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Re: HSC 2013 4U Marathon Right yep my bad. (I don't think i needed that conjugate stuff after all) But I think it yields x=sint+cost instead. Can someone post a question?

Re: HSC 2013 4U Marathon Let r=x(cost+isint) and use the fact that r^k+(conjugate of r)^k=2x^k(cos(kt)) and -i(r^k-(conjugate of r)^k)=2x^ksin(kt) and then sum using the geometric progression sum for infinite series (there are probably some fiddly convergence issues to deal with tho) to...

Re: HSC 2013 4U Marathon Yep that's probably the best way to do it. Q: Show that z1,z2,z3 form an equilateral triangle if and only if (z1)^2+(z2)^2+(z3)^2=z1z2+z2z3+z3z1

Re: HSC 2013 4U Marathon So, for the case with one root, you can just use the fact that a1^2-2a1x+x is identically equal to zero. So, the constant term is zero. But I don't think you can extrapolate this for larger degrees of P.

Re: HSC 2013 4U Marathon Find all real polynomials p(x) such that p(x)p(x+1)=p(x^2) for all integers x.

Re: HSC 2013 4U Marathon Prove that for all non-negative a,b,c a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b) >= 0

Re: HSC 2013 4U Marathon Let the four points be A,B,C,D. Let the incentre of ABC be I and the incentre of DBC be J. Let the midpoint of the arc BC be M. Sketch: 1) Prove IM=CM=BM and similarly JM=CM=BM. 2) Hence BCIJ is cyclic 3) Then fill some angles in to get the right angles

Re: HSC 2013 4U Marathon I think there's a fairly subtantial problem with my proof in that I assume that X+1/X is irrational where X is the cube root of a non-cube rational. However, I'm not entirely sure whether or not this is provable within the syllabus.

Re: HSC 2013 4U Marathon Assume that x>y since the problem is symmetric in x and y. We can manipulate the inequality into wanting to prove that x^y(x^(x-y)-1)/y^y >y^(x-y)-1. Now, x^a>y^a if a,x,y are positive since x^a is increasing for positive x (f'(x)=ax^(a-1)). So, x^(x-y)-1>y^(x-y)-1...

Re: HSC 2013 4U Marathon There's a slight error in that the area of ABD is equal to h/2(BD). But that doesn't affect the proof at all, which is really good apart from that.

Re: HSC 2013 4U Marathon Yep, thats probably the best way to do it. You can also an induction sort of thing.

Re: HSC 2013 4U Marathon Consider a triangle ABC with points D, E, F on BC, CA and AB respectively. Suppose that AD, BE and CF are concurrent. Prove that (BD/DC)*(CE/EA)*(AF/FB)=1

Re: HSC 2013 4U Marathon I'm a student. The answer is 1/2. What methods did people use?

Re: HSC 2013 4U Marathon So, first of all suppose there is a monic rational quadratic with the root X+1/X with X as the cube root of a non-cube rational. From now on this root will be known as T. Let this quadratic be x^2+ax+b. So...

Re: HSC 2013 4U Marathon If Alice is allowed to toss a coin 10 times and Bob 9 times, then what is the probability that Alice gets more heads than Bob?

Re: HSC 2013 4U Marathon Prove that the number of subsets of {1,2,...,n } with even cardinality is equal to the number of subsets of odd cardinality of the same set.

Re: HSC 2013 4U Marathon Even though you set x=1, I think differentiability might have to be proven first to reach that conclusion.

Re: HSC 2013 4U Marathon y \leq 1 x+y \leq 1+x \frac{y}{1+x} \leq \frac{y}{y+x} Similarly, \frac{x}{1+y} \leq \frac{x}{y+x} Adding these two produce the required inequality.

Re: HSC 2013 4U Marathon So, substitute x=y=1 to get that f(1)=0 Set y=1+t/x into the identity to get f(x+t)=f(x)+f(1+t/x). So, f(x+t)-f(x)=f(1+t/x)-f(1) So, x(f(x+t)-f(x))/t=(f(1+t/x)-f(1))/(t/x) So, as t goes to 0, the RHS is f'(1), while the LHS is the limit of x((f(x+t)-f(x))/t as...

Re: HSC 2013 4U Marathon Multiply it by a sufficiently large integer to make it an integer polynomial Q(x) which is also good. Let the leading coefficient of Q be A and the constant be C. Now, considering the graph of Q(x)-C, we can make a prime p large enough so that Q(x)+p-c is above the...