On Blaschke’s Theorem
Published by Angel on Wednesday, April 14, 2010 - 02:32:23 - Filed under Convex Sets
I know I havent been around lately, but Ive been very busy with school and stuff and I think thats a good thing, even for the blog, I have a few more theorems I want to post, Ill try to catch up as fast as possible.
I liked this one a lot, this is again on convex sets, I was going to post Jung’s theorem but the proof I have is all on Rober Webster’s Convexity book, I havent really checked if this one is in there but as far as I know it is not.
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Lemma
Let 
be a convex and compact set, then there is a point x in S such that every chord which includes x is divided by x leaving on each side more than one third of the chord.
Proof
Let y in the boundary of S, then we define the set Sy as the points on every chord that has as a boundary the point y, and that the distance to y its less than 2 thirds of the lenght of the chord.
Let y,z,w be on the boundary of S and lets define Sw,Sz and Sy as before.
Using the triange with vertices wzy we build the medians and define x as the centroid of the triangle, now we’ll show that 
.
This is because 
but
then
then
, using the same argument we can show that 
with w’,z’ and y’ the boundary of the median with vertex w,y,z and a the midpoint of yz.
Now we have that
Then 
using Helly’s theorem
Let bb’ be a chord that includes x0, then 
Then, 
This proofs our lemma.
Theorem ( Blaschke’s ) :
Every convex, compact set S with width 1 contains a circle of radius 1/3.
Proof:
By the last lemma there exists on S.
Lets show that 
.
Let y in the boundary of 
and let ww’ be a chord that includes x0, then, since the width of the set is 1, all chords have lenght greather or equal than 1.
Now 
, then 
. QED.
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I hope you comment, Ill try to upload about other topics, maybe some algebra or number theory.
‘Till next time :)
