HSC 2013 MX2 Marathon (archive) (2 Viewers)

Sy123 · Oct 18, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
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

First, we will prove this is true for distinct a,b,c

$$Without loss of generality, take$ \ \ a>b>c$

$a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) =$

$(a-b)(a-c)(b-c) \left(\frac{a}{b-c} - \frac{b}{a-c} + \frac{c}{a-b} \right)$

$The multiplier at the front is positive due to the initial condition$

$a > b$

$a-c > b-c$

$\therefore \ \frac{1}{a-c} < \frac{1}{b-c}$

$\therefore \ \ \frac{b}{a-c} < \frac{a}{b-c}$

$\therefore \ \frac{a}{b-c} - \frac{b}{a-c} + \frac{c}{a-b} > 0$

$\therefore \ \ (a-b)(a-c)(b-c) \left(\frac{a}{b-c} - \frac{b}{a-c} + \frac{c}{a-b} \right ) > 0$

-----

$$Case 1: For non-distinct$ \ \ a,b,c \ \ $without loss of generality take$ \ \ a=b$

$a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) = c(c-a)^2 > 0$

$$Case 2:$ \ \ a=b=c \ \ \Rightarrow \ \ a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) = 0$

Hence we account for all cases and inequality holds each time

RealiseNothing · Oct 18, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
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

Let $a \geq b \geq c$

Now out of all 3 terms the only negative one is $b(b-a)(b-c)$

Change this to $b(b-a)(a-c)$ as $a-c \geq b-c$ and so this term is an even larger negative number now.

Now consider the first 2 terms together as the last term is positive:

$a(a-b)(a-c) + b(b-a)(a-c)$

$(a-c)[a(a-b) + b(b-a)]$

$(a-c)[a(a-b) - b(a-b)]$

$(a-c)(a-b)^2 \geq 0$

ColdLipstick · Oct 18, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
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

gustavo28 · Oct 18, 2013

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.

Sy123 · Oct 19, 2013

Re: HSC 2013 4U Marathon

$$Let$ \ x> 1$

$Evaluate$

$\frac{x}{x+1} + \frac{x^2}{(1+x)(1+x^2)} + \frac{x^4}{(1+x)(1+x^2)(1+x^4)} + \dots$

Galactic Jam · Oct 19, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Let$ \ x> 1$

$Evaluate$

$\frac{x}{x+1} + \frac{x^2}{(1+x)(1+x^2)} + \frac{x^4}{(1+x)(1+x^2)(1+x^4)} + \dots$

$\frac{x+1-1}{x+1} + \frac{x^2+1-1}{(1+x)(1+x^2)} + \frac{x^4+1-1}{(1+x)(1+x^2)(1+x^4)} + \dots$

$= 1 - \frac{1}{x+1} + \frac{1}{x+1} - \frac{1}{(1+x)(1+x^2)} + \frac{1}{(1+x)(1+x^2)} - \frac{1}{(1+x)(1+x^2)(1+x^4)} \dots$

$= 1$

I hope that's right and you can just assume that the cancelling out continues for the entire sum o.o

Sy123 · Oct 19, 2013

Re: HSC 2013 4U Marathon

Galactic Jam said:
$\frac{x+1-1}{x+1} + \frac{x^2+1-1}{(1+x)(1+x^2)} + \frac{x^4+1-1}{(1+x)(1+x^2)(1+x^4)} + \dots$

$= 1 - \frac{1}{x+1} + \frac{1}{x+1} - \frac{1}{(1+x)(1+x^2)} + \frac{1}{(1+x)(1+x^2)} - \frac{1}{(1+x)(1+x^2)(1+x^4)} \dots$

$= 1$

I hope that's right and you can just assume that the cancelling out continues for the entire sum o.o

Yep that is correct, I would think that you'd need to mention:

$\lim_{n \to \infty} \frac{1}{\prod_{k=0}^n (1+x^{2^k})}} = 0$

There are some cases where the sum would go to infinity and there would be mass cancellation (telescoping), but you still have a limiting value at the end, i.e.

$\sum_{n=1}^{\infty} \tan^{-1}(n+1) - \tan^{-1}(n)$

ColdLipstick · Oct 19, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
Find all real polynomials p(x) such that p(x)p(x+1)=p(x^2) for all integers x.

I'm not sure if this is correct but I guess i found one polynomial solution?

EDIT:
(there should be a plus minus sign after the quadratic is solved)
When I solved the quadratic, I should get two solutions
so the 2nd polynomial would be
p(x) = (-1)^(2n) x^n (x-1)^n
= x^n (x-1)^n
That should justify why I took either solution

gustavo28 · Oct 19, 2013

Re: HSC 2013 4U Marathon

ColdLipstick said:
I'm not sure if this is correct but I guess i found one polynomial solution?

EDIT:
(there should be a plus minus sign after the quadratic is solved)
When I solved the quadratic, I should get two solutions
so the 2nd polynomial would be
p(x) = (-1)^(2n) x^n (x-1)^n
= x^n (x-1)^n
That should justify why I took either solution

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.

ColdLipstick · Oct 19, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
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.

I guess I'll have to continue it for two roots, three roots.... rather than extrapolation?

