Australian Maths Competition (1 Viewer)

Mongoose528 · Jul 25, 2017

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?

Mongoose528 · Jul 25, 2017

jathu123 said:
http://imgur.com/a/uDxq1

you found the diameter of a circle with area 16pi, not the diameter of the tube

Thanks, I should definitely draw a picture next time so I don't get it wrong. Can't make mistakes like this in the competition on Thursday.

dan964 · Jul 25, 2017

Mongose528 said:
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

Mongoose528 · Jul 25, 2017

Here:

1729 · Jul 25, 2017

Mongose528 said:
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?

$$\noindent The area of the gasket is $A = y^2 - 9x^2$ \\ The perimeter of the gasket is $P = 4(y-3x) + 24x = 4y + 12x$. So, \\ \begin{align*} y^2 - 9x^2 &= 4y + 12x \\ (y-3x)(y+3x) &= 4(y+3x) \\ y - 3x &= 4 \text{ since } y+3x \neq 0 \end{align*} \\ If $x = 2$, $y = 10$ which is not prime but if $x = 3$, $y = 13$ which is prime so the least area is trivially $A = 13^2 - 9(3)^2 = 88$ cm$^2$

Mongoose528 · Jul 25, 2017

1729 said:
$$\noindent The area of the gasket is $A = y^2 - 9x^2$ \\ The perimeter of the gasket is $P = 4(y-3x) + 24x = 4y + 12x$. So, \\ \begin{align*} y^2 - 9x^2 &= 4y + 12x \\ (y-3x)(y+3x) &= 4(y+3x) \\ y - 3x &= 4 \text{ since } y+3x \neq 0 \end{align*} \\ If $x = 2$, $y = 10$ which is not prime but if $x = 3$, $y = 13$ which is prime so the least area is trivially $A = 13^2 - 9(3)^2 = 88$ cm$^2$

Nicely done, want to ask a new question?

1729 · Jul 25, 2017

Mongose528 said:
Nicely done, want to ask a new question?

From where?

(Don't have access to the AMC problems)

Mongoose528 · Jul 25, 2017

1729 said:
From where?

(Don't have access to the AMC problems)

Here's a link to some past papers:

Link removed due to copyright

Good luck!

si2136 · Jul 25, 2017

@Mongose @1729

Are you both doing AMC in the senior division?

dan964 · Jul 25, 2017

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 · Jul 25, 2017

si2136 said:
@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 · Jul 25, 2017

pikachu975 · Jul 25, 2017

Mongose528 said:
View attachment 34149

The bottom of the circle travels half a circumference, i.e. pi*r

Then you add on 2 radii (bottom of circle to point X for both sides) so 2r

distance = pi*r + 2r
= r(pi+2) (B) I guess

Mongoose528 · Jul 26, 2017

Mongoose528 · Jul 26, 2017

si2136 said:
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 · Jul 26, 2017

Mongose528 said:
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

Mongoose528 · Jul 26, 2017

Mongose528 said:
View attachment 34150

Here are 3 more to keep you going through the day:

seanieg89 · Jul 26, 2017

Mongose528 said:
View attachment 34150

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.

Jonothancroller · Jul 26, 2017

Hey, I've got some questions that I haven't been able to do. Would anyone care to do these questions as practice and offer an explanation? Thanks.

http://imgur.com/a/YfigO

Jonothancroller · Jul 26, 2017

Mongose528 said:
View attachment 34153

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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