HSC 2016 MX1 Marathon (archive) (1 Viewer)

Paradoxica · Jan 23, 2016

Re: HSC 2016 3U Marathon

DatAtarLyfe said:
im pree sure i did it wrong, coz he posted the same question in the 4u marathon but made us consider e^itheta, so am not sure how you can use 3u to solve this.
My ways soo abstract its gonna baffle even him XD

bring it.

also, complementary angles.

DatAtarLyfe · Jan 23, 2016

Re: HSC 2016 3U Marathon

Yeh i was gonna use sum and product but he wanted to see another way..

If wu345 answer is correct, then my answer is only half correct. I got root 2 - 1 on the numerator and 7 on the denominator....
brb, gonna check my working

leehuan · Jan 23, 2016

Re: HSC 2016 3U Marathon

DatAtarLyfe said:
Yeh i was gonna use sum and product but he wanted to see another way..

If wu345 answer is correct, then my answer is only half correct. I got root 2 - 1 on the numerator and 7 on the denominator....
brb, gonna check my working

I put it on wolfram and it was correct

Paradoxica · Jan 23, 2016

Re: HSC 2016 3U Marathon

wu345 said:
What about this one?
View attachment 32820

Also, if anyone is looking for a challenge (use products to sums)
$$\noindent Prove $64\sin{10^{\circ}}\sin{20^{\circ}}\sin{30^{\circ}}\sin{40^{\circ}}\sin{50^{\circ}}\sin{60^{\circ}} \sin{70^{\circ}}\sin{80^{\circ}}\sin{90^{\circ}}= \frac{3}{4}$

$$\noindent Once again, the mid-point is the vertex of the quadratic equations in $x$ and $y$ obtained through substitution. We have $x(x+1)^2 = 4a(x+1) , \, (y+1)^2 = 4ay^2(y+1)$. We are intentionally ignoring the obvious solution for the sake of the question, so remove the common factors from both equations to leave $x(x+1)=4a , \, 4ay^2=y+1$. By inspection, the midpoint is $(-\frac{1}{2},\frac{1}{8a})$, which is a vertical line.$ \\ a $ must be positive for the intersections to exist, so the locus of the midpoint is the $y$-positive ray $x = -\frac{1}{2}.$

$$\noindent$ \frac{\sin{3A}}{4} \equiv \sin{A}\sin{(60^\circ -A)}\sin{(60^\circ +A)} \\ \frac{\sin{30^\circ}}{4} = \sin{10^\circ}\sin{50^\circ}\sin{70^\circ} \\ \frac{\sin{60^\circ}}{4} = \sin{20^\circ}\sin{40^\circ}\sin{80^\circ} \\ LHS = 64 \cdot \frac{\sin^2{30^\circ}}{4} \cdot \frac{\sin^2{60^\circ}}{4} \cdot \sin{90^\circ} = 4 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 \cdot \left(\frac{1}{2}\right)^2 =\frac{3}{4} = RHS$

Drsoccerball · Jan 23, 2016

Re: HSC 2016 3U Marathon

How many way can I climb stairs of 12 If I can either take 1 or 2 steps at a time? Eg: 222121

wu345 · Jan 23, 2016

Re: HSC 2016 3U Marathon

is this it?
http://i.imgur.com/yEUxJjG.jpg

Drsoccerball · Jan 23, 2016

Re: HSC 2016 3U Marathon

wu345 said:
is this it?
http://i.imgur.com/yEUxJjG.jpg

Correct!!Also if you write it in terms of $\binom{n}{r}$ you can see the pattern and don't have to do all terms of the expression. Now how about n stairs?

wu345 · Jan 23, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
$$\noindent Once again, the mid-point is the vertex of the quadratic equations in $x$ and $y$ obtained through substitution. We have $x(x+1)^2 = 4a(x+1) , \, (y+1)^2 = 4ay^2(y+1)$. We are intentionally ignoring the obvious solution for the sake of the question, so remove the common factors from both equations to leave $x(x+1)=4a , \, 4ay^2=y+1$. By inspection, the midpoint is $(-\frac{1}{2},\frac{1}{8a})$, which is a vertical line.$ \\ a $ must be positive for the intersections to exist, so the locus of the midpoint is the $y$-positive ray $x = -\frac{1}{2}.$

$$\noindent$ \frac{\sin{3A}}{4} \equiv \sin{A}\sin{(60^\circ -A)}\sin{(60^\circ +A)} \\ \frac{\sin{30^\circ}}{4} = \sin{10^\circ}\sin{50^\circ}\sin{70^\circ} \\ \frac{\sin{60^\circ}}{4} = \sin{20^\circ}\sin{40^\circ}\sin{80^\circ} \\ LHS = 64 \cdot \frac{\sin^2{30^\circ}}{4} \cdot \frac{\sin^2{60^\circ}}{4} \cdot \sin{90^\circ} = 4 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 \cdot \left(\frac{1}{2}\right)^2 =\frac{3}{4} = RHS$

cheers again, solving these qs really helps mate

- two more questions though about those 2 -
1. what if the parabola is (y+1)^2=4a(x+1), wouldn't the locus be y=-1/2 then?
2. how did you know that identity sin 3a/4 =... ?

