HSC 2016 MX2 Marathon ADVANCED (archive) (1 Viewer)

lita1000 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
second part

Alternatively, prove that if n is true, n+6 is true, then show trivially that it is true for the first six natural numbers and draw your conclusion from there.

$$oh okay, you just want the second bit$\\ii)\ Let L= \sqrt{s(s-a)(s-b)(s-c)}.\ $(x-alt_a)(x-alt_b)(x-alt_c)=0.-->$ x^3-2L(1/a+\1/b+1/c)x^2+4L^2(1/ab+1/ac+1/bc)x-{8L^3/abc}=0.$moving things over,we get that$ x^3+4L^2(1/ab+1/bc+1/ca})x=2L(1/a+1/b+1/c)x^2+8L^3/abc=L(2(1/a+1/b+1/c)x^2+8L^2/abc).$Squaring both sides gives us a degree 6 polynomial with rational coefficients, since s(s-a)(s-b)(s-c)is a rational number due to the condition that a polynomial of the form a(x-a)(x-b)(x-c).......(Call this Y))\\ where a is a rational number has rational coefficients, and s is the sum of roots of a,b,c (which is rational) divided by 2, and hence it is also a rational. Furthermore, 1/a+1/b+1/c and 1/ab+1/bc+1/ca are both rational due to similar reasons (they can be both expressed in terms of the rational coefficients of the polynomial Y).$

Paradoxica · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$$For any set of non-negative real numbers $ \{a_1, a_2, \dots , a_{n-1}, a_n \}$

$$Prove $(1+a_1)(1+a_2)\dots (1+a_{n-1})(1+a_n) \geq 1+ a_1 + a_2 + \dots + a_{n-1} + a_n$

$State the condition required for equality to hold.$

InteGrand · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
$$For any set of non-negative real numbers $ \{a_1, a_2, \dots , a_{n-1}, a_n \}$

$$Prove $(1+a_1)(1+a_2)\dots (1+a_{n-1})(1+a_n) \geq 1+ a_1 + a_2 + \dots + a_{n-1} + a_n$

Direct expansion.

Paradoxica · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$$A recurrence relation is defined as follows: $T_{n+2} = 5T_{n+1} - 6T_n$ for $n \geq 0$

$T_0 = 0, T_1 = 1.$

$$Prove $T_n = 3^n - 2^n$ for all $T_n$

$Hence, determine the first term in the sequence greater than The Answer to the Ultimate Question of Life, the Universe, and Everything multiplied by 10 million.$

Sy123 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

lita1000 said:
$$oh okay, you just want the second bit$\\ii)\ Let L= \sqrt{s(s-a)(s-b)(s-c)}.\ $(x-alt_a)(x-alt_b)(x-alt_c)=0.-->$ x^3-2L(1/a+\1/b+1/c)x^2+4L^2(1/ab+1/ac+1/bc)x-{8L^3/abc}=0.$moving things over,we get that$ x^3+4L^2(1/ab+1/bc+1/ca})x=2L(1/a+1/b+1/c)x^2+8L^3/abc=L(2(1/a+1/b+1/c)x^2+8L^2/abc).$Squaring both sides gives us a degree 6 polynomial with rational coefficients, since s(s-a)(s-b)(s-c)is a rational number due to the condition that a polynomial of the form a(x-a)(x-b)(x-c).......(Call this Y))\\ where a is a rational number has rational coefficients, and s is the sum of roots of a,b,c (which is rational) divided by 2, and hence it is also a rational. Furthermore, 1/a+1/b+1/c and 1/ab+1/bc+1/ca are both rational due to similar reasons (they can be both expressed in terms of the rational coefficients of the polynomial Y).$

Yep, well done

My solution was pretty much the same (I'll write it out in case people can't read your solution)

$\\ $Let the altitudes of the triangle, intersecting sides$ \ a, b, c \ $be$ \ \alpha, \beta, \gamma \ $respectively$$

$\\ $Considering the classic formula for the area of the triangle, we see that$ \\ A = \frac{1}{2} a \alpha = \frac{1}{2} b \beta = \frac{1}{2}c \gamma = \sqrt{s(s-a)(s-b)(s-c)}$

$\\ $So, considering the polynomial$ \ p(x) = (x-a)(x-b)(x-c)$

$\\ $Consider the new polynomial$ \\\\ q(x) = x^3 p\left(\frac{2A}{x} \right) = (2A - ax)(2A - bx)(2A - cx)$

