HSC 2016 MX2 Marathon (archive) (1 Viewer)

glittergal96 · Dec 23, 2015

Re: HSC 2016 4U Marathon

Created a new marathon thread for undergrad stuff: http://community.boredofstudies.org/1003/maths/346314/undergraduate-mathematics-marathon.html Hopefully some of the uni students here will make use of it!

InteGrand · Dec 28, 2015

Re: HSC 2016 4U Marathon

$\noindent \textbf{NEW QUESTION FOR 2016'ERS}$

$\noindent Let $\mathcal{C}: y = C(x)$ be a cubic and $\ell: y = l(x)$ be a straight line in the Cartesian Plane, and suppose they meet at a point $P \left(x^*,y^*\right)$. Show that $x^*$ is a triple root of the equation $C(x) - l(x) = 0$ if and only if $P$ is the point of inflexion of the cubic $\mathcal{C}$ (recall that all cubics have one and only one point of inflexion) and $\ell$ is tangent to the cubic here.$

Paradoxica · Dec 28, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
$\noindent \textbf{NEW QUESTION FOR 2016'ERS}$

$\noindent Let $\mathcal{C}: y = C(x)$ be a cubic and $\ell: y = l(x)$ be a straight line in the Cartesian Plane, and suppose they meet at a point $P \left(x^*,y^*\right)$. Show that $x^*$ is a triple root of the equation $C(x) - l(x) = 0$ if and only if $P$ is the point of inflexion of the cubic $\mathcal{C}$ (recall that all cubics have one and only one point of inflexion) and $\ell$ is tangent to the cubic here.$

$\noindent The point of inflexion of $C(x)$ and the point of inflexion of $C(x) - l(x)$ occur at the same value of $x$, since $l(x)$ is a line, which has a vanishing second derivative. As $C(x)-l(x)$ is a perfect cube, the point of inflexion occurs at the triple root of the curve. By the first statement, the point of inflexion of $C(x)$ also occurs at the same value. Since $C(x)$ meets $l(x)$ at $x^*$, the triple root of $C(x)-l(x)$ occurs at $x=x^*$. Therefore, by the above reasoning, the point of inflexion of $C(x)$ occurs at $x = x^*$, and hence $P(x^*,y^*)$ is the point of inflexion of $\mathcal{C}$.$

Ambility · Dec 28, 2015

Re: HSC 2016 4U Marathon

Can we have some more complex number questions?

Paradoxica · Dec 28, 2015

Re: HSC 2016 4U Marathon

Ambility said:
Can we have some more complex number questions?

$a+b+c$ is an integer.$$

$a, \; b, \; c$, are complex numbers which satisfy the system of equations:$$

$a^2 + b^2 + c^2 = 26$

$a^3 + b^3 + c^3 = -129$

$a^4 + b^4 + c^4 = 650$

$$Find the last three digits of $a^6+b^6+c^6$

Ambility · Dec 28, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
$a+b+c$ is an integer.$$

$a, \; b, \; c$, are complex numbers which satisfy the system of equations:$$

$a^2 + b^2 + c^2 = 26$

$a^3 + b^3 + c^3 = -129$

$a^4 + b^4 + c^4 = 650$

$$Find the last three digits of $a^6+b^6+c^6$

Does this question use things from the polynomials syllabus? It feels like it does...

leehuan · Dec 28, 2015

Re: HSC 2016 4U Marathon

Not bothered to do any question that requires effort. Last three digits? Bit cruel on innocent students.
But this is a potentially useful hint:

We don't know about whether a, b or c themselves are integers (they could be fractions). But we do know that a+b+c is an integer.

Since a, b and c are complex, there are two possibilities:

Either a and b, or either alternative pair, are complex conjugates.
All 3 variables are real.

However, a polynomials approach may be useful simply because of the nature of roots:
Define any arbitrary cubic polynomial with roots a, b and c, then carry from there.

Paradoxica · Dec 29, 2015

Re: HSC 2016 4U Marathon

leehuan said:
Not bothered to do any question that requires effort. Last three digits? Bit cruel on innocent students.
But this is a potentially useful hint:

We don't know about whether a, b or c themselves are integers (they could be fractions). But we do know that a+b+c is an integer.

