Geometry (an interesting question with a hard proof)

In summary, the problem states that there are n points in the plane such that any three of them form a triangle of area less than or equal to 1. The goal is to show that all n points must lie within a triangle of area less than or equal to 4. The solution involves breaking the points into four triangles, each with an area of 1, and showing that the remaining points must lie within the exterior triangle formed by the four smaller triangles. Proving the contrapositive of the statement, that if n points lie within a triangle of area greater than 4, then three of the points must form a triangle of area greater than 1, helps to prove the original statement.
  • #1
rad0786
188
0

Homework Statement



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



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, becuase, 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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  • #2
Four triangles each of area one, it does not sound like a contradiction.
 
  • #3
rad0786 said:

Homework Statement



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



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, becuase, 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 [tex] \leq 1[/tex]".

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?
 
  • #4
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
 
  • #5
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 [tex] \leq 1[/tex]".

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.
 
  • #6
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
 
  • #7
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.
 

1. What is the Pythagorean theorem and how is it used in geometry?

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This theorem is used in geometry to solve for unknown sides or angles in right triangles.

2. What is the difference between a line segment and a line?

A line segment is a straight path that has a defined length, while a line is an infinitely long and straight path with no defined length. A line can be thought of as a collection of infinite points, while a line segment is a finite set of points.

3. Can two parallel lines ever intersect?

No, by definition, parallel lines are lines that never intersect. They have the same slope and will continue on in the same direction without ever crossing each other.

4. What is the difference between a regular and an irregular polygon?

A regular polygon is a shape with equal sides and equal angles, while an irregular polygon has sides and angles of varying lengths and measures. Regular polygons are also symmetrical, while irregular polygons are not.

5. What is the proof for the sum of angles in a triangle?

The proof for the sum of angles in a triangle is based on the fact that a triangle can be divided into two smaller triangles by drawing a line from one vertex to the opposite side. This creates two interior angles that are supplementary (adding up to 180 degrees). By extending this logic to all three sides of the triangle, it can be proven that the sum of the angles in a triangle is always 180 degrees.

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