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One of the equal sides, not the sum of them. If this were the case then the lengths of the sides are not integers (think of a famous Pythagorean triad)

1) Here is a neat way to do it rather than differentiating and stuff We know that since y=10x+10 is a tangent, it must have only one solution to the equation: Ax^2=10x+10 That means the discriminant of this equation will be 0. Ax^2-10x-10=0 Find the discriminant \triangle =...

Re: HSC 2012 Marathon :) It is asking for what values of theta does it make the normal at Q a tangent to the circle

Q9 is a very good question, it doesnt require any mechanical interaction at all. Think Pythagorean Triads.

Re: HSC 2012 Marathon :) Correct! Except for part v. Use the fact that theta is the inclined angle to x-axis and circle properties to find theta such that it is true (thats how I did it)

How much is that question worth? My only bet is that depending on the velocity the angle theta can be configured in a way such that it does hit the target? So in the cartesian form with h and d. As V approaches infinity h approaches d tan theta \therefore \theta=\tan^{-1}h/d And it doesnt...

Re: HSC 2012 Marathon :) Here is a neat question that I just made (and solved, so it is solvable) It is a 5 part question, so here we go: Difficulty: Easy -> Hard $Given that the angle PO makes with the x-axis is$ \ \ \theta $i) Find the cartesian locus of point P where$ \ \ P(\cos \theta...

Re: 2012 Year 9 &10 Mathematics Marathon This is correct, also Demento I havent recieved your email yet if something is wrong

Re: 2012 Year 9 &10 Mathematics Marathon The problem is you cant factorise it, take the discriminant: 1^2-4(2)=-7 So that means the discriminant is negative and means that there are no real roots. If you expand your answer again and you get something different to the question it means you...

Re: 2012 Year 9 &10 Mathematics Marathon (x+2)(x-1)=0 x^2-x+2x-2=0 x^2+x-2=0 The question had positive 2 not negative 2

Re: 2012 Year 9 &10 Mathematics Marathon Expand your answer, you get a different constant

I guess. The way I thought of it was: S(n)=n+(n-1)+...+(n-n+2)+(n-n+1) Lets take out all the paranthesis and rearrange our sum: S(n)=n+n+..+n - 1-2-..-(n-1) Notice that there are an n number of of n's in the first part of our sum, then the next part of our sum is simply S(n)-n...

Geogebra is a good alternative and is easily adjustable compared to wolframalpha, but its a little less powerful

Re: 2012 Year 9 &10 Mathematics Marathon There is really no need to know any series properties, I dont think its a suitable question for y10 though but it can be attempted by y10s if they think outside the box (or outside the parenthesis, HINT HINT).

Re: 2012 Year 9 &10 Mathematics Marathon Yeah, considering the series in descending order S(n)=n+(n-1)+...+3+2+1 Which is the same thing as writing S(n)=n+(n-1)+...+(n-n+2)+(n-n+1)

You are not considering the series in reverse, if it said otherwise, sure post your answer but its wrong. Try this: S(n)=n+(n-1)+...+(n-n+2)+(n-n+1) Give one crucial observation and the rest will fall into place

The approach to the question graphically The logic behind gradients, smoothness and discontinuity, Note the the difference between the third and second case What I mean when I say there is a possibility that the straight line intersecting with the curve turns out to be smooth, note turning...

HINT for a) Express all the terms in the series in the same form as you would express the first few terms of the descending series, b should be straight forward once you prove the first formula

Ah ok, thank you for the response :)

Try your hand at some proofs yourself, just give yourself the definitions of sine, cosine, and tangent functions. Try to prove simple ones like sin^2 x + cos^2 x =1 or sinx=cos(90-x) etc Then move up to the extension identities (if you do 3U) (and if needed give yourself the first part of the...