Reply With Quote*set on the field of real numbers.Originally Posted by Drongoski
The set of natural numbers is not a field.
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*set on the field of real numbers.
Bachelor of Science (Adv. Mathematics) - University of Sydney:
Yes that is what I meant, but I mistook the N for a Z, that is why I added a N plus.Integers can be negative. Natural numbers are always non-negative.
So you can writeor
for {1, 2, 3, ....}
Theres basically no agreed standard for what "N" means... I usually just write N_0 or N^+ (dunno how to do blackboard bold on this), in a way that would be pretty unambiguous.
Someone give a problem!
Not really something that would appear in the HSC (perhaps a trial if some school likes weird questions).
Consider a unit circle. Inscribed within this unit circle is an equilateral triangle such that all 3 vertices touch the circle. Inscribed within this triangle is another circle such that all 3 sides of the triangle are in contact with the circle.
This process repeats infinitely.
Find the limiting ratio Red : Green as the number of iterations approaches infinity.
Bachelor of Science (Adv. Mathematics) - University of Sydney:
Sorry, I should have made some parts that lead to the final answer.
But I love questions where you're just told what to find, and you can use any method to come to a solution. This is where you start seeing originality.
Bachelor of Science (Adv. Mathematics) - University of Sydney:
DudeHere's the solution to largarithmic's question:
Here's the solution to spiralfex's question:
Here's the next question:
Holy
Crap
Lots of writing!
How long did that take you?
Revived from BoS death, god wants me to continue on my mission....
It's not as bad if you use eiθ notation. (Here I'm referring to spiralflex's question).
Also, for part (iii) you are solving for n, not θ and you should get n=2k.
Last edited by tywebb; 9 Jan 2012 at 10:19 AM.
Just write (without spaces)
[ tex ]\mathbb{N}_0[ /tex ]
or
[ tex ]\mathbb{N}^+[ /tex ]
and you should get
or
For positive integers though it's more common to use
cool problem!!!Not really something that would appear in the HSC (perhaps a trial if some school likes weird questions).
Consider a unit circle. Inscribed within this unit circle is an equilateral triangle such that all 3 vertices touch the circle. Inscribed within this triangle is another circle such that all 3 sides of the triangle are in contact with the circle.
This process repeats infinitely.
Find the limiting ratio Red : Green as the number of iterations approaches infinity.
Anyway here's how I'd do it.
Observation: each circle has half the radius as the one before it. Proof: Clearly all the circles, since the diagram has rotational symmetry, have the same centre; call this O. Now let a vertex of say the biggest triangle by A, and the second biggest circle be tangent to a side of the biggest triangle at say P where one of the endpoints of the side P is on is A. Thenwhere O,I,R,r are the circumcentre, incentre, circumradius, inradius respectively.
We now find the ratio of the red area to the green area in just one annulus. So say take the biggest circle and triangle, and cut a hole where the next circle is. Say the radius of the big circle is one. Then the area of the triangle including the hole is 3*1/2*1*1*sin120 = (3root3)/4. Then, the red area is R = pi - (3root3)/4 and the green area is G = (3root3)/4 - pi/4.
We now find the final answer. All the red and green areas form a geometric series with ratio 1/4 since each circle is half the radius of the one before so areas are reduced to a quarter; so the total red area = R(1+1/4+1/16+...) = R/1-(1/4) = 4R/3. Similarly total green area = 4G/3. Then the final ratio is the same as R/G, i.e. it is
What if the radius isn't one? When I did the question, I got a similar answer but I used a variable r and couldn't simplify past a point, I am sure I did something wrong but when I got a similar looking anwser.cool problem!!!
Anyway here's how I'd do it.
Observation: each circle has half the radius as the one before it. Proof: Clearly all the circles, since the diagram has rotational symmetry, have the same centre; call this O. Now let a vertex of say the biggest triangle by A, and the second biggest circle be tangent to a side of the biggest triangle at say P where one of the endpoints of the side P is on is A. Then
We now find the ratio of the red area to the green area in just one annulus. So say take the biggest circle and triangle, and cut a hole where the next circle is. Say the radius of the big circle is one. Then the area of the triangle including the hole is 3*1/2*1*1*sin120 = (3root3)/4. Then, the red area is R = pi - (3root3)/4 and the green area is G = (3root3)/4 - pi/4.
We now find the final answer. All the red and green areas form a geometric series with ratio 1/4 since each circle is half the radius of the one before so areas are reduced to a quarter; so the total red area = R(1+1/4+1/16+...) = R/1-(1/4) = 4R/3. Similarly total green area = 4G/3. Then the final ratio is the same as R/G, i.e. it is
Last edited by lolcakes52; 9 Jan 2012 at 11:50 AM.
Doesn't matter because of dimensionality. Like, what does "radius one" mean? It means, "radius one UNIT" where that unit is something you chose. You could have, the outer radius is radius "one inch" for instance, or could be "two centimetres": but if its radius "two centimetres" you just let see, a wombiunit = two centimetres, and call is "radius one wombiunit". When dealing with areas you need to adjust because conversion factors are squared. But it should definitely work out if the radius is R instead of 1, you just literally add a factor of R onto every length and a factor of R^2 onto every area and its still correct.
