Solutions to RedStar123456789's problem:

4r + 2c = 36
r + c = 13, so c = 13 - r

Then, 4r + 26 - 2r = 36, so 2r = 10, r = 5, c = 8.

Also, you could assume that all the animals were rabbits which would make 52 legs, and for each rabbit replaced with a chicken, we would lose two legs, so there are (52-36)/2 = 8 chickens and 13 - 8 = 5 rabbits.

Solution to Jaleb's problem:

I will assume that "knowing someone" is symmetric - otherwise this wouldn't be true.

Assume the converse - everyone knows a different number of people at the party. This means that there exists a person that knows 0 people at the party, one that knows 1 person, and so on, until n-1 (there are n people, so this is the only possibility). That means the person that knows n-1 people must know everyone else at the party, including the person who knows 0 people - however, that is a contradiction, so our original statement must be true.

Here's a nice problem:

Points A_1, B_1, C_1 are chosen on the sides BC, CA, AB, respectively of a triangle ABC. Denote by G_a, G_b, G_c are the centroids of triangles AB_1C_1, BC_1A_1, CA_1B_1, respectively. Prove that the lines AG_a, BG_b, CG_c are concurrent if and only if lines AA_1, BB_1, CC_1 are concurrent.

Here's a hint - Ceva or Trig. Ceva, this is the question!

Here is a classic problem you would have to prove in graph theory:

At a party of $n$ people in which each person knowing anywhere between 0 to $n-1$ other people at the party. Prove that at least two people know the same number of people at the party.

RedStar123456789 says

Those maths questions are so hard.

I got a good question: There are some rabbits and some chickens. There are 13 heads and 36 legs. How many of each animal are there?

Yeay, middle school word problem. No guess and check please.

Those maths questions are so hard.

I got a good question: There are some rabbits and some chickens. There are 13 heads and 36 legs. How many of each animal are there?

noooo i suck at it omg

Jaleb says

KillHimNotMe says

there is no such thing as infinity i am an atheist and only believe things that can be proven

Infinity is not a number. It is an term defined to mean an upper limit does not exist. http://youtu.be/23I5GS4JiDg

Link to a Vi Hart video ...... So much respect.

KillHimNotMe says

there is no such thing as infinity i am an atheist and only believe things that can be proven

Infinity is not a number. It is an term defined to mean an upper limit does not exist. http://youtu.be/23I5GS4JiDg

there is no such thing as infinity i am an atheist and only believe things that can be proven

I love math, but I only have limited grasp of what you all are talking about . . .

prcsmath says

Jaleb says

prcsmath says

Given that a, b, c > 1, prove that (a^2+a+1)(b^2+b+1)...(z^2+z+1) is much greater than ab...z - 1. (Variant solved by DT)

Why bother with a,b,c>1 when this is true in all reals?

hence why I said MUCH GREATER

Much greater really only works if you can classify as 2 different infinities. Since both of the functions diverge to the same infinity, it doesn't really classify as much greater.

Well the a,b,c > 1 version is completely trivial by expansion; the LHS contains an ab..z term so we are done..

Jaleb says

prcsmath says

Given that a, b, c > 1, prove that (a^2+a+1)(b^2+b+1)...(z^2+z+1) is much greater than ab...z - 1. (Variant solved by DT)

Why bother with a,b,c>1 when this is true in all reals?

hence why I said MUCH GREATER

prcsmath says

Given that a, b, c > 1, prove that (a^2+a+1)(b^2+b+1)...(z^2+z+1) is much greater than ab...z - 1. (Variant solved by DT)

Why bother with a,b,c>1 when this is true in all reals?

1+1=1 That's all I know

Here is a fun one from last year's Putnam:

Let C(n) be the function such that C(2n)=C(n), C(2n+1)=(-1^n)C(n), and C(1)=1. Find, with proof, the value of

sum_1;^2013 C(n)C(n+2).

Given that a, b, c > 1, prove that (a^2+a+1)(b^2+b+1)...(z^2+z+1) is much greater than ab...z - 1. (Variant solved by DT)

Yeah it is T3 lemma, also a direct consequence of holders
do you want to propose a problem?

T3 Lemma

are you attempting to OPI because it's not working

that's it i'm in

well, PIE is in combinatorics

does it have anything to do with cake?

Disgusting.

Yeah
Heres a well-known warmup
Show that $\dfrac{a^3}{x^2} + \dfrac{b^3}{y^2} + \dfrac{c^3}{z^2} \ge \dfrac{(a+b+c)^3}{(x+y+z)^2}$ for positive reals $a,b,c,x,y,z$.

Like IMO Olympiad math?