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]]>Since triangle has 3 sides it’s odd….so not an even number of segments.

What am I missing….? ]]>

I think Cristi’s solution is closest, but needs a little more. Here is my attempt. Let the square grid be the usual integer grid, generated by (1,0) and (0,1). Let the common length of the segments in the polygonal chain be sqrt(N). Then N is either 0, 1, or 2 (mod 4). Let us consider these cases in reverse order.

If N is 2 (mod 4), then each segment has odd x- and y-components, so both coordinates alternate in parity as we travel around the chain. Thus the number of segments in the chain is even.

If N is 1 (mod 4), then each segment must have one odd and one even component, so the sum of coordinates alternates in parity. Again, the chain must consist of an even number of segments. (The N=2 and N=1 cases are similar, but the checkerboards colorings apply to different grids!)

If N is 0 (mod 4), then each segment has even x- and y-components, so all points visited lie on a subgrid of the original integer grid. We may divide a common power of 2 out of all lengths so as to pass to one of the cases already considered.

]]>All odd squares are 1 (mod 4). So any two odd squares sum to a number that is 2 (mod 4). But all even squares are 0 (mod 4). Therefore, two odd squares can never sum to an even square.

]]>no, thats what the trick is to the problem

if you draw the triangle in a circle, the hypotenuse will be a diameter, and the largest the altitude can be for the hypotenuse of length 10 is 5 (a radius)

so that means that the altitude of 6 isn’t for the hypotenuse, but is the length of one of the legs, (right triangle), so the other length must be 8 (B^2 = C^2- A^2), so the area is 6*8/2 = 24

Marcial Fonseca ]]>

Suppose my polygon has the following vertices:

(0,0)

(4,3)

(4,8)

(0,11)

(-4,8)

(-4,3)

These six points are the vertices of an equilateral hexagon, each of whose segments has length five. This hexagon will serve as a counterexample for both Simon’s and Brenda’s proofs, though it will not disprove the hypothesis (since it has an even number of sides).

Simon claims that as you move from one vertex to the next on the polygon, the Manhattan distance to the origin increases or decreases by a+b or a-b. In my hexagon, the Manhattan distances from the origin to the second and third vertices are 7 and 12. Their difference is 5, which is neither 4+3 nor 4-3. So my polygon meets the conditions of Simon’s construct but contradicts his claim regarding Manhattan distances.

Brenda claims to have proven that a polygon satisfying the problem description must have a number of sides that is a multiple of four. My hexagon has six sides, which is not a multiple of four but satisfies the conditions of the problem.

So neither proof is sufficient.

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