HSC 2016 MX1 Marathon (archive) (1 Viewer)

leehuan · Nov 3, 2015

Re: HSC 2016 3U Marathon

Arucerious said:
I don't understand how to do this.
A form has ten questions in order, each of which requires the answer 'Yes' or 'No'. Find the number of ways the form can be filled in:
(g) if exactly three answers are 'Yes', and they occur together,

_ _ _ _ _ _ _ (_ _ _)
The 3 yes's are always together, so we arrange them as one quantity (let's call it y). Note that the only possible ways of arranging (Y Y Y) in a line is 1.

N N N N N N N y
This is now a typical arrangement (making 'words'). Traditionally, the answer here is now 8!/(7!) = 8 ways.
----
Alternatively, using stars and bars there is obviously 8 ways of doing so, because you have 7 bars (the no's) and only have to arrange one star (the y) anywhere between them or outside them.
----
It's quite easy to check this. If we just list all the outcomes out, they are
Y Y Y N N N N N N N
N Y Y Y N N N N N N
N N Y Y Y N N N N N
N N N Y Y Y N N N N
N N N N Y Y Y N N N
N N N N N Y Y Y N N
N N N N N N Y Y Y N
N N N N N N N Y Y Y
Total: 8 ways

Sien · Nov 3, 2015

Re: HSC 2016 3U Marathon

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leehuan · Nov 3, 2015

Re: HSC 2016 3U Marathon

Sien said:
View attachment 32639

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Is this meant to be the next question in the marathon or a SOS?

Sien · Nov 3, 2015

Re: HSC 2016 3U Marathon

leehuan said:
Is this meant to be the next question in the marathon or a SOS?

Both :')

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Ambility · Nov 3, 2015

Re: HSC 2016 3U Marathon

Sien said:
View attachment 32639

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This is what I think.

$$8 Players in each pool. Each player needs to vs all other players. First player will vs 7 other people, next will vs 6 until the last will vs 0 as that person has already vs'd everyone. The total games in each pool is $8+7+6+5+4+3+2+1 = 36$ games. There are two pools and one extra game is played from the champions of either pool. That's $2\times36+1=73$ games in total. The second question is tricky because I'm not sure what it is asking for. Is it the total time elapsed while people are playing tennis or is it the change in time from the beginning of the tournament to the end? If it is the former, the answer is $\frac{73}{3}\times2 = \frac{146}{3}$. If it is the latter, only 12 games can be played per day, so the total time will be 8 days plus 2 hours for one more game.$

I'm not very confident with this answer, but if it is correct, here's a question to continue on with.

$$Prove the identity: $\csc{2A}-3\cot{2A}=2\tan{A}-\cot{A}$

Sien · Nov 3, 2015

Re: HSC 2016 3U Marathon

Ambility said:
This is what I think.

$$8 Players in each pool. Each player needs to vs all other players. First player will vs 7 other people, next will vs 6 until the last will vs 0 as that person has already vs'd everyone. The total games in each pool is $8+7+6+5+4+3+2+1 = 36$ games. There are two pools and one extra game is played from the champions of either pool. That's $2\times36+1=73$ games in total. The second question is tricky because I'm not sure what it is asking for. Is it the total time elapsed while people are playing tennis or is it the change in time from the beginning of the tournament to the end? If it is the former, the answer is $\frac{73}{3}\times2 = \frac{146}{3}$. If it is the latter, only 12 games can be played per day, so the total time will be 8 days plus 2 hours for one more game.$

I'm not very confident with this answer, but if it is correct, here's a question to continue on with.

$$Prove the identity: $\csc{2A}-3\cot{2A}=2\tan{A}-\cot{A}$

The answers a) 57 b)4days 8hrs

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Drsoccerball · Nov 3, 2015

Re: HSC 2016 3U Marathon

Sien said:
The answers a) 57 b)4days 8hrs

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2(7+6+5+...+1)+1 = total matches

57/12 = length of competition (Assuming they minimise time of tournament)

Sien · Nov 4, 2015

Re: HSC 2016 3U Marathon

Drsoccerball said:
2(7+6+5+...+1)+1 = total matches

57/12 = length of competition (Assuming they minimise time of tournament)

Thanks,i forgot the 2x

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leehuan · Nov 4, 2015

Re: HSC 2016 3U Marathon

This is what I got...