KingOfActing · Jan 23, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
Correct!!Also if you write it in terms of $\binom{n}{r}$ you can see the pattern and don't have to do all terms of the expression. Now how about n stairs?

$Let $f(n)$ denote the possible combinations of 1 or 2 step moves on n stairs. $f(1) = 1, f(2) = 2$. Now consider the final move of n stairs. You either move one step forward from the $(n - 1)th$ step, or move two steps forward from the $(n-2)th$ step. The total possible combinations for $n$ steps is the sum of the $n-1$ possible combinations and the $n-2$ possible combinations. Hence, $f(n) = f(n-1) + f(n-2)$. Alternatively, this may be written as $F_{n+1}$ where $F_n$ represents the nth Fibonacci number.$

Paradoxica · Jan 23, 2016

Re: HSC 2016 3U Marathon

wu345 said:
cheers again, solving these qs really helps mate - two more questions though about those 2 -
1. what if the parabola is (y+1)^2=4a(x+1), wouldn't the locus be y=-1/2 then?
2. how did you know that identity sin 3a/4 =... ?

1. Exchanging the variables flips the entire geometry of the parabola about the line y=x, so it doesn't change what has been proven.

2. Magic.

InteGrand · Jan 25, 2016

Re: HSC 2016 3U Marathon

$\noindent \textbf{NEW QUESTION}$

$\noindent Let $n$ and $k$ be positive integers with $k\leqslant n$. Suppose there are a total of $n$ marbles in an urn and exactly one of these is red, and the others are all green. Alice randomly selects $k$ marbles from the urn without replacement. What is the probability (in terms of $n$ and $k$) that one of her selected marbles is the red one?$

Drsoccerball · Jan 26, 2016

Re: HSC 2016 3U Marathon

$Use the graph $ f(x)= \frac{x^e}{e^x} $ to prove that :$

$e^{\pi}>\pi ^e$

leehuan · Jan 26, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
$Use the graph $ f(x)= \frac{x^e}{e^x} $ to prove that :$

$e^{\pi}>\pi ^e$

That's a bit tough. I would give the hint to treat the graphing like a normal stationary point etc. graph first. And maybe identify the asymptote.

Also extra hint: f(x)=x^e.e^(-x) to anyone who would overkill the quotient rule here

wu345 · Jan 27, 2016

Re: HSC 2016 3U Marathon

http://i.imgur.com/CQjSJHO.jpg
If someone could please help me with these two that'd be awesome

http://i.imgur.com/oXnNdG3.png
Doesn't seem like people are too keen on probability/perms and combs though haha

Paradoxica · Jan 27, 2016

Re: HSC 2016 3U Marathon

wu345 said:
http://i.imgur.com/CQjSJHO.jpg
If someone could please help me with these two that'd be awesome
http://i.imgur.com/oXnNdG3.png
Doesn't seem like people are too keen on probability/perms and combs though haha

The graph of x^e e^-x cannot cross the y-axis.

InteGrand · Jan 27, 2016

Re: HSC 2016 3U Marathon

wu345 said:
http://i.imgur.com/CQjSJHO.jpg
If someone could please help me with these two that'd be awesome
http://i.imgur.com/oXnNdG3.png
Doesn't seem like people are too keen on probability/perms and combs though haha

$\noindent 26. (a) Label the colours 1,2,3,4,5,6. We can put Colour 1 anywhere on the cube by symmetry. Now fix Colour 1's position, then the face opposite Colour 1 can have its colour chosen in \boxed{5} ways (any of the remaining colours), and each of these choices corresponds to a unique colouring of the cube. Now fix the colour opposite Colour 1's face (call it Colour 2), then for the four faces that are in the ``middle'' of the cube (between the faces of Colour 1 and its opposite), we can choose any colour for the first one of these faces by rotational symmetry, call it Colour 3.$

$\noindent Now the colour of the face opposite Colour 3 can be chosen in \boxed{3} ways (three colours left), and each of these corresponds to a differently coloured cube. Once that face has been coloured, for the remaining two faces, there are $2!=\boxed{2}$ ways to colour them (with Colours 5 and 6), and each colouring corresponds to a differently coloured cube, because in one of them, Colour 5 will be ``left'' of Colour 3, and in the other, ``right''. So the answer is $5\times 3 \times 2 = 30$.$

$\noindent 26. (b) Let our die have its faces be fixed with its numbers 1 to 6. There are \boxed{6} choices of what face to put on top (facing the lid). For each of these choices, we can rotate the cube so that there are \boxed{4} choices of what face is facing the ``front'' of the box. So the answer is $6\times 4 = 24$.$