$\\ $It is clear that$ \ q(\alpha) = q(\beta) = q(\gamma) = 0 \ $and so they are roots of$ \ q$

$\\ $Construct the polynomial$ \\\\ r(x) = q(x) \cdot q(-x) = (2A - ax)(2A - bx)(2A - cx)(2A + ax)(2A+bx)(2A+cx) \\\\ \Rightarrow r(x) = (4A^2 - a^2x^2)(4A^2 - b^2x^2)(4A^2-c^2x^2)$

$\\ \alpha, \beta , \gamma \ $are all roots of$ \ r $, also it is a 6th degree polynomial with rational coefficients, as$ \ A^2 = s(s-a)(s-b)(s-c) \ $is rational (since$ \ a,b,c \ $are rational)$$

Sy123 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
$$A recurrence relation is defined as follows: $T_{n+2} = 5T_{n+1} - 6T_n$ for $n \geq 0$

$T_0 = 0, T_1 = 1.$

$$Prove $T_n = 3^n - 2^n$ for all $T_n$

$Hence, determine the first term in the sequence greater than The Answer to the Ultimate Question of Life, the Universe, and Everything multiplied by 10 million.$

$\\ 5(3^{n+1} - 2^{n+1}) - 6(3^{n} - 2^n) = (3+2)(3^{n+1} - 2^{n+1}) - 3 \cdot 2 \cdot (3^n - 2^n) \\ \\ = 3^{n+2} - 3 \cdot 2^{n+1} + 2 \cdot 3^{n+1} - 2^{n+2} - 2 \cdot 3^{n+1} + 3 \cdot 2^{n+1} = 3^{n+2} - 2^{n+2} \\ \\ \Rightarrow 5(3^{n+1} - 2^{n+1}) - 6(3^{n} - 2^n) = 3^{n+2} - 2^{n+2}$

$\\ $Thus, taking$ \ u_n = 3^n - 2^n \ $it is clear from the above that$ \ u_{n+2} = 5u_{n+1} - 6u_{n}$

$\\ $Also,$ \ u_0 = 1 - 1 = 0 =T_0 \ $and$ \ u_1 = 3 -2 =1 = T_1$

$\\ $So,$ \ u_n \ $and$ \ T_n \ $have the same starting conditions and same recurrence, so they are the same sequence$$

$\therefore \ T_n = 3^n - 2^n$

lita1000 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
Yep, well done

My solution was pretty much the same (I'll write it out in case people can't read your solution)

$\\ $Let the altitudes of the triangle, intersecting sides$ \ a, b, c \ $be$ \ \alpha, \beta, \gamma \ $respectively$$

$\\ $Considering the classic formula for the area of the triangle, we see that$ \\ A = \frac{1}{2} a \alpha = \frac{1}{2} b \beta = \frac{1}{2}c \gamma = \sqrt{s(s-a)(s-b)(s-c)}$

$\\ $So, considering the polynomial$ \ p(x) = (x-a)(x-b)(x-c)$

$\\ $Consider the new polynomial$ \\\\ q(x) = x^3 p\left(\frac{2A}{x} \right) = (2A - ax)(2A - bx)(2A - cx)$

$\\ $It is clear that$ \ q(\alpha) = q(\beta) = q(\gamma) = 0 \ $and so they are roots of$ \ q$

$\\ $Construct the polynomial$ \\\\ r(x) = q(x) \cdot q(-x) = (2A - ax)(2A - bx)(2A - cx)(2A + ax)(2A+bx)(2A+cx) \\\\ \Rightarrow r(x) = (4A^2 - a^2x^2)(4A^2 - b^2x^2)(4A^2-c^2x^2)$

$\\ \alpha, \beta , \gamma \ $are all roots of$ \ r $, also it is a 6th degree polynomial with rational coefficients, as$ \ A^2 = s(s-a)(s-b)(s-c) \ $is rational (since$ \ a,b,c \ $are rational)$$

I know that this is a mere typo, but ur last statement about a,b, and c being necessarily rational isn't quite right. Perhaps a different notation would fix the problem

Just really nit picking here

Sy123 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

lita1000 said:
I know that this is a mere typo, but ur last statement about a,b, and c being necessarily rational isn't quite right. Perhaps a different notation would fix the problem

Just really nit picking here

Right, my bad, instead:

