HSC 2016 Maths Marathon (archive) (3 Viewers)

wu345 · Feb 6, 2016

Re: HSC 2016 2U Marathon

$R=h\tan { x } \\ With\quad pythagoras\quad in\quad the\quad smaller\quad triangle,\\ { r }^{ 2 }={ R }^{ 2 }+{ (h-r) }^{ 2 }\\ { r }^{ 2 }=h^{ 2 }\tan ^{ 2 }{ x } +{ h }^{ 2 }-2hr+{ r }^{ 2 }\\ { h }^{ 2 }\sec ^{ 2 }{ x } -2hr=0\\ h(h\sec ^{ 2 }{ x } -2r)=0\\ h>0,\quad thus\quad h=2r\cos ^{ 2 }{ x } \\ R=h\tan { x } \\ R=2r\cos ^{ 2 }{ x } *\frac { \sin { x } }{ \cos { x } } \\ R=2r\sin { x } \cos { x } \\ sub\quad into\quad Vcone=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\\ \\$

leehuan · Feb 6, 2016

Re: HSC 2016 2U Marathon

wu345 said:
$R=h\tan { x } \\ With\quad pythagoras\quad in\quad the\quad smaller\quad triangle,\\ { r }^{ 2 }={ R }^{ 2 }+{ (h-r) }^{ 2 }\\ { r }^{ 2 }=h^{ 2 }\tan ^{ 2 }{ x } +{ h }^{ 2 }-2hr+{ r }^{ 2 }\\ { h }^{ 2 }\sec ^{ 2 }{ x } -2hr=0\\ h(h\sec ^{ 2 }{ x } -2r)=0\\ h>0,\quad thus\quad h=2r\cos ^{ 2 }{ x } \\ R=h\tan { x } \\ R=2r\cos ^{ 2 }{ x } *\frac { \sin { x } }{ \cos { x } } \\ R=2r\sin { x } \cos { x } \\ sub\quad into\quad Vcone=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\\ \\$

This looks much more like what I had in my mind but was lazy to write.

Nailgun · Feb 6, 2016

Re: HSC 2016 2U Marathon

wu345 said:
$R=h\tan { x } \\ With\quad pythagoras\quad in\quad the\quad smaller\quad triangle,\\ { r }^{ 2 }={ R }^{ 2 }+{ (h-r) }^{ 2 }\\ { r }^{ 2 }=h^{ 2 }\tan ^{ 2 }{ x } +{ h }^{ 2 }-2hr+{ r }^{ 2 }\\ { h }^{ 2 }\sec ^{ 2 }{ x } -2hr=0\\ h(h\sec ^{ 2 }{ x } -2r)=0\\ h>0,\quad thus\quad h=2r\cos ^{ 2 }{ x } \\ R=h\tan { x } \\ R=2r\cos ^{ 2 }{ x } *\frac { \sin { x } }{ \cos { x } } \\ R=2r\sin { x } \cos { x } \\ sub\quad into\quad Vcone=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\\ \\$

Ah I see now, thanks!

wu345 · Feb 6, 2016

Re: HSC 2016 2U Marathon

A good challenge question:
A parallelogram is inscribed within a triangle so that they have one vertex in common. The other vertices lie along the sides of the triangle. Prove that the maximum area of the parallelogram is half that of the triangle.

Drsoccerball · Feb 6, 2016

Re: HSC 2016 2U Marathon

wu345 said:
$R=h\tan { x } \\ With\quad pythagoras\quad in\quad the\quad smaller\quad triangle,\\ { r }^{ 2 }={ R }^{ 2 }+{ (h-r) }^{ 2 }\\ { r }^{ 2 }=h^{ 2 }\tan ^{ 2 }{ x } +{ h }^{ 2 }-2hr+{ r }^{ 2 }\\ { h }^{ 2 }\sec ^{ 2 }{ x } -2hr=0\\ h(h\sec ^{ 2 }{ x } -2r)=0\\ h>0,\quad thus\quad h=2r\cos ^{ 2 }{ x } \\ R=h\tan { x } \\ R=2r\cos ^{ 2 }{ x } *\frac { \sin { x } }{ \cos { x } } \\ R=2r\sin { x } \cos { x } \\ sub\quad into\quad Vcone=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\\ \\$

When you're writing text in latex use $ signs to make it look clean.

$Without dollar signs'$

$With dollar signs$
Make sure to leave a space between dollar signs and the next word and previous word. Click "Reply with Quote" to see exactly what I did.

leehuan · Feb 6, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
When you're writing text in latex use $ signs to make it look clean.

$Without dollar signs'$

$With dollar signs$
Make sure to leave a space between dollar signs and the next word and previous word. Click "Reply with Quote" to see exactly what I did.

When I first started using LaTeX I only used Daum, and that is precisely what daum does.

A space is interpreted using the \quad command.

But of course, after I somewhat learnt how to use LaTeX (wu take note) I just typed in normal text with the $ signs.

