HSC 2016 MX2 Marathon (archive) (4 Viewers)

hedgehog_7 · Jul 10, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
$\noindent If we want to do it algebraically, note since $w = \mathrm{cis}\left( \frac{2\pi}{7}\right)$, we have $w^2 = \mathrm{cis}\left(\frac{4\pi}{7}\right)\Rightarrow \mathrm{Im}\left(w^2\right) = \sin \frac{4\pi}{7} = \sin \frac{3\pi}{7}$ (supplementary angles trig. identity for sine).$

$\noindent Also, $w^4 = \mathrm{cis} \left(\frac{8\pi}{7}\right) \Rightarrow \mathrm{Im}\left(w^4\right) = \sin \frac{8\pi}{7} = -\sin \frac{\pi}{7}$.$

$\noindent So the given sum of sines equals $\mathrm{Im}\left( w \right) + \mathrm{Im}\left(w^2\right) +\mathrm{Im} \left(w^4\right) = \mathrm{Im}\left(w + w^2 + w^4 \right) = \mathrm{Im}\left( z_1 \right)$.$

My issue is why was the answer the sum of Im(z1) and not the sum of Im(z2). They provide different answers so what was the reasoning in choosing one over the other

InteGrand · Jul 11, 2016

Re: HSC 2016 4U Marathon

hedgehog_7 said:
My issue is why was the answer the sum of Im(z1) and not the sum of Im(z2). They provide different answers so what was the reasoning in choosing one over the other

$\noindent I showed above why it is $z_1$. It can't be $z_2$ because $z_2= \overline{z_1}$ (you can show this as an exercise; use $\overline{w^k} = w^{7-k}$ and the fact that the conjugate of a sum is the sum of conjugates to do this), so they have opposite imaginary parts.$

InteGrand · Jul 18, 2016

Re: HSC 2016 4U Marathon

$\noindent Let $n$ be a positive integer. For integers $k$, let $\omega_k =\mathrm{c}i\mathrm{s} \left(\frac{\left(2k+1\right)\pi}{n}\right)$ (this is a complex $n^{\text{th}}$ root of $-1$). Show that $\omega_{j}=\overline{\omega_k}$, where $j=n-k-1$.$

KingOfActing · Jul 18, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
$\noindent Let $n$ be a positive integer. For integers $k$, let $\omega_k =\mathrm{c}i\mathrm{s} \left(\frac{\left(2k+1\right)\pi}{n}\right)$ (this is a complex $n^{\text{th}}$ root of $-1$). Show that $\omega_{j}=\overline{\omega_k}$, where $j=n-k-1$.$

$\overline{\omega_k} = \omega_j \iff \omega_k \omega_j = \omega_k\overline{\omega_k} = \lvert \omega_k \rvert^2 = 1\\\\\begin{align*}\text{L.H.S.} &= \omega_k\omega_j \\&=\operatorname{cis}\left(\frac{(2k+1)\pi}{n} \right )\operatorname{cis}\left(\frac{(2n-2k-1)\pi}{n} \right)\\&=\operatorname{cis}\left (\frac{\pi(2k+1+2n-2k-1)}{n} \right )\\&=\operatorname{cis}\left(2\pi\right) \\&= 1 \\&= \text{R.H.S.} \end{align*}$

LMBPFEK · Jul 18, 2016

Re: HSC 2016 4U Marathon

Does anyone have any difficult questions?

Drsoccerball · Jul 18, 2016

Re: HSC 2016 4U Marathon

LMBPFEK said:
Does anyone have any difficult questions?

Somewhat difficult if you don't know the 'guitar method.'

$$ Solve : $ |x| - |7-x| +2|x-2| \leq 4$

InteGrand · Jul 18, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
Somewhat difficult if you don't know the 'guitar method.'

$$ Solve : $ |x| - |7-x| +2|x-2| \leq 4$

Is there any story behind the naming of this 'guitar method' (like there was for the 'burning lady log law' thing)?

Drsoccerball · Jul 18, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
Is there any story behind the naming of this 'guitar method' (like there was for the 'burning lady log law' thing)?

Yes.

Paradoxica · Jul 18, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
Somewhat difficult if you don't know the 'guitar method.'

$$ Solve : $ |x| - |7-x| +2|x-2| \leq 4$

By inspection....

LMBPFEK · Jul 18, 2016

Re: HSC 2016 4U Marathon

I'd hardly call that absolute value question a "difficult" question, does anyone have any hard questions that they couldn't solve?

leehuan · Jul 18, 2016

Re: HSC 2016 4U Marathon

LMBPFEK said:
I'd hardly call that absolute value question a "difficult" question, does anyone have any hard questions that they couldn't solve?

Prove Fermat's Last Theorem by induction.

Paradoxica · Jul 18, 2016

Re: HSC 2016 4U Marathon

leehuan said:
Prove Fermat's Last Theorem by induction.

Prove the Poincaré conjecture using elementary methods.

KingOfActing · Jul 19, 2016

Re: HSC 2016 4U Marathon

LMBPFEK said:
I'd hardly call that absolute value question a "difficult" question, does anyone have any hard questions that they couldn't solve?

Paradoxica and I proved this one together a while back. It's a bit out of scope for 4U I think, but if you're determined you can prove it using just 4U methods.

$Prove that $2^n + 3^n = x^3$ has no integral solutions.$

leehuan · Jul 21, 2016

Re: HSC 2016 4U Marathon

$\text{Give at least 3 methods to finding the exact value of }\cos \frac{\pi}{5}$

Paradoxica · Jul 21, 2016

Re: HSC 2016 4U Marathon

leehuan said:
$\text{Give at least 3 methods to finding the exact value of }\cos \frac{\pi}{5}$

1. Inspection
2. Inspection
3. Inspection

RenegadeMx · Jul 21, 2016

Re: HSC 2016 4U Marathon

leehuan said:
$\text{Give at least 3 methods to finding the exact value of }\cos \frac{\pi}{5}$

google
maple
this forum

hunter22 · Jul 27, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
Somewhat difficult if you don't know the 'guitar method.'

$$ Solve : $ |x| - |7-x| +2|x-2| \leq 4$

Can someone explain this one? I can't do it.

InteGrand · Jul 27, 2016

Re: HSC 2016 4U Marathon

hunter22 said:
Can someone explain this one? I can't do it.

You can analyse it separately in the intervals x < 0, 0 ≤ x ≤ 2, x > 2; this lets us get rid of absolute value signs in each interval.

wu345 · Aug 27, 2016

Re: HSC 2016 4U Marathon

$$If$ \quad a_{r} $denotes the coefficient of $ x^{r} \quad $in the expansion of$ (1+x+x^{2})^{n} \\ $show that a_{r}=a_{2n-r}$

InteGrand · Aug 27, 2016

Re: HSC 2016 4U Marathon

wu345 said:
$$If$ a_{r} $denotes the coefficient of $ x^{r} $in the expansion of$ (1+x+x^{2})^{2} \\ $show that a_{r}=a_{2n-r}$

Is the exponent meant to be n? You wrote 2.

HSC 2016 MX2 Marathon (archive) (4 Viewers)

Member

Well-Known Member

Well-Known Member

lukewarm mess

New Member

Well-Known Member

Well-Known Member

Well-Known Member

-insert title here-

New Member

Well-Known Member

-insert title here-

lukewarm mess

Well-Known Member

-insert title here-

Kosovo is Serbian

New Member

Well-Known Member

Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)