{x^3}{3} + \frac{x^2}{2} + x + c $(c is a constant)$ $If$ \ \ \ c > \frac{37}{60}

Re: HSC 2013 4U Marathon Yeah oops :s

Re: HSC 2013 4U Marathon This problem boils down to: \tan(3x) = \frac{\tan x + \tan(2x)}{1-\tan x \tan(2x)} Rearranging: \tan(3x) - \tan x \tan(2x) \tan(3x) = \tan(x) + \tan(2x) \tan(x) \tan(2x) \tan(3x) = \tan(x) + \tan(2x) + \tan(3x) Which is easily integratable

Re: HSC 2013 4U Marathon Very much so.

Re: HSC 2013 4U Marathon So how did you integrate: \int \tan x \cdot \frac{2\tan x}{1-\tan^2 x} \cdot \frac{3\tan x - \tan^3 x}{1-3\tan^2 x} \ dx ?