HSC 2016 MX1 Marathon (archive) (2 Viewers)

Paradoxica · Apr 16, 2016

Re: HSC 2016 3U Marathon

For completeness' sake...

$$\noindent$a^2 + b^2 \geq 2ab \Rightarrow 2a^2 + 2b^2 \geq a^2 + 2ab + b^2 \\ 2(a^2 + b^2) \geq (a+b)^2 \equiv 2c^2geq (a+b)^2 \\ a+b \leq \sqrt{2}c $ with equality iff $a=b.$

Paradoxica · Apr 16, 2016

Re: HSC 2016 3U Marathon

Alternatively, the result is instantly true by the Quadratic-Arithmetic Mean Inequality.

$$\noindent$ \forall \, \{a_1, a_2, a_3, \cdots, a_n\}\in \mathbb{R}^+ \\ \sqrt{\frac{a_1^2+a_2^2+a_3^2+\cdots+a_n^2}{n}} \geq \frac{a_1+a_2+a_3+\cdots+a_n}{n} \\ $with equality iff $a_1 = a_2 = a_3 = \cdots = a_n \\ $The desired result then follows by setting $n=2.$

leehuan · Apr 16, 2016

Re: HSC 2016 3U Marathon

What would be the fastest way to prove this?

$\arcsin{\frac{3}{5}}+\arctan{\frac{7}{24}}=\arccos{\frac{3}{5}}$

InteGrand · Apr 16, 2016

Re: HSC 2016 3U Marathon

leehuan said:
What would be the fastest way to prove this?

$\arcsin{\frac{3}{5}}+\arctan{\frac{7}{24}}=\arccos{\frac{3}{5}}$

That's equivalent to arcsin(3/5) + pi/2 – arctan(24/7) = pi/2 – arcsin(3/5) <==> 2arcsin(3/5) = arctan(24/7).

The above can be proved by taking the sines of both sides (you can draw a triangle to assist with finding sin(RHS) and use double-angle formula and Pythagorean identity for sin(LHS)). Then argue that LHS and RHS are both acute, which means that if their sines are equal, then they are equal.

The RHS is clearly an acute angle (i.e. is in the range (0, pi/2)). For the LHS, note that 0 < arcsin(3/5) < arcsin(sqrt(2)/2) = pi/4, since 3/5 < sqrt(2)/2, since (3/5)^2 < 1/2, and 3/5 and sqrt(2)/2 are both positive. Hence LHS = 2*arcsin(3/5) is in the range (0, pi/2).

Paradoxica · Apr 16, 2016

Re: HSC 2016 3U Marathon

leehuan said:
What would be the fastest way to prove this?

$\arcsin{\frac{3}{5}}+\arctan{\frac{7}{24}}=\arccos{\frac{3}{5}}$

$\noindent$ 2\sin^{-1}{\frac{3}{5}}=\frac{\pi}{2}-\tan^{-1}{\frac{7}{24}} \\ 2\sin^{-1}{\frac{3}{5}} = \tan^{-1}{\frac{24}{7}} \\ $Using the identity $2\tan^{-1}{\frac{x}{y}} \equiv \tan^{-1}{\frac{2xy}{y^2-x^2}}$\\ which is valid for $\text{sgn}(xy)=1$, we then have$ \\ 2\sin^{-1}{\frac{3}{5}} = 2\tan^{-1}{\frac{3}{4}}\\$The result then follows by considering the inscribed triangle as above.$

DatAtarLyfe · Apr 18, 2016

Re: HSC 2016 3U Marathon

$$If\, cos(2sinx) = sin(2cosx), show\, that$\\ \ $sinx$\,\pm\, $cosx$ = n\pi \pm\frac{\pi}{4}$

Jeff_ · Apr 18, 2016

Re: HSC 2016 3U Marathon

Permutations question: 6 men and 3 women have a choice of two tables - a 4-seated round table and a 5-seated round table. In how many ways can they be seated if there is at least one woman sitting at the 5-seated round table?

wu345 · Apr 21, 2016

Re: HSC 2016 3U Marathon

DatAtarLyfe said:
$$If\, cos(2sinx) = sin(2cosx), show\, that$\\ \ $sinx$\,\pm\, $cosx$ = n\pi \pm\frac{\pi}{4}$

$\cos { (2\sin { x) } } =\sin { (2\cos { x } ) } \\ \cos { (2\sin { x) } } =\cos { (\frac { \pi }{ 2 } } -2\cos { x } )\\ $By,\quad the\quad general\quad solution\quad of\quad the\quad cosine\quad function$,\\ 2\sin { x } =2n\pi \pm (\frac { \pi }{ 2 } -2\cos { x } )\\ $Rearranging\quad and\quad dividing\quad by\quad two$,\\ \sin { x } \pm \cos { x } =n\pi \pm \frac { \pi }{ 4 }$

For Jeff's question:
$n(at\quad least\quad one\quad women\quad on\quad table\quad of\quad 5)\\ =n(total)-n(no\quad women\quad on\quad table\quad of\quad 5)\\ =n(total)-n(all\quad men\quad on\quad table\quad of\quad 5)\\ =\frac { 9P5 }{ 5 } *\frac { 4P4 }{ 4 } -\frac { 6P5 }{ 5 } *\frac { 4P4 }{ 4 } \\ =17,\quad 280$

Ambility · Apr 21, 2016

Re: HSC 2016 3U Marathon

Jeff_ said:
Permutations question: 6 men and 3 women have a choice of two tables - a 4-seated round table and a 5-seated round table. In how many ways can they be seated if there is at least one woman sitting at the 5-seated round table?