$\\ $(a) Round robin seeding - Every player is played against once, so outcomes = 7+6+5+4+3+2+1+0=28, noting that no player can play against himself/herself. Given that we have 2 sets of round robin games, plus the final game, $2\times 28+1=57$ games.$$

$\\ $(b) An assumption is made in this question that no player is playing two games at once. However, this seems valid as we have 16 players altogether. The question is asking for the time taken for the tournament to finish. \\ We deduce this by simply realising firstly that the maximum amount of games we can have per day is $4\times 3=12$ and we then note $\frac { 57 }{ 12 } =4+\frac { 3 }{ 4 } $. So we definitely need 4 days. Then we add on the fraction of a day: $ \frac { 3 }{ 4 } \times 12=8 $ which gives us the hours.$

KingOfActing · Nov 4, 2015

Re: HSC 2016 3U Marathon

Ambility said:
$$Prove the identity: $\csc{2A}-3\cot{2A}=2\tan{A}-\cot{A}$

$\begin{align*}\csc{2A}-3\cot{2A} &= \frac{1}{\sin{2A}} - \frac{3\cos{2A}}{\sin{2A}} \\&= \frac{1-3\cos{2A}}{\sin{2A}} \\&= \frac{1-3(\cos^2{A} - \sin^2{A})}{2\sin{A}\cos{A}} \qquad \fbox{Double Angle Identity} \\&= \frac{\sin^2{A} + \cos^2{A} -3\cos^2{A} + 3\sin^2{A}}{2\sin{A}\cos{A}} \qquad \fbox{Pythagorean Identity}\\&= \frac{4\sin^2{A} - 2\cos^2{A}}{2\sin{A}\cos{A}} \\&= \frac{2(2\sin^2{A} - \cos^2{A})}{2\sin{A}\cos{A}} \\&= \frac{2\sin^2{A} - \cos^2{A}}{\sin{A}\cos{A}} \\&= \frac{2\sin^2{A}}{\sin{A}\cos{A}} - \frac{cos^2{A}}{\sin{A}\cos{A}} \\&= \frac{2\sin{A}}{\cos{A}} - \frac{cos{A}}{\sin{A}} \\&= 2\tan{A} - \cot{A} = R.H.S. \\\text{Q.E.D.}\end{align*}$

Here's a question:

$\text{Find the equation of the tangent to the curve } x^2 + 3xy + 2y^2 + x = 0 \text{ at the point } P(-1,0)$

Drsoccerball · Nov 4, 2015

Re: HSC 2016 3U Marathon

KingOfActing said:
$\begin{align*}\csc{2A}-3\cot{2A} &= \frac{1}{\sin{2A}} - \frac{3\cos{2A}}{\sin{2A}} \\&= \frac{1-3\cos{2A}}{\sin{2A}} \\&= \frac{1-3(\cos^2{A} - \sin^2{A})}{2\sin{A}\cos{A}} \qquad \fbox{Double Angle Identity} \\&= \frac{\sin^2{A} + \cos^2{A} -3\cos^2{A} + 3\sin^2{A}}{2\sin{A}\cos{A}} \qquad \fbox{Pythagorean Identity}\\&= \frac{4\sin^2{A} - 2\cos^2{A}}{2\sin{A}\cos{A}} \\&= \frac{2(2\sin^2{A} - \cos^2{A})}{2\sin{A}\cos{A}} \\&= \frac{2\sin^2{A} - \cos^2{A}}{\sin{A}\cos{A}} \\&= \frac{2\sin^2{A}}{\sin{A}\cos{A}} - \frac{cos^2{A}}{\sin{A}\cos{A}} \\&= \frac{2\sin{A}}{\cos{A}} - \frac{cos{A}}{\sin{A}} \\&= 2\tan{A} - \cot{A} = R.H.S. \\\text{Q.E.D.}\end{align*}$

Here's a question:

$\text{Find the equation of the tangent to the curve } x^2 + 3xy + 2y^2 + x = 0 \text{ at the point } P(-1,0)$

I do not think implicit differentiation is tested 3U.

