geometry - triangles and subsets of the plane

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mbbx3adr

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A subset S of the plane has the property that the area of any triangle ABC with vertices A, B and C all in S is less than 1.

Prove that there exists a triangle of area 4 which contains the whole set S.
 
problem

I could be on wrong track, but here goes.
Let b1 and h1 be base and height of triangle in subset S.
And let b4 and h4 be base and height of triangle with area 4.
Therefore (b1*h1)/2<1 or b1*h1<2 and (b4*h4)/2=4 or b4*h4=8.

In order for subset S to be in triangle with area 4, I think that's all required is b1<b4 and h1<h4 or b1*h1<b4*h4.

Now since b4*h4=8, then we just need b1*h1<8.

Now since b1*h1<2, then it has to be true that b1*h1<8.
 
oh, thank you. could you please show me how to do it using a rectangle for an arbitrary set S.
 
It should be clear from the given that S is a bounded set. We may assume that S is a compact subset of the plane. Therefore, S has a diameter, the greatest distance between any two points of S, call it M. Now because of compactness, there are two points in S, A & B, such that d(A,B)=M. The line l(AB) determines two half-planes, H<SUB>1</SUB> & H<SUB>2</SUB>. Any point of S in H<SUB>1</SUB> can be no more than a distance of 2/M from l(AB) [by symmetry same is true for H<SUB>2</SUB>]. Thus construct a rectangle with points A & B as midpoints of sides with length 4/M and the other sides have length M. The area of the rectangle is (4/M)(M)=4 and it encloses S.
 
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