Cool problem of the day! (1 Viewer)

nightweaver066

Well-Known Member
Joined
Jul 7, 2010
Messages
1,585
Gender
Male
HSC
2012
Very nice Realise!

nightweaver, you are very close, but not quite.

A few things:

1. It's called the Triangle Inequality, not the 'Axiom of Triangle'. It is something prove-able.

2. 2 is indeed a lower bound for x, but it isn't necessarily the LARGEST lower bound. Consider the limiting cases when the triangle is right angled.
Updated. :)
 

barbernator

Active Member
Joined
Sep 13, 2010
Messages
1,439
Gender
Male
HSC
2012
sorry I phrased the inequality completely wrongly. I don't think it's really an inequality I have to solve as an upper bound of seaniegs question now i think of it. The lower bound was just x^2 + y^2 +1 > 2xy, but i've got no idea for how to prove that z<4. It's probably something to do with the discriminant being a perfect square when i solve the quadratic for x possibly, but I don't really know how to prove that anyway. Oh well.
 

Demento1

Philosopher.
Joined
Dec 23, 2011
Messages
866
Location
Sydney
Gender
Male
HSC
2014
If only I was capable enough to do the questions which Carrotsticks posts up daily. I must be missing out.
 

barbernator

Active Member
Joined
Sep 13, 2010
Messages
1,439
Gender
Male
HSC
2012
If only I was capable enough to do the questions which Carrotsticks posts up daily. I must be missing out.
in time you will :) btw, i have a good one which i will suggest to carrot for the next round of q's
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
sorry I phrased the inequality completely wrongly. I don't think it's really an inequality I have to solve as an upper bound of seaniegs question now i think of it. The lower bound was just x^2 + y^2 +1 > 2xy, but i've got no idea for how to prove that z<4. It's probably something to do with the discriminant being a perfect square when i solve the quadratic for x possibly, but I don't really know how to prove that anyway. Oh well.
Proving that z>2 is trivial, but proving z<4 is not...Assume solutions with z>3 exists and look for a contradiction. I will post a solution tomorrow.
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Hint for the Triangular Number question:

First prove that the sum of two consecutive triangular numbers is a perfect square....

 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Sweet!

Here is the next set of problems:

Geometry (courtesy of barbernator):

Consider a circular paddock with radius R such that R > 5 metres.

Along the circumference of the paddock, a post hammered into the ground and a sheep (poor sheep) is tied to the post by a 10 metre long rope.

In terms of R, what percentage of the paddock can the sheep roam?

Note: To check if your answer is correct or not, you should be able to show that when R --> Infinity, the percentage approaches 0 and when R --> 5, the percentage approaches 100.

Algebra:

Prove that the sum of the reciprocals of INFINITE Triangular Numbers is 2.

Number Theory:

seanieg89 will be providing the Number Theory questions from now on. Much appreciated =)
 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Algebra:

Prove that the sum of the reciprocals of INFINITE Triangular Numbers is 2.
I've noticed that the triangular numbers are either multiples of 3, or you can make a series of the non-multiples of 3 which differ by increasing factors of 9.

This have anything to do with it?
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007


New number theory questions, ordered roughly by increasing difficulty:

1. Prove that the product of 5 consecutive positive integers cannot be the square of a positive integer.

2. The positive integers m and n satisfy: . Prove that m-n is a perfect square.

3. An odd positive integer n is said to be 'sexy' if n divides: Prove that for any twin prime pair (p,q), EXACTLY one of p,q is sexy.

4. Let a be an arbitrary irrational number. Prove that there are irrational numbers b and b' such that:

-a+b, ab' are both rational,

-a+b', ab are both irrational.
 

qwerty44

Member
Joined
Oct 12, 2011
Messages
557
Gender
Male
HSC
2013
I never saw the day coming where you would prove something is sexy...
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Please revive this!
I would love more questions to come along, some of them albeit hard are fun to do, and it excercise obscure math techniques.
 

AAEldar

Premium Member
Joined
Apr 5, 2010
Messages
2,246
Gender
Male
HSC
2011
Please revive this!
I would love more questions to come along, some of them albeit hard are fun to do, and it excercise obscure math techniques.
I've got an idea for one, just have to go about typing it up >.< I'll try and have it up by tomorrow - Carrot will hopefully have his bunch up as well, but he is a busy man!
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Please revive this!
I would love more questions to come along, some of them albeit hard are fun to do, and it excercise obscure math techniques.
Sorry, been a bit busy lately. I might be able to do something tonight.

I've got an idea for one, just have to go about typing it up >.< I'll try and have it up by tomorrow - Carrot will hopefully have his bunch up as well, but he is a busy man!
That would be great! Please PM me =)
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Geometry:

Consider a circular ring of radius R.

A stick of length k such that is placed within this circle so it becomes a 'chord'.

The stick is then 'spun' around inside the circle such that the two ends of the stick are always touching the circumference of the circle.

Find the area of the shape created by the midpoint of the stick once it's made a full rotation.



Algebra:

Determine all real values of X and Y satisfying the following system of equations:



University students/Postgrads: Please refrain from answering, as tempting as it may be.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top