KingOfActing · Nov 4, 2015

Re: HSC 2016 3U Marathon

Drsoccerball said:
I do not think implicit differentiation is tested 3U.

Really? Our class has been going over it for the past few lessons, so I assumed it would be.

Drsoccerball · Nov 4, 2015

Re: HSC 2016 3U Marathon

KingOfActing said:
Really? Our class has been going over it for the past few lessons, so I assumed it would be.

It's an extension 2 technique... And never will you use it in extension 1.

KingOfActing · Nov 4, 2015

Re: HSC 2016 3U Marathon

Drsoccerball said:
It's an extension 2 technique... And never will you use it in extension 1.

Good to know, I guess. Here's a different question then:

$\text{Find } \frac{d}{dx} x^x \text{ using the fact } e^{\ln{x}} = x \text{. Hence find } \frac{d}{dx} x^{x-2}$.$$

leehuan · Nov 4, 2015

Re: HSC 2016 3U Marathon

$\frac { d }{ dx } \left( { x }^{ x } \right) \\ =\frac { d }{ dx } { e }^{ x\ln { x } }\\ ={ e }^{ x\ln { x } }\left( \frac { x }{ x } +\ln { x } \right) \\ ={ x }^{ x }\left( 1+\ln { x } \right)$

$\\ $Similarly$ \\ \frac { d }{ dx } \left( { x }^{ x-2 } \right) \\ =\frac { d }{ dx } { e }^{ \left( x-2 \right) \ln { x } }\\ ={ e }^{ \left( x-2 \right) \ln { x } }\left( \frac { x-2 }{ x } +\ln { x } \right) \\ ={ x }^{ x-2 }\left( 1+\ln { x } -\frac { 2 }{ x } \right) \\$

NEXT QUESTION:
Ok, let's see how many 3U students will get thrown off by this...

$\\ $Given any cyclic quadrilateral whose side lengths are known, Brahmagupta's formula allows us to determine the area of said quadrilateral. \\ For this question, you may use the diagram supplied. (The centre of the circle is irrelevant and was only used to construct a circle.$$

$\\$ (a) With extensive use of algebra, prove that $4{ \left( ab+cd \right) }^{ 2 }-{ \left( { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }-{ d }^{ 2 } \right) }^{ 2 }=\left( a+b+c-d \right) \left( a+b-c+d \right) \left( a-b+c+d \right) \left( -a+b+c+d \right) \\$You may manipulate any difference of two squares to find shortcuts.$

$\\ $ (b) Let $s=\frac { a+b+c+d }{ 2 } $, i.e., s is the semi-perimeter of the cyclic quadrilateral $PQRS$. Prove that $ \\ \left( a+b+c-d \right) \left( a+b-c+d \right) \left( a-b+c+d \right) \left( -a+b+c+d \right) = 16\left( s-a \right) \left( s-b \right) \left( s-c \right) \left( s-d \right)$

$\\ $ (c) If $\angle P=\theta $, what is $\angle R $?$$

$\\ $ (d) By equating expressions for the square of the diagonal, $l$, prove that $\cos { \theta } =\frac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }-{ d }^{ 2 } }{ 2(ab+cd) }$

$\\ $ (e) Hence, show that $\sin { \theta } =\frac { 2\sqrt { \left( s-a \right) \left( s-b \right) \left( s-c \right) \left( s-d \right) } }{ ab+cd }$

$\\ $ (f) Give the area of $PQRS$ in simplest form.$

$\\ $ (g) Heron's formula can be generalised using this process, and is used to find the area of any triangle without information on height or angle given. Deduce an expression for the area of a triangle using only side lengths and it's semi-perimeter, and explain why the supposed condition of being a 'cyclic triangle' is unnecessary.$$

As the question is long, feel free to only answer a few parts of it. However, don't hog all the easy parts like (c).