Paradoxica · Feb 6, 2016

Re: HSC 2016 2U Marathon

$$\noindent Given $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=-2$ and $\frac{x}{z}+\frac{z}{y}+\frac{y}{x}=0\\ $Find the value of $\frac{x^3}{y^3}+\frac{y^3}{z^3}+\frac{z^3}{x^3}$ and $\frac{x^3}{z^3}+\frac{y^3}{x^3}+\frac{z^3}{y^3}\\ $Try one of the two following identities: $\\a^3+b^3+c^3 -3abc\equiv (a+b+c)[(a+b+c)^2 -3(ab+bc+ca)] \\ (a+b+c)^3 \equiv a^3+b^3+c^3 +3(a+b)(b+c)(c+a)$

wu345 · Feb 6, 2016

Re: HSC 2016 2U Marathon

Ceebs posting full working this late, hopefully this is right:
$$Let$\quad a=\frac { x }{ y } ,\quad b=\frac { y }{ z } ,\quad c=\frac { z }{ x } \\ $Using\quad identity$\quad { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }={ (a+b+c) }^{ 3 }+3(a+b+c)(ab+ac+bc)-3abc,\\ \frac { { x }^{ 3 } }{ { y }^{ 3 } } +\frac { { y }^{ 3 } }{ { z }^{ 3 } } +\frac { { z }^{ 3 } }{ { x }^{ 3 } } =-5\\ $Using\quad the\quad same\quad identity,\quad we\quad get$\quad \frac { { x }^{ 3 } }{ { z }^{ 3 } } +\frac { { z }^{ 3 } }{ { y }^{ 3 } } +\frac { { y }^{ 3 } }{ { x }^{ 3 } } =3\\$

Paradoxica · Feb 6, 2016

Re: HSC 2016 2U Marathon

wu345 said:
Ceebs posting full working this late, hopefully this is right:
$$Let$\quad a=\frac { x }{ y } ,\quad b=\frac { y }{ z } ,\quad c=\frac { z }{ x } \\ $Using\quad identity$\quad { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }={ (a+b+c) }^{ 3 }+3(a+b+c)(ab+ac+bc)-3abc,\\ \frac { { x }^{ 3 } }{ { y }^{ 3 } } +\frac { { y }^{ 3 } }{ { z }^{ 3 } } +\frac { { z }^{ 3 } }{ { x }^{ 3 } } =-5\\ $Using\quad the\quad same\quad identity,\quad we\quad get$\quad \frac { { x }^{ 3 } }{ { z }^{ 3 } } +\frac { { z }^{ 3 } }{ { y }^{ 3 } } +\frac { { y }^{ 3 } }{ { x }^{ 3 } } =3\\$

I can confirm the -5 but not the 3.

wu345 · Feb 7, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:
I can confirm the -5 but not the 3.

well i checked my working and i can't find anything wrong - its probably because you have z^3/x^3 as your last term where as i have y^3/x^3

davidgoes4wce · Feb 7, 2016

Re: HSC 2016 2U Marathon

I can confirm I got the same answer as Paradoxica:

I think the trick part is to obtain the ac+ab+bc=0 equation, I've also shown that in my working:

davidgoes4wce · Feb 7, 2016

Re: HSC 2016 2U Marathon

Oh wait, I didnt see that wu345 got the 1st part correct.

davidgoes4wce · Feb 7, 2016

Re: HSC 2016 2U Marathon

redoing part b) of Paradoxica's question

davidgoes4wce · Feb 7, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Circle geometry is not assumed knowledge for mathematics 2 unit as the three LOBES of 10 5.3 are ONLY RECOMMENDED for those who intended to take Extension 1 mathematics.

Those that do 5.3 withOUT the LOBES are highly recommended to have sticked with 2 unit to begin with.
The LOBES were recommended for those that specifically wanted to do 3 unit.

I bumped across this thread again but what is the acronym of LOBES?

davidgoes4wce · Feb 7, 2016

Re: HSC 2016 2U Marathon

wu345 said:
nono you can do it without any 3u i just did a dumb way hahaha

edit: Wait nvm you need 3u identity to transform cos 2x if youre using cosine rule oops

$Let\quad s=slant\quad side\quad of\quad cone\\ { s }^{ 2 }={ r }^{ 2 }+{ r }^{ 2 }-2{ r }^{ 2 }cos(180°-2x)\\ { s }^{ 2 }=2{ r }^{ 2 }+2{ r }^{ 2 }cos(2x)\\ { s }^{ 2 }=2{ r }^{ 2 }+2{ r }^{ 2 }(2\cos ^{ 2 }{ x } -1)\\ { s }^{ 2 }=2{ r }^{ 2 }+4{ r }^{ 2 }\cos ^{ 2 }{ x } -2{ r }^{ 2 }\\ s=4{ r }^{ 2 }\cos ^{ 2 }{ x } \\ s=2rcosx\\ R=s\sin { x } ,h=s\cos { x } \\ sub\quad into\quad Vcone=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\\ \\$

it seems like you have the equation but I am struggling to visualise this question. I am getting the feeling there is a cone within a sphere?(I don't think the cone is touching the sphere, Im assuming we make that assumption). I also think the big R, in this example refers to the radius of the cone if I stand to be corrected? I drew something along the lines of this:

Nailgun · Feb 7, 2016

Re: HSC 2016 2U Marathon

I think it's assumed it's touching the sphere
This was the diagram wu345 made

davidgoes4wce · Feb 7, 2016

Re: HSC 2016 2U Marathon

Ok thanks, this isnt a 2U question, i pinched the question from a miscellaneous book.

Nailgun · Feb 7, 2016

Re: HSC 2016 2U Marathon

davidgoes4wce said:
Ok thanks, this isnt a 2U question, i pinched the question from a miscellaneous book.

Ahahah well there was a 2U solution, so really I should have got it. Still need heaps of work lol

Paradoxica · Feb 7, 2016

Re: HSC 2016 2U Marathon

$\noindent Given $p$ and $q$ are integers that satisfy $p+q = 198$ , find all possible integer roots of the quadratic equation $x^2+px+q=0 $ Hint: 199 is prime.$

Nailgun · Feb 7, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:
$\noindent Given $p$ and $q$ are integers that satisfy $p+q = 198$ , find all possible integer roots of the quadratic equation $x^2+px+q=0 $ Hint: 199 is prime.$

Okay question, do I have to use this Euclid's lemma thing

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