# Theorem tennis match!

• Jun 8th 2010, 05:40 PM
Bruno J.
Theorem tennis match!
Okay, so this is a non-standard problem. The game is to play tennis with theorems. Gameplay goes as follows :

(1) A theorem comes your way.
(2) You respond to the theorem with another theorem, which you must relate to the first.

If you play, try to keep the game open, by stating theorems which can easily be related to by others.

I'll begin with a well-known theorem from high-school geometry : the medians of a triangle intersect.
• Jun 8th 2010, 06:14 PM
tonio
Quote:

Originally Posted by Bruno J.
Okay, so this is a non-standard problem. The game is to play tennis with theorems. Gameplay goes as follows :

(1) A theorem comes your way.
(2) You respond to the theorem with another theorem, which you must relate to the first.

If you play, try to keep the game open, by stating theorems which can easily be related to by others.

I'll begin with a well-known theorem from high-school geometry : the medians of a triangle intersect.

I'm not sure I completely grasp the goal of such a game, but let's play a little:

Theorem: The intersection point of the three medians in any triangle divides each median in a proportion 1:2, where the largest part is always on the vertex's side.

Tonio
• Jun 8th 2010, 06:51 PM
Bruno J.
If $X, Y, Z$ are identically distributed real-valued random variables, then $\frac{E(X)}{E(X+Y+Z)}:\frac{E(Y+Z)}{E(X+Y+Z)} = 1/2$.
• Jun 10th 2010, 03:40 AM
Moo
Quote:

Originally Posted by Bruno J.
If $X, Y, Z$ are identically distributed real-valued random variables, then $\frac{E(X)}{E(X+Y+Z)}:\frac{E(Y+Z)}{E(X+Y+Z)} = 1/2$.

Hmm you have to make asumptions about integrability, and about the fact that E[X+Y+Z] is not 0.
• Jun 10th 2010, 03:59 AM
tonio
Quote:

Originally Posted by Moo
Hmm you have to make asumptions about integrability, and about the fact that E[X+Y+Z] is not 0.

Not to mention that it'd be nice to know how this relates to the first two geometric theorems, though I can guess a probable rather tenuous link...

Tonio
• Jun 10th 2010, 05:34 AM
Quote:

Originally Posted by tonio
Not to mention that it'd be nice to know how this relates to the first two geometric theorems, though I can guess a probable rather tenuous link...

Tonio

I'm guessing perhaps he meant you can use parts of the previously posted theorem in your one.

Like if I said

The cardinality of the integers is $\aleph_0$

Your next theorem could be about cardinality, the integers or something to do with $\aleph_0$.

Is that what's going on in here..?

(Although that does seem a little simple for a maths challenge question...)
• Jun 10th 2010, 07:39 AM
Bruno J.
Quote:

Originally Posted by Moo
Hmm you have to make asumptions about integrability, and about the fact that E[X+Y+Z] is not 0.

Absolutely right! Make all necessary assumptions. (Wink)

Tonio : here's how I thought it related to the geometric theorem. The point of intersection of the medians is the center of gravity of the three vertices, assuming they have equal mass. With this in mind, the theorem you stated is a specific case of the following more general theorem : the center of mass of $m+n$ points of equal (nonzero (Giggle)) mass, $n$ of which are blue and $m$ red, lies $n/m$ of the way along the segment joining the centers of mass of the red points and of the blue points. But we can also view the points as events in a real probability space, with the expectation replacing the center of mass...

Deadstar : that's essentially it. The point is just to have fun and to see what kind of unexpected theorems people will pull out of their hats.
• Jun 10th 2010, 10:32 AM
tonio
Quote:

Originally Posted by Bruno J.
Absolutely right! Make all necessary assumptions. (Wink)

Tonio : here's how I thought it related to the geometric theorem. The point of intersection of the medians is the center of gravity of the three vertices, assuming they have equal mass. With this in mind, the theorem you stated is a specific case of the following more general theorem : the center of mass of $m+n$ points of equal (nonzero (Giggle)) mass, $n$ of which are blue and $m$ red, lies $n/m$ of the way along the segment joining the centers of mass of the red points and of the blue points. But we can also view the points as events in a real probability space, with the expectation replacing the center of mass...

Deadstar : that's essentially it. The point is just to have fun and to see what kind of unexpected theorems people will pull out of their hats.

Hmmm...center of mass smell big time to physics mingling into our mathematics. Don't like that (Giggle)

Would the game perhaps be more interesting if we stick to only maths?

Tonio
• Jun 10th 2010, 01:27 PM
Moo
Maybe the expression "center of mass" can equivalently be replaced by "centroid", or "barycenter" ? I think it doesn't change the true meaning of what Bruno says, and it's a real mathematical notion ! :D
• Jun 10th 2010, 01:47 PM
Bruno J.
Quote:

Originally Posted by Moo
Maybe the expression "center of mass" can equivalently be replaced by "centroid", or "barycenter" ? I think it doesn't change the true meaning of what Bruno says, and it's a real mathematical notion ! :D

Haha! What do you guys have against physics? (Giggle)

Come on guys, the ball is in your court. Feel free to reset the thread if you are not inspired by the current discussion!
• Jun 10th 2010, 02:26 PM
chiph588@
Quote:

Originally Posted by Bruno J.
Haha! What do you guys have against physics? (Giggle)

Come on guys, the ball is in your court. Feel free to reset the thread if you are not inspired by the current discussion!

Let $A$ be a complex $n\times n$ matrix. Let $R_i=\sum_{j\neq i}|a_{ij}|$ and $D(a_{ii},R_i)$ be the closed disc centered at $a_{ii}$ and radius $R_i$.
Then each eigenvalue of $A$ lies in at least one $D(a_{ii},R_i)$.
This one is from Putnam a few years back. Let $B$ be a finite collection of open discs in the plane (possibly overlapping). Then there is a subcollection $B'$ of them, such that they are pairwise disjoint, and such that $\displaystyle\cup_{D \in B'}3D$ covers $\cup_{D \in B}D$, where $3D$ denotes the disc with the same center as $D$ but with three times the radius.