KingOfActing · Nov 4, 2015

Re: HSC 2016 3U Marathon

$$a)$\\\begin{align*}4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2&=(2(ab+cd))^2-(a^2+b^2-c^2-d^2)^2\\&=(2(ab+cd)+a^2+b^2-c^2-d^2)(2(ab+cd)-a^2-b^2+c^2+d^2)\\&=(a^2+2ab+b^2-c^2+2cd-d^2)(c^2+2cd+d^2-a^2+2ab-b^2)\\&=((a+b)^2-(c+d)^2)((c+d)^2-(a+b)^2)\\&=(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)&=R.H.S\end{align*}$

$$b)$\\\begin{align*}s = \frac{a+b+c+d}{2} \Longleftrightarrow 2s = a + b + c + d\\\\(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)&=(a+b+c+d-2d)(a+b+c+d-2c)(a+b+c+d-2b)(a+b+c+d-2a)\\&=(2s-2d)(2s-2c)(2s-2b)(2s-2a)\\&=16(s-a)(s-b)(s-c)(s-d)=RHS\end{align*}$

I've done the rest but I think I should let other people have a go at doing them. These were really fun.

kawaiipotato · Nov 5, 2015

Re: HSC 2016 3U Marathon

leehuan said:
$\frac { d }{ dx } \left( { x }^{ x } \right) \\ =\frac { d }{ dx } { e }^{ x\ln { x } }\\ ={ e }^{ x\ln { x } }\left( \frac { x }{ x } +\ln { x } \right) \\ ={ x }^{ x }\left( 1+\ln { x } \right)$

$\\ $Similarly$ \\ \frac { d }{ dx } \left( { x }^{ x-2 } \right) \\ =\frac { d }{ dx } { e }^{ \left( x-2 \right) \ln { x } }\\ ={ e }^{ \left( x-2 \right) \ln { x } }\left( \frac { x-2 }{ x } +\ln { x } \right) \\ ={ x }^{ x-2 }\left( 1+\ln { x } -\frac { 2 }{ x } \right) \\$

NEXT QUESTION:
Ok, let's see how many 3U students will get thrown off by this...

$\\ $Given any cyclic quadrilateral whose side lengths are known, Brahmagupta's formula allows us to determine the area of said quadrilateral. \\ For this question, you may use the diagram supplied. (The centre of the circle is irrelevant and was only used to construct a circle.$$

$\\$ (a) With extensive use of algebra, prove that $4{ \left( ab+cd \right) }^{ 2 }-{ \left( { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }-{ d }^{ 2 } \right) }^{ 2 }=\left( a+b+c-d \right) \left( a+b-c+d \right) \left( a-b+c+d \right) \left( -a+b+c+d \right) \\$You may manipulate any difference of two squares to find shortcuts.$

$\\ $ (b) Let $s=\frac { a+b+c+d }{ 2 } $, i.e., s is the semi-perimeter of the cyclic quadrilateral $PQRS$. Prove that $ \\ \left( a+b+c-d \right) \left( a+b-c+d \right) \left( a-b+c+d \right) \left( -a+b+c+d \right) = 16\left( s-a \right) \left( s-b \right) \left( s-c \right) \left( s-d \right)$

$\\ $ (c) If $\angle P=\theta $, what is $\angle R $?$$

$\\ $ (d) By equating expressions for the square of the diagonal, $l$, prove that $\cos { \theta } =\frac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }-{ d }^{ 2 } }{ 2(ab+cd) }$

$\\ $ (e) Hence, show that $\sin { \theta } =\frac { 2\sqrt { \left( s-a \right) \left( s-b \right) \left( s-c \right) \left( s-d \right) } }{ ab+cd }$

$\\ $ (f) Give the area of $PQRS$ in simplest form.$

$\\ $ (g) Heron's formula can be generalised using this process, and is used to find the area of any triangle without information on height or angle given. Deduce an expression for the area of a triangle using only side lengths and it's semi-perimeter, and explain why the supposed condition of being a 'cyclic triangle' is unnecessary.$$

As the question is long, feel free to only answer a few parts of it. However, don't hog all the easy parts like (c).

