I thought that there would be more....
On the xy-plane consider a square grid with corners (0,0), (0,10), (10,10) and (10,0). Let S be the set of all (non-degenerate) triangles that can be formed with vertices that have integer coordinates on this grid. (For example, the triangle with vertices (0,1), (5,5) and (9,2) would be one element of S).
How many distinct area values are there among the elements of S?
The answer is 100.
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1 solution
Wow, I find it interesting that all areas can be achieved. How do we get an area of 4 9 2 1 or 4 9 ?
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I think that the easiest way of finding all the values is to use the vector cross-product. Assume that the three vertices are ( 0 , 0 ) , ( a , b ) , ( c , d ) . Then the area of the triangle is ∣ a d − b c ∣ / 2 . So with vertices ( 0 , 0 ) , ( 1 0 , 1 ) , ( 1 , 1 0 ) we get an area of 9 9 / 2 , and with vertices ( 0 , 0 ) , ( 1 0 , 2 ) , ( 1 , 1 0 ) we get an area of 4 9 .
We can check that ∣ a d − b c ∣ takes on all integer values from 1 to 1 0 0 by taking a = d or a = d + 1 , setting b = 0 or b = 1 and then playing around with c. This is the tedious part, but it does fill in the gaps eventually. I still wonder if there is a more elegant approach, but nevertheless it is an interesting result. Thanks for "liking" my problem; it encourages me to post more questions. :)
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PLEASE KEEP POSTING :) ....... enjoyed solving this
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@Abhinav Raichur – Thanks for the encouragement, Abhinav. I'm glad that you enjoyed this problem. :)
U sing points: (0,0), (10,1), (a,b) will do the work giving Area = 0.5 (10b - a). Any integer 1-100 can be represented as 10b - a. As a generalization to this problem, it is possible to get all possible multiples of 0.5 by using the following points: (0,0), (10,-1), (a,b). Then Area = 0.5 * (10* b + a).
can u please provide the link for Pick's theorem
Pick's Theorem guarantees that all the area values will be either integers or half-integers, (e.g., 15/2). It then becomes a matter of showing that each value from 1/2 to 50, in steps of 1/2, corresponds to the area of at least one element of S. Having done that, we can then conclude that there a total of 100 possible area values.
There are several ways of establishing that all of the aforementioned area values are possible, all of them pretty tedious. If you have an elegant way of doing it then please share. I'll try and post one approach later when I'm less tired. :)