Right, so I did it again and I still get a different answer. Your method is definitely shorter as I am using infinite series and a lot more algebra.
You dont actually need to use an infinite series. You can use the following fact (prove it for yourself its easy):
If, then both these are also equal to
.
Then if R_i and G_i denote the red area/green area within the ith annulus, clearly R1/G1 = R2/G2 = R3/G3 =... by similarity so applying the above identity the total ratio = R1+R2+.../G1+G2+... = R1/G1. I computed R1 and G1 above (and I'm pretty sure theyre both correct), giving another way to get my answer.
Post your whole method?
When you say 'infinite series', are you referring to the infinite geometric series? If the algebra is crazy, it implies that you are going down the wrong track. There is always a simpler solution.
Nice work larg on your solution. Perhaps you can post a similar 'fun' question?
Bachelor of Science (Adv. Mathematics) - University of Sydney:
Yes, infinite geometric series. I think I am just confusing myself.
This is from the unsw school mathematics comp, it is interesting and I would consider it harder 3 unit.
Screen shot 2012-01-09 at 6.53.45 PM.png
Last edited by lolcakes52; 9 Jan 2012 at 6:58 PM.
----------------------------------------------------------------------------------------------
I already posted up a question before but maybe people forgot, so here it is again:
\text{ If }y=\textstyle{1\over x}+\sum_{n=1}^\infty{2x\over x^2-n^2}\text{ for }0<x<1\text{ use the fact that }\pi^2=\sum_{n=1}^\infty{6\over n^2}\text{ to show }\newline\ \newline{\hskip0.25in} \pi^2+y'+y^2=0\newline\ \newline (b)\text{ Hence show }y=\pi\cot\pi x\text{ for }0<x<1\newline\ \newline{\hskip0.25in}\text{ and }\cot x=\textstyle{1\over x}+\sum_{n=1}^\infty{2x\over x^2-\pi^2n^2}\text{ for }0<x<\pi\newline\ \newline (c)\text{ Find }\int_0^x(\cot t-{1\over t})\ \!dt\text{ and hence show }\newline\ \newline{\hskip0.25in}\sin x=\textstyle x\prod_{n=1}^\infty(1-{x^2\over\pi^2n^2})\text{ for }0<x<\pi\newline\ \newline (d)\text{ Let }x=\textstyle{\pi\over k}\text{ to show }\prod_{n=1}^\infty{(kn)^2\over(kn)^2-1}={\pi/k\over\sin(\pi/k)}\text{ for }k>1\newline\ \newline (e)\text{ Let }k=2\text{ to prove Wallis' Product, }\prod_{n=1}^\infty{(2n)^2\over (2n)^2-1}={\pi\over2})
Last edited by tywebb; 9 Jan 2012 at 7:54 PM.
tywebb, I have already done it, but I didn't want to post a solution because I knew it would be a crazy amount of typing
I think somebody else should try posting a solution, this is a thread aimed at MX2 students after all.
Bachelor of Science (Adv. Mathematics) - University of Sydney:
No. If you've got a solution, post it up. Don't wait for others to do it or we might be waiting forever.
You don't need to type it. Just scan and upload will do.
Last edited by tywebb; 9 Jan 2012 at 8:20 PM.
Funny how this thread is titled "2012 HSC MX2 Marathon" yet there are hardly any current HSC students participating. How about toning down the difficulty and topic coverage of the questions to encourage participation from HSC students who've only done a term of the course? I'd imagine the questions above would be very intimidating for most of them.
You are right.
Note: People who have posted in this thread are NOT ALLOWED to answer this. I want some fellow extension 2 students to feel welcomed.
 Find the area of the quadrilateral formed by these roots.$)
lol
I don't know how to plot them on here but the third part:
c) All diagonals are equal length (all have modulus 1). They are perpendicular to each other (separation of pi degrees). Therefore the quadrilateral is a square with diagonal length 2 units. Each side has length root 2 so the area of the quad is 2 units squared.
Hooray, first time using latex.
A few questions: How do you type text without it becoming bunched up and how do you align your equals signs? I tried following the guide but it wasn't working...
How do you type text without it becoming bunched up and how do you align your equals signs?
To align, type without spaces for [ tex ] and [ /tex ]:
[ tex ]\begin{aligned}a)\ z^4&=-1\\ &=\mathrm{c}i\mathrm{s}\; (\pi+2k\pi)\\ &=\mathrm{c}i\mathrm{s}\; (\pi/4),\ \mathrm{c}i\mathrm{s}\; (3\pi/4),\ \mathrm{c}i\mathrm{s}\; (-\pi/4),\ \mathrm{c}i\mathrm{s}\; (-3\pi/4)\end{aligned}[ /tex ]
to get
Although your third line is actually incorrect and so it should be
And suppose I want to type this sentence in latex without it being bunched up.
I'd do it like this:
[ tex ]\text{And suppose I want to type this sentence in latex without it being bunched up.}[ /tex ]
to get
Last edited by tywebb; 10 Jan 2012 at 11:16 AM.
Thanks for that
complex.png
I could do it but the second part took me a bit of time whilst in the answers they just wrote it down with no working (past school paper). Am I missing something ridiculously simple?
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