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#1 26-09-2014 19:10:11

mona123456
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problem in topologie

hello could someone please help me solving this problem:

Let E be a compact set in R2 containing at least three noncollinear points. Show that
there is a triangle T such that the vertices of T are contained in E and the area of T is
greater than or equal to the area of any triangle whose vertices are contained in E.

#2 27-09-2014 01:09:13

MOHAMED_AIT_LH
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Re : problem in topologie

Hello
Here is an indication (Hint) :
Since [tex]E[/tex] is compact the set [tex]E^3[/tex] is compact too.
The map : [tex]f: E^3 \to \mathbb R[/tex] such that [tex]f(A,B,C)[/tex] is the area of the triangle [tex]ABC[/tex] is contiuous: to show that,  you can for  example use  the Heron formula who  gives  [tex]f(A,B,C)=\sqrt{p(p-a)(p-b)(p-c)}[/tex] where [tex]p=\frac 12 (AB+BC+CA)[/tex]  and  [tex]a=BC, b=CA, c=AB[/tex]  all of which are continuous functions of points [tex]A,B,C[/tex] coordinates.
Now, using this  compactness theorm:

A continuous real-valued function on a nonempty compact space is bounded above and attains its supremum

since [tex]E^3 \neq  \emptyset[/tex] you can conclude.

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