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This is a question for my math seminar class.

Let $T$ be a right triangle with sides having lengths $3$, $4$, and $5$. A point $P$ is called "special" is $P$ is the center of a parallelogram whose vertices all lie on the boundary of $T$. What is the area of the set of special points?

I've tried drawing a couple parallelograms within this triangle, but since there are a couple different options, I unsure of how to get an exact area related to point $P$. I've tried to see what these parallelograms have in common, and if a proof can be written based on this to find the area of the special points. I have noticed the center tends to be close the to 3 side of the triangle, and am trying to see what I can do with that. I've also tried case work (drawings points within) to see if I can average them.

Any ideas? Any and all help is appreciated.

user978757
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    Can you show that the area is not zero? What did you try? – markvs Dec 18 '21 at 23:35
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    You can split cases based on which side has two parallelogram vertices – user_194421 Dec 18 '21 at 23:42
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    @user_194421: the funny thing is that each of the three cases produces exactly the same special points (if we accept a single line through one of the vertices as a "degenerate" parallelogram). – Troposphere Dec 19 '21 at 00:55

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