-
Looking for HSC notes and resources? Check out our Notes & Resources page
Australian Maths Competition (2 Viewers)
- Thread starter maths94
- Start date
dan964
what
1 2 3
4 5 6
7 8 9
Note: you can only pick max of 2 points on any given line/diagonal. You must pick 3 points.
Pick (1) as a fixed point:
Pick (5) = 2,3,4,6,7,8 = 6
Remove (5), Pick (9) = 2,3,4,6,7,8 = 6
Remove (9), Pick (2) = 4,6,7,8 = 4 or Pick (3) = 4,6,7,8 = 4
Remove (2,3), Pick (4) = 6,8 = 2 or Pick (7) = 6,8 = 2
Remove (4,7), Pick (6) = 8, = 1
= 25
Remove (1), Pick (9) as a fixed point
Pick (5) = 2,3,4,6,7,8 = 6
Remove (5), Pick (3) = 2,4,7,8 = 4 or Pick (6) = 2,4,7,8 = 4
Remove (3,6), Pick (7) = 2,4 = 2 or Pick (8) = 2,4 = 2
Remove (7,8), Pick (2) = 4, = 1
= 19
Remove (9), Pick (3) as a fixed point
Pick (5) = 2,4,6,8 = 4
Remove (5), Pick (7) = 2,4,6,8 = 4
Remove (7), Pick (2) = 4,6,8 = 3
Remove (2), Pick (4) = 6,8 = 2
Remove (4), Pick (6) = 8, = 1
= 14
Remove (3), Pick (7) as a fixed point
Pick (5) = 2,4,6,8 = 4
Remove (5), Pick (2) = 4,6,8 = 3
Remove (2), Pick (4) = 6,8 = 2
Remove (4), Pick (6) = 8, = 1
= 10
Remove (7), Pick (2) as a fixed point
Pick (5) = 4,6 = 2
Remove (5), Pick (4) = 6,8 = 2
Remove (4), Pick (6) = 8, = 1
= 5
Remove (2), Pick (8) as a fixed point
Pick (5) = 4,6 = 2
Remove (5), Pick (4) = 6, = 1
= 3
Remove (8), note that only (4,5,6) remain which no valid triangles can be formed.
1 2 3
4 5 6
7 8 9
Pick (1) as a fixed point:
Pick (5) = 6,8 = 2
Remove (5), Pick (9) = 2,4,6,8 = 4
Remove (9), Pick (2) = 6,7,8 = 3 or Pick (3) = 4,6 = 2
Remove (2,3), Pick (4) = 6,8 = 2 or Pick (7) = 8, = 1
= 14
Remove (1), Pick (9) as a fixed point
Pick (5) = 2,4 = 2
Remove (5), Pick (3) = 2,8 = 2 or Pick (6) = 2,4,7 = 3
Remove (3,6), Pick (7) = 4, = 1 or Pick (8) = 2,4 = 2
= 10
Remove (9), Pick (3) as a fixed point
Pick (5) = 4,8 = 2
Remove (5), Pick (7) = 2,4,6,8 = 4
Remove (7), Pick (2) = 4,8 = 2
Remove (2), Pick (4) = 6, = 1
Remove (4), Pick (6) = 8, = 1
= 10
Remove (3), Pick (7) as a fixed point
Pick (5) = 2,6 = 2
Remove (5), Pick (2) = 4,8 = 2
Remove (2), Pick (4) = 6, = 1
Remove (4), Pick (6) = 8, = 1
= 6
Note no scalene triangles remain
Total of triangles = 76
Total of scalene triangles = 40
4 5 6
7 8 9
Note: you can only pick max of 2 points on any given line/diagonal. You must pick 3 points.
Pick (1) as a fixed point:
Pick (5) = 2,3,4,6,7,8 = 6
Remove (5), Pick (9) = 2,3,4,6,7,8 = 6
Remove (9), Pick (2) = 4,6,7,8 = 4 or Pick (3) = 4,6,7,8 = 4
Remove (2,3), Pick (4) = 6,8 = 2 or Pick (7) = 6,8 = 2
Remove (4,7), Pick (6) = 8, = 1
= 25
Remove (1), Pick (9) as a fixed point
Pick (5) = 2,3,4,6,7,8 = 6
Remove (5), Pick (3) = 2,4,7,8 = 4 or Pick (6) = 2,4,7,8 = 4
Remove (3,6), Pick (7) = 2,4 = 2 or Pick (8) = 2,4 = 2
Remove (7,8), Pick (2) = 4, = 1
= 19
Remove (9), Pick (3) as a fixed point
Pick (5) = 2,4,6,8 = 4
Remove (5), Pick (7) = 2,4,6,8 = 4
Remove (7), Pick (2) = 4,6,8 = 3
Remove (2), Pick (4) = 6,8 = 2
Remove (4), Pick (6) = 8, = 1
= 14
Remove (3), Pick (7) as a fixed point
Pick (5) = 2,4,6,8 = 4
Remove (5), Pick (2) = 4,6,8 = 3
Remove (2), Pick (4) = 6,8 = 2
Remove (4), Pick (6) = 8, = 1
= 10
Remove (7), Pick (2) as a fixed point
Pick (5) = 4,6 = 2
Remove (5), Pick (4) = 6,8 = 2
Remove (4), Pick (6) = 8, = 1
= 5
Remove (2), Pick (8) as a fixed point
Pick (5) = 4,6 = 2
Remove (5), Pick (4) = 6, = 1
= 3
Remove (8), note that only (4,5,6) remain which no valid triangles can be formed.