$\\ A^2 = \frac{1}{16} (a+b+c) p(a+b+c) = \frac{1}{32} p''(0) p\left(\frac{p''(0)}{2} \right)$

$\\ $Since$ \ p \ $is a rational polynomial, so must$ \ p'' \ $and thus$ \ p''(0) \ $is rational$$

$\\ $So,$ \ p\left(\frac{p''(0)}{2} \right) \ $must be rational, and so$ \ A^2 \ $must be rational from the original expression$$

(or, -(a+b+c) is the co-efficient of x^2 in p which is rational, and so (a+b+c) is rational, and so p(a+b+c) is rational and so is the above expression)

lita1000 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy, can you post some difficult questions like you usually do? I know that you stepped the difficulties down a little for your last couple of questions lol

Sy123 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

This is a good one that I just did:

$\\ $Find all polynomials with real coefficients that satisfy:$ \ p(x) p(x+1) = p(x^2)$

I don't know if you consider this difficult enough

VBN2470 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
This is a good one that I just did:

$\\ $Find all polynomials with real coefficients that satisfy:$ \ p(x) p(x+1) = p(x^2)$

I don't know if you consider this difficult enough

I got one solution to be $p(x)=x^{k} (x-1)^k$ for $k\in\mathbb{Z}^+$ , but I don't know if that's all the possible solutions.

Sy123 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

VBN2470 said:
I got one solution to be $p(x)=x^{k} (x-1)^k$ for $k\in\mathbb{Z}^+$ , but I don't know if that's all the possible solutions.

I got that too (with p(x) = 0) but I managed to show that it was the only possible solution

Paradoxica · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

VBN2470 said:
I got one solution to be $p(x)=x^{k} (x-1)^k$ for $k\in\mathbb{Z}^+$ , but I don't know if that's all the possible solutions.

Try contradiction. Everything you need to know is given to you, all you need to do is make a few assumptions about P(x)

Sy123 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Something a little easier:

$\\ $Show that no number of the form$ \ 4k+2 \ $or$ \ 4k+3 \ $for any positive integer$ \ k \ $can be a perfect square$$

VBN2470 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
Something a little easier:

$\\ $Show that no number of the form$ \ 4k+2 \ $or$ \ 4k+3 \ $for any positive integer$ \ k \ $can be a perfect square$$

For $4k+2$ The square of an even number is always divisible by 4 and so no number of the form $4k+2=4(k+\frac{1}{2})$ , where $k$ is a positive integer, can be a perfect square.

For $4k+3$ , since this always odd for any positive integer $k$ , for it to be square it must be the square of another odd number (say) $2n+1$ for some positive integer $n$ . Suppose that $4k+3$ is a perfect square. Then $4k+3=(2n+1)^2 \Rightarrow 4k+3=4n^2+4n+1 \Rightarrow k=(n^2+n)-\frac{1}{2}$ which is a contradiction since $k$ and $n$ are always positive integers. Therefore no number of that form can be expressed as a perfect square.

simpleetal · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
This is a good one that I just did:

$\\ $Find all polynomials with real coefficients that satisfy:$ \ p(x) p(x+1) = p(x^2)$

I don't know if you consider this difficult enough

$$Suppose the polynomial had a root$ \alpha $, then $\alpha ^2 \ $would be another root. This would lead to a chain of infinite roots, implying that the polynomial is a constant function, unless $\alpha^2=\alpha .$ the only such$ \alpha $ satisying this is 0 or 1. You can do something similar with complex numbers to show that for the function to have a complex root, its imaginary part must be 0. So p(x) is in the form$ (x)^m(x-1)^n.\\$ Subbing this into our original equation,$ x^m(x-1)^n(x+1)^m(x)^n=x^{2m}(x^2-1)^n.\\$ At x=0, LHS=RHS regardless of the values of m and n, and we are trying to find n and m such that LHS =RHS for all values of x. So we may as well assume that x isn't 0 or 1 and divide both sides by $x^m(x-1)^n. $This gives us$ (x+1)^m(x)^n=x^m(x+1)^n,$ and comparing the coefficients of the term with power m+n-1, we get that n=m. If however the polynomial is a constant function, then the only solutions are 0,1. Thus, the only solutions are 0,1,$ (x)^n(x-1)^n.$

Paradoxica · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Show that no integer of the form 9k + 4 or 9k + 5 can be expressed as the sum of three integer cubes. (Taken from a recent Numberphile video)

