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2012 Year 9 &10 Mathematics Marathon (4 Viewers)

Fawun

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Re: 2012 Year 9 &10 Mathematics Marathon

Apologies for not seeing this question earlier:

I let the radius of the larger circle be x and the radius of the smaller circle be r,



Rearrange for x and eventually you get:



Therefore, we are given that the radius of the circle is a unit, therefore the radius of the larger circle is units.
Demendog, can u pls stawp?

you're too smart for these things.
 

Demento1

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Re: 2012 Year 9 &10 Mathematics Marathon

The stick divides the grey region into 2 separate areas. What is the area of the smaller one?
I will show you the working out if you want although I am slightly too lazy to type it all out. Essentially:

Construct a trapezium from the stick to the radius of the large circle and find area. Add the area of the large semicircle. After this, find the area of the large circle and subtract the previous trapezium and semicircle working out.

In terms of pi, I got the smaller grey area divided by the stick as:

units squared.
 

Carrotsticks

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Re: 2012 Year 9 &10 Mathematics Marathon

Apologies for not seeing this question earlier:

I let the radius of the larger circle be x and the radius of the smaller circle be r,



Rearrange for x and eventually you get:



Therefore, we are given that the radius of the circle is a unit, therefore the radius of the larger circle is units.
Yep =) or by Realise's method, it's simply the diagonal of the square circumscribed by the larger circle.

I will show you the working out if you want although I am slightly too lazy to type it all out. Essentially:

Construct a trapezium from the stick to the radius of the large circle and find area. Add the area of the large semicircle. After this, find the area of the large circle and subtract the previous trapezium and semicircle working out.

In terms of pi, I got the smaller grey area divided by the stick as:

units squared.
Nice =) This is also an interesting proof for the fact that pi > 1+sqrt(2) (as trivial as it may be using a calculator).

And indeed the stick has length 2.

I'll post up some more random questions as I think of them.
 

Demento1

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Re: 2012 Year 9 &10 Mathematics Marathon

Yep =) or by Realise's method, it's simply the diagonal of the square circumscribed by the larger circle.

Nice =) This is also an interesting proof for the fact that pi > 1+sqrt(2) (as trivial as it may be using a calculator).

And indeed the stick has length 2.

I'll post up some more random questions as I think of them.
Also a valid way as I found. Some of your questions are quite interesting. Post some more in the future and I'll be happy to give them a go.
 

Carrotsticks

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Re: 2012 Year 9 &10 Mathematics Marathon

Seriously, making up questions is heaps fun.

Just take a peaceful walk somewhere with some music + headphones, and look for potential questions everywhere.

Geometry ones are easiest to make especially when dealing with shapes with some sort of symmetry.

Here's another one, which is quite appropriate considering the recent romances floating around these forums.

Consider a 'love heart' made from three semi-circles, two of which are identical and having radius exactly half of the larger one.

The point where the two smaller ones intersect is called a 'cusp'.

Prove that if we construct an arbitrary line through that cusp, then the perimeter is always bisected.

(One could go all philosophical about this being similar to a relationship break-up etc but I won't bother)

 

RealiseNothing

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Re: 2012 Year 9 &10 Mathematics Marathon

Seriously, making up questions is heaps fun.

Just take a peaceful walk somewhere with some music + headphones, and look for potential questions everywhere.

Geometry ones are easiest to make especially when dealing with shapes with some sort of symmetry.

Here's another one, which is quite appropriate considering the recent romances floating around these forums.

Consider a 'love heart' made from three semi-circles, two of which are identical and having radius exactly half of the larger one.

The point where the two smaller ones intersect is called a 'cusp'.

Prove that if we construct an arbitrary line through that cusp, then the perimeter is always bisected.

(One could go all philosophical about this being similar to a relationship break-up etc but I won't bother)

I got it, but I use circle geo lol.
 

Carrotsticks

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Re: 2012 Year 9 &10 Mathematics Marathon

Really? My proof only used methods all learnt in Year 9.

1. Vertically opposite angles are equal.

2. Arc length (which can be deduced as a fraction of the circumference depending on the angle).
 

Demento1

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Re: 2012 Year 9 &10 Mathematics Marathon

I might have an intuitive method to solve this but I'm going to sleep and think about it for tomorrow.
 

RealiseNothing

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Re: 2012 Year 9 &10 Mathematics Marathon

Really? My proof only used methods all learnt in Year 9.

1. Vertically opposite angles are equal.

2. Arc length (which can be deduced as a fraction of the circumference depending on the angle).
1) Vertically opposite angles equal, let them be theta.

2) Arc length of larger circle is 2r(theta) since the radius is 2r and angle is theta.

3) Angle at centre is twice the angle at circumference, hence the angle made by the arc of the smaller circle is 2theta, but has radius 1. So the arc length is also 2r(theta).

4) They have equal arc lengths. Therefore, the perimeter is always bisected (after testing and showing true when the arbitrary line is horizontal, quite trivial).
 

Fawun

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Re: 2012 Year 9 &10 Mathematics Marathon

Seriously, making up questions is heaps fun.

Just take a peaceful walk somewhere with some music + headphones, and look for potential questions everywhere.

Geometry ones are easiest to make especially when dealing with shapes with some sort of symmetry.

Here's another one, which is quite appropriate considering the recent romances floating around these forums.

Consider a 'love heart' made from three semi-circles, two of which are identical and having radius exactly half of the larger one.

The point where the two smaller ones intersect is called a 'cusp'.

Prove that if we construct an arbitrary line through that cusp, then the perimeter is always bisected.

(One could go all philosophical about this being similar to a relationship break-up etc but I won't bother)

Can year 10s besides Demento answer this question? nope.

out.
 

Carrotsticks

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Re: 2012 Year 9 &10 Mathematics Marathon

I don't even understand what the question is asking.
The heart shaped thing has some perimeter, we don't know.

We're just proving that if we draw ANY RANDOM line through that 'cusp' point, then that line cuts the perimeter exactly in half.

And here's the cool part.

It cuts the perimeter in half NO MATTER how we draw the line, as long as it goes through that point.
 

RealiseNothing

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Re: 2012 Year 9 &10 Mathematics Marathon

I don't even understand what the question is asking.
See where the two smaller circles join together? That is called a cusp. You have to show that if you draw a straight line that goes through that cusp, the perimeter of the whole shape is always cut in half.
 

enoilgam

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Re: 2012 Year 9 &10 Mathematics Marathon

I don't even understand what the question is asking.
A year 10 can do it, but it is something which you just need to see. I think this question is definitely up there on the difficulty scale.
 

Fawun

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Re: 2012 Year 9 &10 Mathematics Marathon

See where the two smaller circles join together? That is called a cusp. You have to show that if you draw a straight line that goes through that cusp, the perimeter of the whole shape is always cut in half.
How the hell are you suppose to answer that without any numbers
 

Carrotsticks

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Re: 2012 Year 9 &10 Mathematics Marathon

How the hell are you suppose to answer that without any numbers
That's the beauty of it.

No mindless and tedious number crunching or computations.

Just proofs.
 

RealiseNothing

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Re: 2012 Year 9 &10 Mathematics Marathon

How the hell are you suppose to answer that without any numbers
Well the way I did it was:

1) Show that it works for one simple case (which is often trivial).

2) Show that by rotating the line, the two arc lengths made are equal in length. Hence each "half" of the perimeter is neither gaining or losing anything.
 

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