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Australian Maths Competition (1 Viewer)

Mongoose528

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NEW Q: . Small squares of side x cm have been removed from the corners, sides and centre of a square of side y cm to form the gasket shown.
If x and y are prime numbers and the sum of the inside and outside perimeters of the gasket, in centimetres, is equal to the area of the gasket, in square centimetres, what is the smallest possible value of the area of the gasket?
 

dan964

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NEW Q: . Small squares of side x cm have been removed from the corners, sides and centre of a square of side y cm to form the gasket shown.
If x and y are prime numbers and the sum of the inside and outside perimeters of the gasket, in centimetres, is equal to the area of the gasket, in square centimetres, what is the smallest possible value of the area of the gasket?
needs a diagram
 

1729

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NEW Q: . Small squares of side x cm have been removed from the corners, sides and centre of a square of side y cm to form the gasket shown.
If x and y are prime numbers and the sum of the inside and outside perimeters of the gasket, in centimetres, is equal to the area of the gasket, in square centimetres, what is the smallest possible value of the area of the gasket?
 

si2136

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@Mongose @1729

Are you both doing AMC in the senior division?
 

dan964

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Just a friendly reminder, that AMC papers are copyrighted. While we permit posting of individual questions at a time, posting sites that host AMC (Australian Maths Competition), AIMO papers is not allowed.

Please remember this for future.
 

Mongoose528

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@Mongose @1729

Are you both doing AMC in the senior division?
Nah, I'm doing the intermediate. I don't think they vary too much in difficulty though as I've noticed that question 30 in the intermediate is around question 28/29 in senior. I've also noticed that their are quite a few common questions each year.
 

Mongoose528

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20.

Q: If x and y are positive integers which satisfy x^2 - 8x -1001y^2 = 0, what is the smallest possible value of x + y?
Could you please post your working si2136.

I wasn't sure on how to do that question, thanks.
 

1729

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Could you please post your working si2136.

I wasn't sure on how to do that question, thanks.
Exactly six of these 'unusually shaped portions' fit into the regular octahedron by symmetry (one at each vertex) so the volume of one portion is one-sixth the volume of the octahedron, ie. 1/6 * 120 = 20
 
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seanieg89

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Let f(n) be the quantity for a given n. It is pretty easy to show that f(mn)=f(m)f(n) for coprime m and n (a and b each uniquely factorise into the product of something that divides m and something that divides n), and also that f(p^n)=(n+1)(n+2)/2 by manual counting.

Hence f(6^6)=f(2^6)f(3^6)=28^2=784.

Note that this method makes it easy to compute f(n) for an arbitrary n that we know the prime factorisation of.
 
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Notice that the unshaded area must also be divided in half. By drawing a diagram with the centres of the circles and the perpendiculars from the centres to the dividing line, it can be shown that the line must pass through the midpoint between the centres of the spheres. Letting P be the origin and then have a normal x-y plane, M = (5/2, 1) and also passes through C = (6,2). Then, find the line MN and substitute x=0 to obtian y = 2/7, which is A.
 

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