Where'd you get that question from?

leehuan · Nov 5, 2015

Re: HSC 2016 3U Marathon

kawaiipotato said:
Where'd you get that question from?

From a textbook

calamebe · Nov 5, 2015

Re: HSC 2016 3U Marathon

leehuan said:
$\frac { d }{ dx } \left( { x }^{ x } \right) \\ =\frac { d }{ dx } { e }^{ x\ln { x } }\\ ={ e }^{ x\ln { x } }\left( \frac { x }{ x } +\ln { x } \right) \\ ={ x }^{ x }\left( 1+\ln { x } \right)$

$\\ $Similarly$ \\ \frac { d }{ dx } \left( { x }^{ x-2 } \right) \\ =\frac { d }{ dx } { e }^{ \left( x-2 \right) \ln { x } }\\ ={ e }^{ \left( x-2 \right) \ln { x } }\left( \frac { x-2 }{ x } +\ln { x } \right) \\ ={ x }^{ x-2 }\left( 1+\ln { x } -\frac { 2 }{ x } \right) \\$

NEXT QUESTION:
Ok, let's see how many 3U students will get thrown off by this...

$\\ $Given any cyclic quadrilateral whose side lengths are known, Brahmagupta's formula allows us to determine the area of said quadrilateral. \\ For this question, you may use the diagram supplied. (The centre of the circle is irrelevant and was only used to construct a circle.$$

$\\$ (a) With extensive use of algebra, prove that $4{ \left( ab+cd \right) }^{ 2 }-{ \left( { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }-{ d }^{ 2 } \right) }^{ 2 }=\left( a+b+c-d \right) \left( a+b-c+d \right) \left( a-b+c+d \right) \left( -a+b+c+d \right) \\$You may manipulate any difference of two squares to find shortcuts.$

$\\ $ (b) Let $s=\frac { a+b+c+d }{ 2 } $, i.e., s is the semi-perimeter of the cyclic quadrilateral $PQRS$. Prove that $ \\ \left( a+b+c-d \right) \left( a+b-c+d \right) \left( a-b+c+d \right) \left( -a+b+c+d \right) = 16\left( s-a \right) \left( s-b \right) \left( s-c \right) \left( s-d \right)$

$\\ $ (c) If $\angle P=\theta $, what is $\angle R $?$$

$\\ $ (d) By equating expressions for the square of the diagonal, $l$, prove that $\cos { \theta } =\frac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }-{ d }^{ 2 } }{ 2(ab+cd) }$

$\\ $ (e) Hence, show that $\sin { \theta } =\frac { 2\sqrt { \left( s-a \right) \left( s-b \right) \left( s-c \right) \left( s-d \right) } }{ ab+cd }$

$\\ $ (f) Give the area of $PQRS$ in simplest form.$

$\\ $ (g) Heron's formula can be generalised using this process, and is used to find the area of any triangle without information on height or angle given. Deduce an expression for the area of a triangle using only side lengths and it's semi-perimeter, and explain why the supposed condition of being a 'cyclic triangle' is unnecessary.$$

As the question is long, feel free to only answer a few parts of it. However, don't hog all the easy parts like (c).

a-f: https://imgur.com/a/bDrUd and for some reason again, it is in the wrong order.

leehuan · Nov 5, 2015

Re: HSC 2016 3U Marathon

NEXT QUESTION:

$\\ $Use mathematical induction to prove the general arithmetic mean-geometric mean inequality: $ \\ \frac { { a }_{ 1 }+{ a }_{ 2 }+\dots +{ a }_{ k } }{ k } \ge \sqrt [ k ]{ { a }_{ 1 }{ a }_{ 2 }\dots { a }_{ k } }$

I'll tone the difficulty back down after this one.

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HSC 2016 MX1 Marathon (archive) (1 Viewer)

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