HSC 2016 MX2 Marathon ADVANCED (archive) (5 Viewers)

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).

HSC 2016 MX2 Marathon ADVANCED (archive) (5 Viewers)

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