ColdLipstick · Oct 19, 2013

Re: HSC 2013 4U Marathon

New question by me.

EDIT: Deduce that Z1,Z3 and Z3 are Cyclic with the centre as k.
Sorry about that

Sy123 · Oct 19, 2013

Re: HSC 2013 4U Marathon

ColdLipstick said:
New question by me.

Are you sure the last line of the question isn't a typo?
Any 3 points are cyclic

ColdLipstick · Oct 19, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Are you sure the last line of the question isn't a typo?
Any 3 points are cyclic

Yep, sorry. Read the EDIT please

seanieg89 · Oct 19, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
Find all real polynomials p(x) such that p(x)p(x+1)=p(x^2) for all integers x.

0. The only constant solutions are p=0,1. Let us now assume that p is nonconstant. By comparing leading coefficients we must have p monic.
1. For any such p, the identity p(x)p(x+1)=p(x^2) must hold for all complex numbers, as nonzero polynomials have only finitely many roots. (Equivalently, the two polynomials on either sides are identical in the sense that all coefficients are equal upon expansion.)
2. The set of roots is closed under the maps z->z^2 and z->(z-1)^2. The former of these maps (and the fact that polys only have finitely many roots) implies that all roots of p must lie on the origin or unit circle. The latter map then implies that p has roots 0 and 1 (and ONLY 0 and 1).
3. Plugging in p=(x^m)((x-1)^n) into the initial functional equation (for m,n > 0), cancelling and equating multiplicity of roots of both sides, we get m=n. All steps in this process are biconditional so this is indeed a solution.

Conclusion: Solutions are p=0,1,[x(x-1)]^n (n>0).

gustavo28 · Oct 19, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
0. The only constant solutions are p=0,1. Let us now assume that p is nonconstant. By comparing leading coefficients we must have p monic.
1. For any such p, the identity p(x)p(x+1)=p(x^2) must hold for all complex numbers, as nonzero polynomials have only finitely many roots. (Equivalently, the two polynomials on either sides are identical in the sense that all coefficients are equal upon expansion.)
2. The set of roots is closed under the maps z->z^2 and z->(z-1)^2. The former of these maps (and the fact that polys only have finitely many roots) implies that all roots of p must lie on the origin or unit circle. The latter map then implies that p has roots 0 and 1 (and ONLY 0 and 1).
3. Plugging in p=(x^m)((x-1)^n) into the initial functional equation (for m,n > 0), cancelling and equating multiplicity of roots of both sides, we get m=n. All steps in this process are biconditional so this is indeed a solution.

Conclusion: Solutions are p=0,1,[x(x-1)]^n (n>0).

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

Sy123 · Oct 19, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
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

First move the complex numbers and make $z_3$ the origin

$z_1 \cdot \left( \cos \frac{\pi}{3} + i\sin \frac{\pi}{3} \right) = z_2$

$Through manipulation we find$

$z_1^2 + z_2^2 = z_1 z_2$

$z_1 \rightarrow z_1 - z_3$

$z_2 \rightarrow z_2 - z_3$

This corresponds to an equilateral triangle translated

Which would mean:

$(z_1 - z_3)^2 + (z_2 - z_3)^2 = (z_1 - z_3)(z_2-z_3) \Rightarrow \ \ z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$

----------

$Find$ \ \ x \ \ $in terms of theta such that$

$\sum_{j=0}^{\infty} x^j (\cos(j \theta) + \sin(j \theta)) = 1$

gustavo28 · Oct 19, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
First move the complex numbers and make $z_3$ the origin

$z_1 \cdot \left( \cos \frac{\pi}{3} + i\sin \frac{\pi}{3} \right) = z_2$

$Through manipulation we find$

$z_1^2 + z_2^2 = z_1 z_2$

$z_1 \rightarrow z_1 - z_3$

$z_2 \rightarrow z_2 - z_3$

This corresponds to an equilateral triangle translated

Which would mean:

$(z_1 - z_3)^2 + (z_2 - z_3)^2 = (z_1 - z_3)(z_2-z_3) \Rightarrow \ \ z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$

----------

$Find$ \ \ x \ \ $in terms of theta such that$

$\sum_{j=0}^{\infty} x^j (\cos(j \theta) + \sin(j \theta)) = 1$

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 eventually get that x=sint-sqrt(1+(sint)^2) or sint+sqrt(1+(sint)^2) depending on which one is between 1 and -1.

Sy123 · Oct 19, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
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 eventually get that x=sint-sqrt(1+(sint)^2) or sint+sqrt(1+(sint)^2) depending on which one is between 1 and -1.

Yea something like that, though I'm getting:

$\frac{1 - x\cos \theta + x\sin \theta}{x^2 - 2x\cos \theta + 1} = 1$

Which yields, x=1 - (cos t + sin t)

gustavo28 · Oct 19, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yea something like that, though I'm getting:

$\frac{1 - x\cos \theta + x\sin \theta}{x^2 - 2x\cos \theta + 1} = 1$

Which yields, x=1 - (cos t + sin t)

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?

RealiseNothing · Oct 20, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
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?

Find:

$\sum_{k=1}^n \frac{n-k+1}{\sin^2(\frac{k\pi}{2n+1})}$

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