Non HSC Maths Marathon (2 Viewers)

[Affinity]

Four distinct points including 0;, 0, z1, z2, z3 lie on a circle. Show that the points 1/z1, 1/z2, 1/z3 are collinear, i.e lie on a straight line.

Taken from 2520 maths book

This is actually quite interesting, reminds me of something called inversion:

http://whistleralley.com/inversion/inversion.htm

 

[KeypadSDM]

Four distinct points including 0;, 0, z1, z2, z3 lie on a circle. Show that the points 1/z1, 1/z2, 1/z3 are collinear, i.e lie on a straight line.

Taken from 2520 maths book

This shouldn't be allowed, it's a 4-unit question. [Saw it in a '98 past paper, I can't find it right now though...]

 

[Affinity]

This shouldn't be allowed, it's a 4-unit question. [Saw it in a '98 past paper, I can't find it right now though...]

So picky... Prove that a topology strictly stronger than a compact hausdorff topology on a given space is not compact, and a strictly weaker topology is not hausdorff.

 

[vulgarfraction]

Prove that there is an explicit way to disect a cube into n cubes (perhaps of different sizes) for say... n>100

7 cubes can be added to the total number by dividing any existing cube into 8. Just prove it's possible for all remainders modulo 7 for a value equal to the lowest number of that remainder above/equal to 100.

You start off with 1 cube, and can add 7, 26, 63, 124 etc. You can also 'remove' 7, 26, etc cubes by refraining from cutting them when you're creating a larger number.

Congruent to (mod 7):

0: 105 = 63+ 6*7
1: 106 = 3*26 + 4*7
2: 100 = 27 + 2*26 + 3*7
3: 101 = 64 + 63 - 26
4: 102 = 27 + 26 + 7*7
5: 103 = 64 + 63 - 26 - 26 + 4*7
6: 104 = 64 + 26 + 2*7

 

[acmilan]

This ones actually easy, i just find it kind of cute:

Prove that, at most, only one of e*pi and e+pi can be rational.

 

[KeypadSDM]

This ones actually easy, i just find it kind of cute:

Prove that, at most, only one of e*pi and e+pi can be rational.

Assume they're both rational:

Note: e - pi = (e+pi)-2pi, hence trancendental (as pi is trancendental)
(e + pi)^2 - 4*e*pi = rational = (e - pi)^2
But as e - pi is trancendental, it's not the root of a rational polynomial
Hence (e - pi)^2 is also trancendental, and hence irrational.
Thus at least one of e + pi and e*pi is irrational.

Is there an easier way which just uses e&pi being irrational rather than trancendental?

 

[acmilan]

Is there an easier way which just uses e&pi being irrational rather than trancendental?

Not as far as i know. The way i wouldve done it is consider f(x) = (x-e)(x-pi) = x² - (e+pi)x + e*pi and use the fact that e and pi are transcendental

 

[KeypadSDM]

Ahh, very nice. Solutions to algebraic polynomials are themselves algebraic. Hence, since the roots are trancendental, the polynomial cannot be algebraic.

 

[KeypadSDM]

Prove e^pi is irrational.

I don't believe there's a simple way, so I'll just go ahead and do it the hard way.

Prove Gelfond's theorem over the complex domain.

Hence (-1)^-i = e^pi is irrational.

 

[KeypadSDM]

It makes the most sense to do it this way, though. As the proof for a^b being trancendental, except when it trivially isn't, is so well known, I feel it's a valid solution.

 

[KeypadSDM]

Alright, I can't be arsed making a cool question up, so I'll just do a boring one.

Prove that given any 3 dimensional body, with arbitrary mass distribution, there is a set of coordinate axes which when defined around the centre of mass of the 3-d body such that one of the axes points in the direction of least moment of inertia, and one of the axes points in the direction of highest moment of inertia.

The x,y,z axes you use must be orthogonal.

Spoiler

I'm not sure if this is correct but maybe:
Start with a generalised tensor for the moments of inertia about an arbitrary set of axes:
[Ixx,Ixy,Ixz]
[Iyx,Iyy,Iyz]
[Izx,Izy,Izz]
And diagonalise

The hint might be completely talking out of my ass. I don't actually know.

 

[Li0n]

This shouldn't be allowed, it's a 4-unit question. [Saw it in a '98 past paper, I can't find it right now though...]

Err, well firstly i didn't do 4unit. And like i said i pulled it out of my complex analysis question booklet. Peter Brown (who sets questions for 2520/2620)writes 4-unit questions too afaik.

 

[Affinity]

Alright, I can't be arsed making a cool question up, so I'll just do a boring one.

Prove that given any 3 dimensional body, with arbitrary mass distribution, there is a set of coordinate axes which when defined around the centre of mass of the 3-d body such that one of the axes points in the direction of least moment of inertia, and one of the axes points in the direction of highest moment of inertia.

The x,y,z axes you use must be orthogonal.

Spoiler

I'm not sure if this is correct but maybe:
Start with a generalised tensor for the moments of inertia about an arbitrary set of axes:
[Ixx,Ixy,Ixz]
[Iyx,Iyy,Iyz]
[Izx,Izy,Izz]
And diagonalise

The hint might be completely talking out of my ass. I don't actually know.

works.. Statisticians use the same in principal component regression.

 

[KeypadSDM]

works.. Statisticians use the same in principal component regression.

Score, that means I'm right about how to figure out the values.