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Prove or disprove: For a set of at least 3 points not all collinear, we can always construct a triangle that contains the points, with the added condition that each of the triangle's edges has a point at its center.

Example: See the figure below. 7 points are scattered about at random, and the black triangle contains all of them. (Note that I consider the red point in the bottom left contained despite being intersected by the edge of the triangle.) Furthermore, the triangle's sides each have a point from the set at their center. I have colored these center points orange for convenient viewing, but there is nothing special about them: I might have selected another three center points when building a triangle to contain this set.

Example Points and Triangle

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    Before attempting a rigorous look into this proof, it is important to note that by definition you would need to have a set of, at least, 3 points for this proposition to hold. Otherwise the counterexample of $\{1\}$ is trivial. – 2017-01-24
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    True. I'll clarify that. – 2017-01-24
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    I don't think this can work for the vertices of a rectangle. – 2017-01-24
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    @OpenBall sure it does. It just leaves the non-medial point on one of the vertices. – 2017-01-24
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    How did you come up with a conjecture as such? I might as well just call this *Archr’s Theorem* :) – 2017-12-16
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    @user477343 Archr's Theorem is quite a stretch. I was trying to solve this problem from the 2016 Rasor-Bareis exam: *There are 2016 points in the plane such that any triangle with the vertices at three of those points has area at most 1. Prove that all these points are contained in a triangle of area 4.* I knew that this conjecture, if true, would give it to me. As it turns out, the official solution to the problem more or less proves this along the way. – 2017-12-23
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    @Archr ahhh ok then – 2017-12-26

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Your conjecture is true, and for a surprisingly (to me) elementary reason. I strongly suggest you draw my solution for yourself - it isn't deep.

A finite set of points $S$ will determine finitely many triangles; take the largest of these in area, $T$. I claim that all points will fall inside the triangle $M(T)$ which has $T$ as its medial triangle (which will clearly satisfy our requirements on each side of the triangle having a midpoint belonging to $S$). Now suppose that some point $s\in S$ fell outside $M(T)$; let $L$ be the side of $M(T)$ that $s$ lies above. (By this I mean that if we extend the sides of $M(T)$ to infinity, there will be three sections of the plane, determined by these lines, which touch the sides of the triangle; if $s$ is in the section that touches $L$, we can visualize it as "lying above $L$.")

Now, take the side $K$ in the original triangle $T$ which is parallel to $L$ in $M(T)$. Because $s$ lies above $L$, it lies above the vertex $v$ in $T$ that falls along $L$, which means - since $K$ and $L$ are parallel - that the altitude of $s$ from $K$ is higher than that of $v$ from $K$. But then the triangle having $s$ as a vertex instead of $v$ would have strictly larger area (take $K$ as the triangle's base to see this). This would contradict our hypothesis that $T$ was the largest triangle we could make from points in $S$!

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    Had to sketch it to understand it, but great answer! – 2017-01-24