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Say we have a convex quadrilateral $ABCD$ and a set of points $S$ (no three of which are collinear) such that all $P \in S$ are in the closed set of $ABCD$, and that the points $E$, $F$, $G$, and $H$ lie on $AB$, $BC$, $CD$, and $DA$, which form the convex quadrilateral $EFGH$. For all $P \in S$, $P$ is not in the closed set of $EFGH$. Prove (or disprove) that there exist four points $P_1, P_2, P_3, P_4$ such that the convex quadrilateral $P_1P_2P_3P_4$ can be formed such that no points in $S$ are in the closed set of $P_1P_2P_3P_4$.

My intuition tells me that yes, a quadrilateral does exist, and it would be points that minimize the Euclidean distance from a point $P$ to one of the sides of $EF$. But I'm not sure how to prove this rigorously. I was also thinking about convex hulls, but I don't have too much knowledge on those.

Rough depiction of an example for the problem in Geogebra

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  • $\begingroup$ Source of this question, please? $\endgroup$ Commented Oct 6 at 0:50
  • $\begingroup$ (1) In your drawing, there's another point near to $E$ that I think you should erase. (2) I think something needs to be said about the minimum cardinality that $S$ should have. (3) What do you think is the "closed set of $EFGH$? This should be clarified because $E, F, G, H$ are in that "closed set." $\endgroup$ Commented Oct 6 at 1:19
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    $\begingroup$ @GerryMyerson a friend proposed it to me. I actually think I got it: pick the closest point $P$ to each side (via perpendicular Euclidean distance), and draw lines parallel to each side of $EFGH$ that pass through $P$. Their intersection forms a quadrilateral that does not contain any points in $S$ and has the same angle measures as $EFGH$. Since the selected points are a sub-polygon of the convex polygon we constructed, it also must be convex. Ataulfo, $S$ is finite. I meant to say that borders are allowed. $\endgroup$ Commented Oct 6 at 3:25
  • $\begingroup$ @Boris Mordvinov: It could be possible there is not "the closest point" but "TWO closest points" (in one or more sides). $\endgroup$ Commented Oct 6 at 11:48

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