1 2 3
4 5 6
7 8 9
Pick (1) as a fixed point:
Pick (5) = 6,8 = 2
Remove (5), Pick (9) = 2,4,6,8 = 4
Remove (9), Pick (2) = 6,7,8 = 3 or Pick (3) = 4,6 = 2
Remove (2,3), Pick (4) = 6,8 = 2 or Pick (7) = 8, = 1
= 14
Remove (1), Pick (9) as a fixed point
Pick (5) = 2,4 = 2
Remove (5), Pick (3) = 2,8 = 2 or Pick (6) = 2,4,7 = 3
Remove (3,6), Pick (7) = 4, = 1 or Pick (8) = 2,4 = 2
= 10
Remove (9), Pick (3) as a fixed point
Pick (5) = 4,8 = 2
Remove (5), Pick (7) = 2,4,6,8 = 4
Remove (7), Pick (2) = 4,8 = 2
Remove (2), Pick (4) = 6, = 1
Remove (4), Pick (6) = 8, = 1
= 10
Remove (3), Pick (7) as a fixed point
Pick (5) = 2,6 = 2
Remove (5), Pick (2) = 4,8 = 2
Remove (2), Pick (4) = 6, = 1
Remove (4), Pick (6) = 8, = 1
= 6
Note no scalene triangles remain
Total of triangles = 76
Total of scalene triangles = 40
Mongoose528
Member
Thought this was a good question:
A regular octahedron has eight triangular faces and all sides the same length. A portion of a regular octahedron of volume 120 cm^3 consists of that part of it which is closer to the top vertex than to any other one. In the diagram, the outside part of this volume is shown shaded, and it extends down to the centre of the octahedron.
What is the volume, in cubic centimetres, of this unusually shaped portion?
A regular octahedron has eight triangular faces and all sides the same length. A portion of a regular octahedron of volume 120 cm^3 consists of that part of it which is closer to the top vertex than to any other one. In the diagram, the outside part of this volume is shown shaded, and it extends down to the centre of the octahedron.
What is the volume, in cubic centimetres, of this unusually shaped portion?
Last edited:
Mongoose528
Member
Around September iirc
30june2016
Active Member
The results were distributed to schools ~ August 28 the past few years
dan964
what
Last edited:
Mongoose528
Member
Yep, I saw that
x^2-8x = 1001y^2
x(x-8) = 1001y^2
So for x+y to be at a minimum, we had to have y as low as possible.
So I substituted y values from 1 onwards until I got a result that worked, Y=3, X=99
(99*91 = 1001*3^2)
Mongoose528
Member
NEW Q: Thanom has a roll of paper consisting of a very long sheet of thin paper tightly rolled around a cylindrical tube. Initially, the diameter of the roll is 12 cm and the diameter of the tube is 4 cm. After Thanom uses half of the paper, the diameter of the remaining roll is closest to what value? (Round to nearest 0.5)
Is the answer 9 cm?
(I did it by finding the area of the annulus which represents the length of the paper, so I halved this area since half the paper was used and then found the new diameter which gives such area)
Last edited:
Mongoose528
Member
Yeah this is right. Could you see what part of my working out is wrong please:
Area of roll = pi*r^2 = 36pi (diameter = 12)
Area of tube = pi*r^2 = 4pi (diameter = 4)
Area of annulus = 32pi
After half has been used: 16pi
Which gives us a radius of 4cm and a diameter of 8cm.
This is incorrect though but could someone please tell me why.
http://imgur.com/a/uDxq1
you found the diameter of a circle with area 16pi, not the diameter of the tube
Mongoose528
Member
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?
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
Member
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
what
needs a diagram
Mongoose528
Member
Here: 
Mongoose528
Member
Nicely done, want to ask a new question? + 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 )
Last edited:
Mongoose528
Member
Last edited by a moderator: