Author Topic: Area of a Lattice Triangle  (Read 16853 times)

MrPuzzle

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Area of a Lattice Triangle
« on: 八月 31, 2004, 12:49:24 pm »
It seems we are in geometry mode this week, here's another geometry related puzzle.

A lattice point on the plane is a point where both x and y coordinates are integers. Prove the area of a triangle formed by any three lattice points is always equal to integer/2.

packman

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Area of a Lattice Triangle
« Reply #1 on: 八月 31, 2004, 03:02:20 pm »
Draw a rectangle that contains this triangle along the lattice (i.e. three points of the triangle are on the lines of rectangle). The area of the triangle is equal to the area of rectangle (integer) minus three 90 degree triangles (half integer).
简单==完美

MrPuzzle

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Area of a Lattice Triangle
« Reply #2 on: 九月 01, 2004, 07:49:30 am »
Since we are talking about any 3 lattice points, you may not be able to construct a lattice rectangle that contains all these 3 points on the edge.  Your idea still works, but need more explaination.

packman

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Area of a Lattice Triangle
« Reply #3 on: 九月 01, 2004, 08:18:47 am »
sure I can construct a rectangle (maybe I didn't explain clearly).
Assume 3 points are on an x-y plain, pick the points with the largest and the smallest y and draw 2 horizontal lines; pick the points with the largest and the smallest x and draw 2 vertical lines. These 4 lines make a rectangle, and it's area is integer x integer.
简单==完美

MrPuzzle

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Area of a Lattice Triangle
« Reply #4 on: 九月 01, 2004, 09:35:22 am »
Yes, you can construct a lattice rectangle to contain the triangle. But your original solution said that the 3 triangle lattice points are on the rectangle edge, from the way of your construction, I cannot see that happening.

packman

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Area of a Lattice Triangle
« Reply #5 on: 九月 01, 2004, 01:55:06 pm »
well, I guess I didn't explain clearly. :wink:
简单==完美

MrPuzzle

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Area of a Lattice Triangle
« Reply #6 on: 九月 01, 2004, 02:33:24 pm »
Which means your proof needs more explaination. It is obvious that your current proof doesn't apply to (0,0), (3,3), (2,1).  However, as I said before, your idea works, just need extra explaination for the details.

The reason I am being so difficult is that I have a solution in mind that doesn't require any rectangle stuff, and I am trying to convince myself there's no easier way. Your proof seems easier, but with extra explaination, it is not as clear as the proof I have in mind. :)

packman

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Area of a Lattice Triangle
« Reply #7 on: 九月 01, 2004, 02:49:50 pm »
Click the link here to see the picture. The area outside the triangle (but inside the rectangle) can be easily proved as half integer.
http://www.geocities.com/packman_at_cnd/math_problems/triangle.html
简单==完美

MrPuzzle

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Area of a Lattice Triangle
« Reply #8 on: 九月 01, 2004, 03:52:54 pm »
The first picture in your link was what I had in mind when I questioning your first post which claim all 3 points were on the rectangle edge.  With all these 3 pictures,  now I think your proof is solid.

Now, here's the proof I had in mind. We all know that the cross product of two vector equals the area of the parallelogram formmed by by these two vectors  A = AxB = |A|*|B|*sin(alpha). The area of the triangle is just half of the area of the parallelgram.  Also, we can always assume one of the points is the origin, thus, the cross product is equal to |x1*y2-x2*y1|, and the area is half of that.

sean

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A theorem related to this problem
« Reply #9 on: 九月 05, 2004, 10:36:39 am »
Mr. Puzzle, your proof is just  the analytic geometry proof.    A famous theorem (called Picard theorem?) in elmentary geometry states that the area of a polygon (convex or not) equals
       
         number of lattice points inside the polygon
     + half of number of lattice points on the boundary of the polygon
     -1


For instance, the area of the triangle formed by <0, 0>, <1, 0>, and <0, 1> is 1/2.

Packman's method can be easily modified to prove this.

万精油

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Area of a Lattice Triangle
« Reply #10 on: 九月 07, 2004, 09:26:22 am »
Picard's theorem is about complex entire functions, very deep and hard, way beyond this forum. The theorem you are talking about is called Pick's theorem.  As you said, Packman's proof can be edited to prove Pick's theorem. it can also be proved using induction.