My attempt:
$\\$Combinations in 5-seated table: $\frac{3\times 8\times 7\times 6\times 5}{5}\\$Combinations in 4-seated table: $\frac{4\times 3\times 2\times 1}{4}\\$Total combinations: $\frac{3\times 8\times 7\times 6\times 5}{5}\times \frac{4\times 3\times 2\times 1}{4}=6048$

wu345 · Apr 21, 2016

Re: HSC 2016 3U Marathon

Ambility said:
My attempt:
$\\$Combinations in 5-seated table: $\frac{3\times 8\times 7\times 6\times 5}{5}\\$Combinations in 4-seated table: $\frac{4\times 3\times 2\times 1}{4}\\$Total combinations: $\frac{3\times 8\times 7\times 6\times 5}{5}\times \frac{4\times 3\times 2\times 1}{4}=6048$

Hmmm.. I wonder what was wrong with my solution :/

InteGrand · Apr 21, 2016

Re: HSC 2016 3U Marathon

wu345 said:
Hmmm.. I wonder what was wrong with my solution :/

What was your solution?

wu345 · Apr 21, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
What was your solution?

It's in the reply above ambility (2nd part)

Jeff_ · Apr 21, 2016

Re: HSC 2016 3U Marathon

wu345 said:
For Jeff's question:
$n(at\quad least\quad one\quad women\quad on\quad table\quad of\quad 5)\\ =n(total)-n(no\quad women\quad on\quad table\quad of\quad 5)\\ =n(total)-n(all\quad men\quad on\quad table\quad of\quad 5)\\ =\frac { 9P5 }{ 5 } *\frac { 4P4 }{ 4 } -\frac { 6P5 }{ 5 } *\frac { 4P4 }{ 4 } \\ =17,\quad 280$

Ambility said:
My attempt:
$\\$Combinations in 5-seated table: $\frac{3\times 8\times 7\times 6\times 5}{5}\\$Combinations in 4-seated table: $\frac{4\times 3\times 2\times 1}{4}\\$Total combinations: $\frac{3\times 8\times 7\times 6\times 5}{5}\times \frac{4\times 3\times 2\times 1}{4}=6048$

Checked the answer... wu345 should be right on this one.

InteGrand · Apr 21, 2016

Re: HSC 2016 3U Marathon

wu345's answer is correct. It would be an instructive exercise to find the flaw with the method posted by Ambility.

Jeff_ · Apr 21, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
wu345's answer is correct. It would be an instructive exercise to find the flaw with the method posted by Ambility.

It's probably because it's supposed to be a permutations question.

EDIT. lemme rethink this lol

InteGrand · Apr 21, 2016

Re: HSC 2016 3U Marathon

Jeff_ said:
It's probably because it's supposed to be a permutations question.

EDIT. Although if I didn't say it was a permutations question, could it have been a combinations question?

EDIT 2. I don't think I've encountered any "round table" questions with combinations.

The term posted for Ambility's no. of ways to seat people in the 5-seat table is incorrect. Can anyone explain what was the reasoning behind Ambility's expression for that, and why it is not quite right? (Just focus on the numerator.)

Jeff_ · Apr 21, 2016

Re: HSC 2016 3U Marathon

Can someone explain why they divided by 5 or 4 in the denominator? I'm having a mind blank here. (Or is this because it's combinations, because I haven't really learned much of that yet?)

InteGrand · Apr 21, 2016

Re: HSC 2016 3U Marathon

Jeff_ said:
Can someone explain why they divided by 5 or 4 in the denominator? I'm having a mind blank here. (Or is this because it's combinations, because I haven't really learned much of that yet?)

The reason for that is to account for the rotational symmetry of the seating in tables. It's standard required knowledge for 3U perms and combs for circular arrangements.

Jeff_ · Apr 21, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
The reason for that is to account for the rotational symmetry of the seating in tables. It's standard required knowledge for 3U perms and combs for circular arrangements.

I'm laughing at myself now

I usually just do (n-1)! eg. 3! for a 4-seated instead of 4!/4

InteGrand · Apr 21, 2016

Re: HSC 2016 3U Marathon

Jeff_ said:
I'm laughing at myself now I usually just do (n-1)! eg. 3! for a 4-seated instead of 4!/4

Yeah, they're equivalent methods. The reason those people did divisions by 5 etc. instead was that they found the ways of getting people for the table from the entire group of people first, and this generally in nPr form, rather than a factorial form, so they divided that by 5 and couldn't just straightaway do a 4!. To do the 4! thing we could first choose people for that table in the appropriate way, and then multiply by 4!.

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