Mathematics marathon

[LewisJ]

It is also simple to find a polynomial in \( \mathbb{R}^2 \) rather than using imaginaries:
\( p(x,y) = \pi*x + y \)
but by requiring coefficients to be integers, this easy solution is also eliminated.

 

[qqwref]

It is also simple to find a polynomial in \( \mathbb{R}^2 \) rather than using imaginaries:
\( p(x,y) = \pi*x + y \)
but by requiring coefficients to be integers, this easy solution is also eliminated.

Right, solutions which use pi, i, square roots, etc. are trivial and boring, and they eliminate the point of the problem.

keemy:

Spoiler

Can you think of a polynomial p(x) that uniquely converts an integer x into a positive integer?

 

[keemy]

oh yeah lol so I guess just square each term I had or w/e

 

[FMC]

An optimization problem,calculus not needed,level:high school.
find the minimum value of sqrt(x^2+1) + sqrt((x-y)^2+1 ) + sqrt((y-z)^2+1) + sqrt((z-10)^2+9), what values of x,y and z give that minimum

 

[qqwref]

Spoiler

oh yeah lol so I guess just square each term I had or w/e

No, it isn't quite that simple.

 

[Rinfiyks]

No, it isn't quite that simple.

Spoiler

\( p(x) = x^2 + x/2 \)

 

[blah]

Prove that for any 5 points placed randomly on the surface of a sphere, that at least four of those points are always contained on the same hemisphere.

Putnam

Hint:

Spoiler

Big round things and birds.

 

[keemy]

Putnam

Hint:

Spoiler

Big round things and birds.

Spoiler

I don't think you even to do that much if you just use the fact that any 3 points are on the same hemisphere

 

[blah]

Spoiler

I don't think you even to do that much if you just use the fact that any 3 points are on the same hemisphere

Spoiler

I think that fact comes from the fact about great circles?

 

[hawkmp4]

Chris':

Spoiler

Any 3 points are on a hemisphere...call that hemisphere A...then the 4th point must either be in A or not in A. If it is in A, 4 points in the same hemisphere. Done. If the 4th point is not in A, then it is in B, C, or D, the hemispheres formed by the 4th point and one of the pairs possible from the original 3 points.

Grr...
Don't know where to go from here...

EDIT:
Michael's:

Spoiler

So we're looking for a antisymmetric polynomial relation, of degree 2 or greater. A linear function won't work since there are no two distinct numbers that don't span the integers. What about \( x^3+y^5 \)? No, there are surely numbers that are both cubes and...quintics? Hm. I'll return to this later.

 

[blakedacuber]

why is pi 3.14 or 22 over 7?

 

[kvaele]

22/7 is an approximation. Pi is irrational, meaning it cannot be put into a fraction.

 

[hawkmp4]

why is pi 3.14 or 22 over 7?

Pi isn't either of those things

 

[blakedacuber]

Pi isn't either of those things

in maths it is

i dont care why its a letter

 

[blakedacuber]

22/7 is an approximation. Pi is irrational, meaning it cannot be put into a fraction.

but why those numbers?

who said PI is 3.14 or 22/7 but more importantly(and my question) why?

 

[nathanajah]

in maths it is

i dont care why its a letter

It is not exactly 22/7 or 3.14, both of it are just approximations.
Pi is an irrational number.

edit: ninja'd, I'll answer the second question.
Pi, counted to 50 decimal digit is 3.14159265358979323846264338327950288419716939937510 (wikipedia.
We could see that 22/7 = 3.142857142857142857.... which is close enough (at least the first 3 digits are close enough) to pi, while 3.14 is clearly the truncation of pi.

http://en.wikipedia.org/wiki/Numerical_approximations_of_π (calculating pi)

 

[hawkmp4]

why is pi 3.14 or 22 over 7?

but why those numbers?

who said PI is 3.14 or 22/7 but more importantly(and my question) why?

You did...

 

[Rinfiyks]

but why those numbers?

who said PI is 3.14 or 22/7 but more importantly(and my question) why?

When you're using pi for whatever reason (working with circles or spheres maybe), you don't usually need it that accurate. So 3.14 is a good approximation. 22/7 is even better, and something you can easily do on a calculator.
But pi isn't 3.14 or 22/7. It's 3.1415926535... digits go on forever with no known pattern.

 

[kinch2002]

Someone just answer properly! It's the ratio between the circumference and the diameter of a circle

 

[JBCM627]

Spoiler

I think that fact comes from the fact about great circles?

What is this about birds?

Spoiler

2 points on a sphere and a point in the center of the sphere define a plane slicing the sphere into hemispheres. If the points coincide, there is some freedom to pick whatever plane you like. Then, at least 2 of the 3 remaining points must lie on either hemisphere (or also on the plane). So 4 points are on that hemisphere. Of course, that hemisphere must include the great circle drawn out by the plane, so the problem becomes somewhat ill defined here, since the hemisphere is not expressly open or closed.

Spoiler

\( p(x) = x^2 + x/2 \)

That doesn't return integers.

Spoiler

\( 4(2x+1/2)^2 \)

I'll need to think of a problem more like these, I guess. Nobody likes my integrals