$\noindent 27. We are simply asking for the number of ways to place 200 indistinguishable dots into 3 distinguishable boxes subject to the restriction that each box has between 0 and 100 (inclusive) dots. Instead of placing 200 dots in, we can consider placing in 100 crosses into the three boxes. Each cross corresponds to a mark \textit{lost}, so that by placing in 100 crosses into the boxes, we lose 100 marks and end up with 200/300. Each box can have up to 100 crosses (as each category is out of 100). Thinking about it this way means we don't need to worry about restrictions on the number of items in each box, since the number of crosses we have to place in total is equal to the maximum number allowed in a box. Using the standard ``stars and bars'' formula, the answer is thus $\binom{102}{2}=5151$.$

wu345 · Jan 27, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
$\noindent 26. (a) Label the colours 1,2,3,4,5,6. We can put Colour 1 anywhere on the cube by symmetry. Now fix Colour 1's position, then the face opposite Colour 1 can have its colour chosen in \boxed{5} ways (any of the remaining colours), and each of these choices corresponds to a unique colouring of the cube. Now fix the colour opposite Colour 1's face (call it Colour 2), then for the four faces that are in the ``middle'' of the cube (between the faces of Colour 1 and its opposite), we can choose any colour for the first one of these faces by rotational symmetry, call it Colour 3.$

$\noindent Now the colour of the face opposite Colour 3 can be chosen in \boxed{3} ways (three colours left), and each of these corresponds to a differently coloured cube. Once that face has been coloured, for the remaining two faces, there are $2!=\boxed{2}$ ways to colour them (with Colours 5 and 6), and each colouring corresponds to a differently coloured cube, because in one of them, Colour 5 will be ``left'' of Colour 3, and in the other, ``right''. So the answer is $5\times 3 \times 2 = 30$.$

$\noindent 26. (b) Let our die have its faces be fixed with its numbers 1 to 6. There are \boxed{6} choices of what face to put on top (facing the lid). For each of these choices, we can rotate the cube so that there are \boxed{4} choices of what face is facing the ``front'' of the box. So the answer is $6\times 4 = 24$.$

$\noindent 27. We are simply asking for the number of ways to place 200 indistinguishable dots into 3 distinguishable boxes subject to the restriction that each box has between 0 and 100 (inclusive) dots. Instead of placing 200 dots in, we can consider placing in 100 crosses into the three boxes. Each cross corresponds to a mark \textit{lost}, so that by placing in 100 crosses into the boxes, we lose 100 marks and end up with 200/300. Each box can have up to 100 crosses (as each category is out of 100). Thinking about it this way means we don't need to worry about restrictions on the number of items in each box, since the number of crosses we have to place in total is equal to the maximum number allowed in a box. Using the standard ``stars and bars'' formula, the answer is thus $\binom{102}{2}=5151$.$

Thanks for help again, I got everything up to the last part of 27 where you said using the 'stars and bars' formula (not sure if i have learnt this yet - is this binomial?). Also sorry my bad, Paradoxica, read calculator wrong hahaha, limit should be 0 not -infinity

leehuan · Jan 27, 2016

Re: HSC 2016 3U Marathon

wu345 said:
Thanks for help again, I got everything up to the last part of 27 where you said using the 'stars and bars' formula (not sure if i have learnt this yet - is this binomial?). Also sorry my bad, Paradoxica, read calculator wrong hahaha

Stars and bars is a clever trick to doing combinatorics. Usually not necessary for 3U (in 4U it has been found to be useful for specific questions), refer to the principles in https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

It does make life easier for 3U students that care to learn it anyway.

InteGrand · Jan 27, 2016

Re: HSC 2016 3U Marathon

wu345 said:
Thanks for help again, I got everything up to the last part of 27 where you said using the 'stars and bars' formula (not sure if i have learnt this yet - is this binomial?). Also sorry my bad, Paradoxica, read calculator wrong hahaha

$\noindent This ``stars and bars'' trick is a classic trick for doing such combinatorial problems. I suggest you read about it online, maybe here for starters:$

https://www.quora.com/How-do-I-use-the-stars-and-bars-combinatorics

$\noindent It's a method that allows us to do questions like ``how many ways can I place $k$ indistinguishable balls into $n$ distinguishable boxes'' in a routine manner. Equivalent types problems are ``how many ways can I select $r$ numbers from the numbers $1,2,\ldots,m$ if the order I select them is unimportant (and we choose the numbers with replacement)'' and ``how many solutions are there to the equation $x_1 + x_2 + \cdots + x_r = N$ if each $x_i$ is a non-negative integer''. (Knowing about stars and bars makes these types of problems almost trivial.)$

Paradoxica · Jan 27, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
$New Question: Paradoxica, Integrand, Ekman and Drsoccerball decide to play a game of 'Bean-boozled.' Taking turns respectively.$

$Assuming there are 4n beans in total and half are barf flavoured, sneeze flavoured, dog food and grass clipping while the other half [are nicer beans(all different).$

$Find the probability that Drsoccerball eats a tainted jelly-bean if when eaten the game ends.$

$$The probability, given some arbitrary $ n \in \mathbb{N}$ is $2n \sum_{r=1}^{\left \lfloor \frac{n}{2} \right \rfloor} \frac{^{2n}\textit{P}_{4r-1}}{^{4n}\textit{P}_{4r}}$

$Now if someone could prove this, that would be nice.$

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