Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Geometry (an interesting question with a hard proof)

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data

    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].

    2. Relevant equations


    3. 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?
    Last edited: Feb 12, 2007
  2. jcsd
  3. Feb 12, 2007 #2
    Four triangles each of area one, it does not sound like a contradiction.
  4. Feb 12, 2007 #3


    User Avatar
    Science Advisor

    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]".

    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?
  5. Feb 12, 2007 #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

    Might be I am guessing wrongly.

    EDIT: sorry for my formatting skills
  6. Feb 12, 2007 #5
    Sorry, I was not understanding the question properly, but now I do.

    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

    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.
  7. Feb 12, 2007 #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
  8. Feb 12, 2007 #7
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook