Easy [contraction on the euclidean plane]

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Omid Hatami

1275 posts

Omid Hatami

#1 h Sep 14, 2004, 6:34 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

$f:\mathbb{R}^2 \to \mathbb{R}^2$ is injective and surjective. Distance of $X$ and $Y$ is not less than distance of $f(X)$ and $f(Y)$ . Prove for $A$ in plane:
$S(A) \geq S(f(A))$
where $S(A)$ is area of $A$

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Sailor

256 posts

Sailor

#2 h Sep 15, 2004, 12:53 PM • 2 Y

Y by Adventure10, Mango247

Maybe it's time to post your solution, since you say it's easy...

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sam-n

793 posts

sam-n

#3 h Sep 15, 2004, 2:40 PM • 2 Y

Y by Adventure10, Mango247

Hint : u should solve it with circle packing

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DusT

297 posts

DusT

#4 h Sep 15, 2004, 7:35 PM • 2 Y

Y by Adventure10, Mango247

Maybe the best idea is to break it in small squares, or any regular figure..., but I think the problem is the function is not continuous.......
Also, here arises a question, is A a contour of the figure, or is A the entire figure, because some continuity problems appear....
What do you think, sam-n ????

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Simo_the_Wolf

106 posts

Simo_the_Wolf

#5 h Sep 15, 2004, 8:51 PM • 2 Y

Y by Adventure10, Mango247

Take a point P inside A.
Take all the distances from P to all points of the perimeter of A.

we take $k=max[d(P,f(B))/d(P,B)]$ where B is a point of the perimeter.

We know that k<1 and all points of f(A) are in A' where A' is A dilatate about P of the coefficient k. So S[f(A)]<=S(A') but S(A')=kS(A)<S(A) and so
S[f(A)]<S(A)

c.v.d.

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Pascual2005

1160 posts

Pascual2005

#6 h Sep 22, 2004, 12:01 AM • 3 Y

Y by Adventure10, Adventure10, Mango247

what is circle packing?

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Omid Hatami

1275 posts

Omid Hatami

#7 h Sep 22, 2004, 11:22 AM • 2 Y

Y by Adventure10, Mango247

Well for any figure $B$ you must first put some cirlces in $B$ that they are distinct and sum of their areas is $S(B)- \epsilon$ .

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Omid Hatami

1275 posts

Omid Hatami

#8 h Sep 28, 2004, 8:30 AM • 2 Y

Y by Adventure10, Mango247

Well consider circles with centers $f(A_1),f(A_2),...,f(A_n)(A_i \in A)$ with radius $r_1,...,r_n$ that are distinct and sum of ther areas is $S(f(A))- \epsilon$
Now draw circles with centers $A_1,...,A_n$ and radius $r_1,...,r_n$ .
Prove they are distinct and are in $A$ so for every $\epsilon$ :
$S(A) \geq S(f(A))-\epsilon$
And this proves the problem

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Omid Hatami

1275 posts

Omid Hatami

#9 h Oct 2, 2004, 7:06 PM • 2 Y

Y by Adventure10, Mango247

For circle packing you must use this that for every $\epsilon$ there are finite squares in $A$ sum of their area is at least $S(A) - \epsilon$

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P-H-David-Clarence

92 posts

P-H-David-Clarence

#10 h Jun 5, 2016, 1:27 PM • 2 Y

Y by Adventure10, Mango247

Simo_the_Wolf wrote:

Take a point P inside A.
Take all the distances from P to all points of the perimeter of A.

we take $k=max[d(P,f(B))/d(P,B)]$ where B is a point of the perimeter.

We know that k<1 and all points of f(A) are in A' where A' is A dilatate about P of the coefficient k. So S[f(A)]<=S(A') but S(A')=kS(A)<S(A) and so
S[f(A)]<S(A)

c.v.d.

But the figure may not be convex so why f(A) is in A'?

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