Sy123 · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

simpleetal said:
$$Suppose the polynomial had a root$ \alpha $, then $\alpha ^2 \ $would be another root. This would lead to a chain of infinite roots, implying that the polynomial is a constant function, unless $\alpha^2=\alpha .$ the only such$ \alpha $ satisying this is 0 or 1. You can do something similar with complex numbers to show that for the function to have a complex root, its imaginary part must be 0. So p(x) is in the form$ (x)^m(x-1)^n.\\$ Subbing this into our original equation,$ x^m(x-1)^n(x+1)^m(x)^n=x^{2m}(x^2-1)^n.\\$ At x=0, LHS=RHS regardless of the values of m and n, and we are trying to find n and m such that LHS =RHS for all values of x. So we may as well assume that x isn't 0 or 1 and divide both sides by $x^m(x-1)^n. $This gives us$ (x+1)^m(x)^n=x^m(x+1)^n,$ and comparing the coefficients of the term with power m+n-1, we get that n=m. If however the polynomial is a constant function, then the only solutions are 0,1. Thus, the only solutions are 0,1,$ (x)^n(x-1)^n.$

Yep that's the idea, but to clean up the solution, it should be noted that if we started with $x =\cos \theta + i \sin \theta$ then we possibly will not get an infinite chain if we kept squaring it.

But, suppose that $x= \cos \theta + i \sin \theta$ is a root, then substitute into the equation: $x = \cos \frac{\theta}{2} + i \sin \frac{\theta}{2}$

Since $p\left(1 + \cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right) \neq 0$ (as the root must have modulus 1), then:

$p\left(\cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right) = 0$

And then we are off again with infinitely many solutions unless we can guarantee no infinite chain results

Therefore, the only roots to the polynomial can be of the form:

$\\ x=0 \ $and$ \ x = \cos \theta + i \sin \theta$

Where also, applying the mapping $(1): x \rightarrow x^2$ and $(2): x\rightarrow \sqrt{x}$ must all form finite sequences

$\\ $If$ \ \theta \ $is non-zero, then either mapping (1) or (2) will provide an infinite distinct sequence, therefore$ \ \theta = 0$

$\\ $In fact, only mapping$ \ (2) \ $is sufficient, because$ \ z_k = \cos \frac{\theta}{2^k} + i \sin \frac{\theta}{2^k} \ $is all distinct for$ \ \theta \neq 0$

$\\ $Thus, the only solutions can be$ \ x=0, 1$

simpleetal · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
Yep that's the idea, but to clean up the solution, it should be noted that if we started with $x =\cos \theta + i \sin \theta$ then we possibly will not get an infinite chain if we kept squaring it.

But, suppose that $x= \cos \theta + i \sin \theta$ is a root, then substitute into the equation: $x = \cos \frac{\theta}{2} + i \sin \frac{\theta}{2}$

Since $p\left(1 + \cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right) \neq 0$ (as the root must have modulus 1), then:

$p\left(\cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right) = 0$

And then we are off again with infinitely many solutions unless we can guarantee no infinite chain results

Therefore, the only roots to the polynomial can be of the form:

$\\ x=0 \ $and$ \ x = \cos \theta + i \sin \theta$

Where also, applying the mapping $(1): x \rightarrow x^2$ and $(2): x\rightarrow \sqrt{x}$ must all form finite sequences

$\\ $If$ \ \theta \ $is non-zero, then either mapping (1) or (2) will provide an infinite distinct sequence, therefore$ \ \theta = 0$

$\\ $In fact, only mapping$ \ (2) \ $is sufficient, because$ \ z_k = \cos \frac{\theta}{2^k} + i \sin \frac{\theta}{2^k} \ $is all distinct for$ \ \theta \neq 0$

$\\ $Thus, the only solutions can be$ \ x=0, 1$

When I said in my solution that you could show that the complex roots satisfying this condition had imaginary part =0, I meant exactly what you had outlined in your post (showing that theta=0)

InteGrand · Nov 8, 2015

Re: HSC 2016 4U Marathon - Advanced Level

simpleetal said:
When I said in my solution that you could show that the complex roots satisfying this condition had imaginary part =0, I meant exactly what you had outlined in your post (showing that theta=0)

I think maybe Sy123 didn't notice your edit to include the complex roots case (or maybe he started typing his reply before the edit).

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