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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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    $\begingroup$ 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. $\endgroup$ – Dylan Jan 24 '17 at 6:29
  • $\begingroup$ True. I'll clarify that. $\endgroup$ – Archr Jan 24 '17 at 6:31
  • $\begingroup$ I don't think this can work for the vertices of a rectangle. $\endgroup$ – user384138 Jan 24 '17 at 7:00
  • $\begingroup$ @OpenBall sure it does. It just leaves the non-medial point on one of the vertices. $\endgroup$ – Joffan Jan 24 '17 at 7:09
  • $\begingroup$ How did you come up with a conjecture as such? I might as well just call this Archr’s Theorem :) $\endgroup$ – Mr Pie Dec 16 '17 at 0:43
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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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  • $\begingroup$ Had to sketch it to understand it, but great answer! $\endgroup$ – Archr Jan 24 '17 at 19:12

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