Geometry (an interesting question with a hard proof)

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SUMMARY

The discussion centers on a geometric problem involving n > 3 points in a plane, where any three points form a triangle with an area ≤ 1. The goal is to demonstrate that all n points lie within a triangle of area ≤ 4. Participants clarify that proving the contrapositive—if n > 3 points lie in a triangle with area > 4, then at least one triangle formed by any three points must have an area > 1—validates the original statement. The conversation highlights the importance of understanding the implications of triangle areas and the relationships between points.

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  • Understanding of basic geometric principles, specifically triangle area calculations.
  • Familiarity with contrapositive reasoning in mathematical proofs.
  • Knowledge of point distribution in a plane and its implications on geometric properties.
  • Ability to visualize geometric configurations and their relationships.
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rad0786
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Homework Statement



There are n > 3 points in the plane such that any three of the points form a triangle of area \leq 1. Show that all n points lie in a triangle of area \leq 4.



Homework Equations





None.


The Attempt at a Solution



This is hard to discribe since it is immposible to "draw" on a computer.

However, I think this problem may not have been stated correctly, because, if we choose our 3 points to be the vertices of a triangle with area = 4, then we found a contradiction.

Although, if I am wrong and this problem is correct, i thought that proving the contrapostivie proves the question.

Can somebody please help me out?
 
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Four triangles each of area one, it does not sound like a contradiction.
 
rad0786 said:

Homework Statement



There are n > 3 points in the plane such that any three of the points form a triangle of area \leq 1. Show that all n points lie in a triangle of area \leq 4.



Homework Equations





None.


The Attempt at a Solution



This is hard to discribe since it is immposible to "draw" on a computer.

However, I think this problem may not have been stated correctly, because, if we choose our 3 points to be the vertices of a triangle with area = 4, then we found a contradiction.
What do you mean by "our 3 points"? The problem requires "N> 3" points- at least 4. Also "any three of the points form a triangle of area \leq 1".

Although, if I am wrong and this problem is correct, i thought that proving the contrapostivie proves the question.

Can somebody please help me out?
Yes, proving the contrapositive proves the question. Here the contrapositive is "If n>3 points in the plane lie in a triangle with area greater than 4 then 3 of the points form a triangle of area greater than 1". Does that help you?
 
Four triangles, each in the plane z = 0 for example, interior triangle i has an area of 1, the exterior triangle and area 4.

x - i - x
i - i
x

Might be I am guessing wrongly.

EDIT: sorry for my formatting skills
 
HallsofIvy said:
What do you mean by "our 3 points"? The problem requires "N> 3" points- at least 4. Also "any three of the points form a triangle of area \leq 1".

Sorry, I was not understanding the question properly, but now I do.


HallsofIvy said:
Yes, proving the contrapositive proves the question. Here the contrapositive is "If n>3 points in the plane lie in a triangle with area greater than 4 then 3 of the points form a triangle of area greater than 1". Does that help you?

For some reason I can still come up with a case that contradicts this.

Consider a triangle with area greater than 4 and choose 3 out of the n points such that tose 3 points are as close as as possible to each other and give an area less than 1.

Perhaps I am interpreting this entire problem wrong

Shade said:
Four triangles, each in the plane z = 0 for example, interior triangle i has an area of 1, the exterior triangle and area 4.

x.-.i.-.x
..i..-..i
...x

Might be I am guessing wrongly.

EDIT: sorry for my formatting skills

I know what you mean, it is hard to draw stuff out on the forums.

I see that triangle i has an area of 1, but i missed what you are saying about the exterior triangle.

I see if we choose 3 arbitrary points i and form a triangle of them, then place any amount of points in the triangle x, we have a solution.
 
One big "exterior" triangle from X0 to X2, while have the area summed of the four sub triangles
triangle t0 = X0,i0,i2 t1 = i0,i2,i1, t2 =i0,X1,i2 and t3 = i1,i2,X2. The exterior triangle X0,X1,X2.

This will give four triangles each with an area of 1, the center triangle i0,i1,i2 will per definition lie within the triangle of area four.

The problem as I see it is not possible to solve with a single triangle but by letting the vertices form indidual triangles and be part of a large mesh, also defined by a single triangle.
X0 - i0 - X1
...i1 - i2
...X2
 
Shade said:
One big "exterior" triangle from X0 to X2, while have the area summed of the four sub triangles
triangle t0 = X0,i0,i2 t1 = i0,i2,i1, t2 =i0,X1,i2 and t3 = i1,i2,X2. The exterior triangle X0,X1,X2.

This will give four triangles each with an area of 1, the center triangle i0,i1,i2 will per definition lie within the triangle of area four.

The problem as I see it is not possible to solve with a single triangle but by letting the vertices form indidual triangles and be part of a large mesh, also defined by a single triangle.



X0 - i0 - X1
...i1 - i2
...X2

Your explanation was very very clear.

When I see "If ..then" problems, I allways try to write it out as "suppose so and so..." and see where I can get...and if i get lost, i go for the contrapositive.

I guess my method fails here.

Thank you for your explanation.
 

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