# Thread: 2012 Year 9 &10 Mathematics Marathon

1. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by SpiralFlex
Who here would like to do a Spiral exam?
me

2. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by SpiralFlex
Who here would like to do a Spiral exam?
sup

3. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by twinklegal19
There are still infinite solutions for x

Let's make it between 0 and 360 degrees then?
$sinx-cosx=0$

$sinx=cosx$

$\frac{sinx}{cosx}=1$

$tanx=1$

$\therefore ~ x=45~or~225$

Originally Posted by kev-
Water after 5 hours
=800x5
=4000L

let a be height/depth
Volume of diving pool
=6x4xa
=24a m^3
=24000a L

equate: 24000a = 4000
a= 1/6 m
Nice work.

4. ## Re: 2012 Year 9 &10 Mathematics Marathon

Here's another one:
Peter stands due south of a pole. Sam stands 10m due east of Peter. The angle of elevations of Peter and Sam to the top of the pole is 25(degrees) and 15(degrees) respectively. Find the height of the Pole.

5. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by kazemagic
Here's another one:
Peter stands due south of a pole. Sam stands 10m due east of Peter. The angle of elevations of Peter and Sam to the top of the pole is 25(degrees) and 15(degrees) respectively. Find the height of the Pole.
3.65 metres?

6. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by kev-kun
3.65 metres?
o.O I got 3.27 for that one. How did you do it?

7. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by kazemagic
o.O I got 3.27 for that one. How did you do it?
Sorry that I didn't see this Q earlier. After about 3/4 of a page of working, I had 3.27 metres. I treated the length of P to the pole as 7.02 (2 d.p.).

8. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by kazemagic
o.O I got 3.27 for that one. How did you do it?
Lol probably because I used dp's too early == yeah did it again got the same

9. ## Re: 2012 Year 9 &10 Mathematics Marathon

I feel like an idiot, can't get my head around this one. drew a 3D diagram and everything

10. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by kat-
I feel like an idiot, can't get my head around this one. drew a 3D diagram and everything
Nah, don't worry too much about it. Never get thrown off by a question simply because it seems new to you. Make an effort to interpret it. If it doesn't hit you, take a break, have fun and relax and get back to the question once your mind is refreshed. Often, the human mind takes longer to comprehend things at times so it's perfectly normal not to get it the first time around.

When you do problem solving, think this: 'I have all this given information and thus I will record it but what do I need?'

11. ## 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.

12. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by Carrotsticks
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.

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

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

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

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

13. ## Re: 2012 Year 9 &10 Mathematics Marathon

damn beat me to it

14. ## 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?

15. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by maths_genius
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.

16. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by Fawun
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

17. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by Fawun
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.

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

18. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by Demento1

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)

Untitled.jpg

19. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by maths_genius
Correct This one is a really hard question - actually from an hsc paper (Question 10)

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

hint: $m=\tan\theta$

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

20. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by maths_genius
Correct This one is a really hard question - actually from an hsc paper (Question 10)

Untitled.jpg
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)

21. ## 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 ?"

22. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by enoilgam
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

23. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by enoilgam
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.

24. ## 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.

25. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by Demento1
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?

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