Australian Maths Competition (1 Viewer)

[30june2016]

Would you need to know things such as modular arithmetic and geometric identities for the intermediate section? Especially for the last 5?

You never know, but for the Australian Intermediate Mathematics Olympiad (similar to AMC, but harder) a few years ago, we got asked a question that could be solved using modular arithmetic/bases etc

 

[Mongoose528]

You never know, but for the Australian Intermediate Mathematics Olympiad (similar to AMC, but harder) a few years ago, we got asked a question that could be solved using modular arithmetic/bases etc

Would you have any tips for the amc (especially last 5) and aimo? Completing them both this year.

 

[Mongoose528]

Terry has invented a new way to extend lists of numbers. To Terryfy a list such
as [1, 8] he creates two lists [2, 9] and [3, 10] where each term is one more than
the corresponding term in the previous list, and then joins the three lists together
to give [1, 8, 2, 9, 3, 10]. If he starts with a list containing one number [0] and
repeatedly Terryfies it he creates the list
[0, 1, 2, 1, 2, 3, 2, 3, 4, 1, 2, 3, 2, 3, 4, 3, 4, 5, 2, 3, 4, . . . ].

What is the 2012th number in this Terryfic list?

 

[Mongoose528]

Continuing the Marathon:

1. A high school marching band can be arranged in a rectangular formation with exactly
three boys in each row and exactly five girls in each column. There are several sizes
of marching band for which this is possible. What is the sum of all such possible
sizes?

2. Around a circle, I place 64 equally
spaced points, so that there are
64×63÷2 = 2016 possible chords
between these points.
I draw some of these chords, but
each chord cannot cut across more
than one other chord.
What is the maximum number of
chords I can draw?

3. A symmetrical cross with equal arms has an area of 2016 cm2 and all sides of integer
length in centimetres. What is the smallest perimeter the cross can have, in
centimetres?

 

[si2136]

The APMO papers are just freaking hard.

American papers have so much probability it's not funny

 

[Paradoxica]

The APMO papers are just freaking hard.

American papers have so much probability it's not funny

hence why we are shit with our technology... everything is probability these days...

 

[si2136]

3. A symmetrical cross with equal arms has an area of 2016 cm2 and all sides of integer
length in centimetres. What is the smallest perimeter the cross can have, in
centimetres?

Let x = Large Square
Let y = Small Square

Therefore 2016 = (x+2y)(x-2y)

Through trial and error, x+2y = 56, x-2y = 36.

Therefore 2x = 92.

Therefore the perimeter is 184cm.

NEW Q:

There are 42 Points P1, P2, P3, ....., P42, placed in order on a straight line so that each distance from Pi to P(i+1) is 1/i, where 1≤i≤41. What is the sum of the distances between every pair of these points?

 

[Mongoose528]

Let x = Large Square
Let y = Small Square

Therefore 2016 = (x+2y)(x-2y)

Through trial and error, x+2y = 56, x-2y = 36.

Therefore 2x = 92.

Therefore the perimeter is 184cm.

NEW Q:

There are 42 Points P1, P2, P3, ....., P42, placed in order on a straight line so that each distance from Pi to P(i+1) is 1/i, where 1≤i≤41. What is the sum of the distances between every pair of these points?

Completed it for 1≤i≤2, 1≤i≤3 ... 1≤i≤5, and the answer was always a triangular number of n-1

For: 1≤i≤2, the sum of distances was 1

For: 1≤i≤3, the sum of distances was 3

For: 1≤i≤5, the sum of distances was 10

So for 1≤i≤42 the sum of distances will be 41*42/2 = 861

New Q: In a 3 × 3 grid of points, many triangles can be formed using 3 of the points as
vertices. Of all these possible triangles, how
many have all three sides of different lengths?

 

[si2136]

@M: Is the answer 68 for the new q?

Is there a faster way to solve 3 system of lines with 4 unknowns than matrices?

 

[Mongoose528]

@M: Is the answer 68 for the new q?

Is there a faster way to solve 3 system of lines with 4 unknowns than matrices?

Unfortuantely that's incorrect @si2136

The answer they've given is 40

 

[dan964]

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

 

[Mongoose528]

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?

 

[Mongoose528]

When are AMC results usually released?

Around September iirc

 

[30june2016]

Hopefully before Sept 30, I need my bonus points lol

The results were distributed to schools ~ August 28 the past few years

 

[dan964]

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?

 

[Mongoose528]

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]

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)

 

[1729]

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)

 

[Mongoose528]

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)

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.

 

[jathu123]

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