Since a, b and c are complex, there are two possibilities:

Either a and b, or either alternative pair, are complex conjugates.
All 3 variables are real.

However, a polynomials approach may be useful simply because of the nature of roots:
Define any arbitrary cubic polynomial with roots a, b and c, then carry from there.

I'll accept that, since the resulting working is incredibly tedious.

$$\noindent Let $x,\; y,\; z,\; $be complex numbers such that the following system of equations is true.$

$x+y+z = 2$

$x^2 + y^2 + z^2 = 3$

$xyz = 4$

$$Evaluate $\sum_{\mathrm{cyc}} \frac{1}{xy+z-1}$

$\mathrm{cyc} $ denotes a cycle over the set of variables involved$$

InteGrand · Dec 29, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
$\noindent The point of inflexion of $C(x)$ and the point of inflexion of $C(x) - l(x)$ occur at the same value of $x$, since $l(x)$ is a line, which has a vanishing second derivative. As $C(x)-l(x)$ is a perfect cube, the point of inflexion occurs at the triple root of the curve. By the first statement, the point of inflexion of $C(x)$ also occurs at the same value. Since $C(x)$ meets $l(x)$ at $x^*$, the triple root of $C(x)-l(x)$ occurs at $x=x^*$. Therefore, by the above reasoning, the point of inflexion of $C(x)$ occurs at $x = x^*$, and hence $P(x^*,y^*)$ is the point of inflexion of $\mathcal{C}$.$

$\noindent For the first part, how did you deduce that $C(x)-l(x)$ was a perfect cube? This was basically equivalent to what we had to prove for that part, so we should probably give a bit more reasoning.$

Ambility · Dec 29, 2015

Re: HSC 2016 4U Marathon

leehuan said:
Since a, b and c are complex, there are two possibilities:

Either a and b, or either alternative pair, are complex conjugates.
All 3 variables are real.

What? Are you sure about this? I thought a b and c could be anything as long as their imaginary parts sum to zero. This means something like 2+4i, 8+19i and 17-23i is a possible solution. These numbers are neither conjugates nor real.

leehuan · Dec 29, 2015

Re: HSC 2016 4U Marathon

Ambility said:
What? Are you sure about this? I thought a b and c could be anything as long as their imaginary parts sum to zero. This means something like 2+4i, 8+19i and 17-23i is a possible solution. These numbers are neither conjugates nor real.

That has quite a simple fallacy of being hard to make a^2+b^2+c^2 an integer as well, and subsequent cases.

But yes, I forgot to address a contradiction. My bad.

Ambility · Dec 29, 2015

Re: HSC 2016 4U Marathon

Can we have a question which only requires knowledge up to the 4U complex numbers syllabus?

InteGrand · Dec 29, 2015

Re: HSC 2016 4U Marathon

Ambility said:
Can we have a question which only requires knowledge up to the 4U complex numbers syllabus?

Here is one:

$\noindent Let $n$ be a positive integer and let $\omega$ be an $n^\text{th}$ root of unity. If $k$ is an integer, show that $\omega ^k$ and $\omega^{n-k}$ are complex conjugates.$

Paradoxica · Dec 29, 2015

Re: HSC 2016 4U Marathon

Ambility said:
Can we have a question which only requires knowledge up to the 4U complex numbers syllabus?

$$What is the maximum value of $|z|$, if $z$ satisfies $\left | z + \frac{2}{z} \right | = 2$

ze- · Jan 5, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$What is the maximum value of $|z|$, if $z$ satisfies $\left | z + \frac{2}{z} \right | = 2$

Can anyone post the solution?

Drsoccerball · Jan 5, 2016

Re: HSC 2016 4U Marathon

ze- said:
Can anyone post the solution?

Triangle inequality.

Paradoxica · Jan 5, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
Triangle inequality.

Technically, yes, but the question requires the use of a corollary of the triangle inequality.

Drsoccerball · Jan 5, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
Technically, yes, but the question requires the use of a corollary of the triangle inequality.

Is the answer |z|=1 btw ?

Paradoxica · Jan 5, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
Is the answer |z|=1 btw ?

No. My sources state the answer is not an integer.

Drsoccerball · Jan 5, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
No. My sources state the answer is not an integer.

are you sure it can even have a maximum value? Its a min parabola ?

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