2012 Year 9 &10 Mathematics Marathon (1 Viewer)

Carrotsticks · Aug 1, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Show that the sum of the two roots of any quadratic ax^2 +bx+c is -b/a.

Demento1 · Aug 1, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Carrotsticks said:
Show that the sum of the two roots of any quadratic ax^2 +bx+c is -b/a.

My attempt to type it all out on LaTeX. It would be quite tedious typing up the quadratic equation but I managed.

$x=\frac{-b\pm \sqrt[]{b^2-4ac}}{2a}$

Add the 2 roots together.

$=\frac{-b+ \sqrt[]{b^2-4ac}}{2a} + \frac{-b- \sqrt[]{b^2-4ac}}{2a}$

$=\frac{-2b}{2a}$

$=\frac{-b}{a}$

kat- · Aug 1, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

damn beat me to it

maths_genius · Aug 8, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

An isosceles triangle has a height that is half the length of its base. If the area of the triangle is 36m^2, what is the perimeter?

Fawun · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

maths_genius said:
An isosceles triangle has a height that is half the length of its base. If the area of the triangle is 36m^2, what is the perimeter?

is the perimeter somewhere around 29m? because that's what I got. I'll post up my working out once I know if I got my answer right or not.

maths_genius · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Fawun said:
is the perimeter somewhere around 29m? because that's what I got. I'll post up my working out once I know if I got my answer right or not.

Yep

Demento1 · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Fawun said:
is the perimeter somewhere around 29m? because that's what I got. I'll post up my working out once I know if I got my answer right or not.

You are correct. Was this your working out?

Let x = height of triangle, hence 2x = base of triangle

By triangle formula:

$\frac{2x^2}{2}=36$

$x^2=36$

$x=6$

Reject - 6.

Now by pythagorean method on a right angle triangle, let y = hypotenuse:

$6^2 + 6^2 = y^2$

$y^2=72$

$y=\sqrt{72}=6\sqrt{2}$

Therefore, perimeter = $6\sqrt{2} + 6\sqrt{2} +12 \approx 29$

maths_genius · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
You are correct. Was this your working out?

Let x = height of triangle, hence 2x = base of triangle

By triangle formula:

$\frac{2x^2}{2}=36$

$x^2=36$

$x=6$

Reject - 6.

Now by pythagorean method on a right angle triangle, let y = hypotenuse:

$6^2 + 6^2 = y^2$

$y^2=72$

$y=\sqrt{72}=6\sqrt{2}$

Therefore, perimeter = $6\sqrt{2} + 6\sqrt{2} +12 \approx 29$

Correct

This one is a really hard question - actually from an hsc paper (Question 10)

twinklegal19 · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

maths_genius said:
Correct This one is a really hard question - actually from an hsc paper (Question 10)

View attachment 26075

I think this is sort of a proof for the "acute angle between two lines" formula in 3U, nice question

hint: $m=\tan\theta$

$$where $\theta$ is any angle that a line makes with the positive direction of the x-axis$

Demento1 · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

maths_genius said:
Correct This one is a really hard question - actually from an hsc paper (Question 10)

View attachment 26075

For i)

OQ = OP (radii of circle)

Therefore, triangle OPQ is isosceles.

$\alpha = 2\beta$

(angle subtended at the centre is twice the angle subtended at the circumference by the same arc)

enoilgam · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Here is one I was doing the other day with a student - it isnt the hardest question but its definitely an interesting one (it had me stumped for a bit).

"In my money box, I have only $1 and $2 coins. I have 240 coins in total worth $318. How many $2 coins do I have ?"

twinklegal19 · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

enoilgam said:
Here is one I was doing the other day with a student - it isnt the hardest question but its definitely an interesting one (it had me stumped for a bit).

"In my money box, I have only $1 and $2 coins. I have 240 coins in total worth $318. How many $2 coins do I have ?"

Ah, I remember those questions. I was so stumped when I came across them in the primary school maths olympiads and the selective test

However once I learnt the beauty of algebra and solving simultaneously, it made everything so much clearer haha

Demento1 · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

enoilgam said:
Here is one I was doing the other day with a student - it isnt the hardest question but its definitely an interesting one (it had me stumped for a bit).

"In my money box, I have only $1 and $2 coins. I have 240 coins in total worth $318. How many $2 coins do I have ?"

Does somebody capable want to do do this question? I have an answer for it but I will let others answer it instead of me for a while.

Demento1 · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

The circle paper (almost forgot about it)

For ii)

Now, $sin(180-2\beta )=sin2\beta$

Apply sine rule:

$\frac{PQ}{sin2\beta }= \frac{1}{sin\beta }$

$PQ = \frac{sin2\beta }{sin\beta }$

I'm not quite sure if this is correct or whether I can continue to simplify. Worth a shot answering.

twinklegal19 · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
The circle paper (almost forgot about it)

For ii)

Now, $sin(180-2\beta )=sin2\beta$

Apply sine rule:

$\frac{PQ}{sin2\beta }= \frac{1}{sin\beta }$

$PQ = \frac{sin2\beta }{sin\beta }$

I'm not quite sure if this is correct or whether I can continue to simplify. Worth a shot answering.

phew, at least you didn't simplify it into 2!

if you want to know,

$\sin2x=2\sin{x}\cos{x}$

although this is a 3U shortcut, I highly doubt that it is actually needed

maybe try a slightly different approach?

iBibah · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
The circle paper (almost forgot about it)

For ii)

Now, $sin(180-2\beta )=sin2\beta$

Apply sine rule:

$\frac{PQ}{sin2\beta }= \frac{1}{sin\beta }$

$PQ = \frac{sin2\beta }{sin\beta }$

I'm not quite sure if this is correct or whether I can continue to simplify. Worth a shot answering.

What you have done there is find the length of PQ, not the equation of the line.

Demento1 · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

iBibah said:
What you have done there is find the length of PQ, not the equation of the line.

My bad. You're already given the gradient for m so can you explain how y-intercept is found please? I haven't done any year 11 maths so forgive me if I am clueless as to how I should approach the question.

iBibah · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
My bad. You're already given the gradient for m so can you explain how y-intercept is found please? I haven't done any year 11 maths so forgive me if I am clueless as to how I should approach the question.

You use y-y1=m(x-x1) (a formula for working out the equation of a line given a point and gradient) and your given Q(-1,0).........

so y1=0
x1=-1

I think you can do the rest

kat- · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

enoilgam said:
Here is one I was doing the other day with a student - it isnt the hardest question but its definitely an interesting one (it had me stumped for a bit).

"In my money box, I have only $1 and $2 coins. I have 240 coins in total worth $318. How many $2 coins do I have ?"

78?

EDIT: my working out
x + 2y = 318 (1)
x = 240 - y (2)

(240 - y) + 2y = 318
y = 318 - 240
= 78

therefore amount of $2 coins is 78.

Where's the latex button? Or do i need to download it separately

Fawun · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
You are correct. Was this your working out?

Let x = height of triangle, hence 2x = base of triangle

By triangle formula:

$\frac{2x^2}{2}=36$

$x^2=36$

$x=6$

Reject - 6.

Now by pythagorean method on a right angle triangle, let y = hypotenuse:

$6^2 + 6^2 = y^2$

$y^2=72$

$y=\sqrt{72}=6\sqrt{2}$

Therefore, perimeter = $6\sqrt{2} + 6\sqrt{2} +12 \approx 29$

Similar working out

I will never be able to understand how to use latex

2012 Year 9 &10 Mathematics Marathon (1 Viewer)

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