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So this is a rather old problem, but I still cannot find a pure constructive solution to it. Please, do not offer me to write a plane equation, etc. I would be grateful, if you offer a solution by only using the means of construction.

Problem. Construct(!) a plane intersection of the cube by three points of that plane. None the pairs of these points lie on the same face of the cube.

enter image description here

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Let $PQR$ be the three points defining the plane, belonging to edges $AE$, $BC$ and $FH$ of the given cube (see picture below). Draw from $Q$ a parallel to $BD$, to meet $DF$ at $Q'$. Let $Q''$ be the symmetric of $Q'$ with respect to the midpoint of $ER$ and let $Q'''$ be the symmetric of $Q$ with respect to the midpoint of $PR$. Then $Q'$ and $Q''$ are the projections of $Q$ and $Q'''$ on plane $EDFH$, so that $QQ'''$ and $Q'Q''$ meet at a point $S'$ belonging both to plane $EDFH$ and to plane $PQRQ'''$. Line $RS'$ is then the intersection of those two planes and it intersects edge $EH$ at a point $S$, which is the first of the three points we must construct.

You could now repeat the same construction starting from $P$ and then from $R$, to get other two points $T$ and $U$ on the edges of the cube, but there is a shortcut: just draw from $Q$ a line parallel to $RS$ and another parallel to $PS$, so that $U$ and $T$ will be the intersections of those lines with $AB$ and $CF$ respectively. Finally join $RSPUQT$ to get a hexagon which is the desired intersection.

enter image description here

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  • $\begingroup$ Thank you! The answer, indeed, very cozy and elegant = ) $\endgroup$ – hayk Sep 11 '15 at 19:01
  • $\begingroup$ Sorry, what software did you use to draw this? $\endgroup$ – Who Save Me Save Entire World Jun 15 at 23:13
  • $\begingroup$ @TheShortestMustacheTheorem I used GeoGebra 5. $\endgroup$ – Intelligenti pauca Jun 16 at 6:38
  • $\begingroup$ Thank you very much! $\endgroup$ – Who Save Me Save Entire World Jun 16 at 7:00
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Another way to solve the problem is the following. Given are the points $A$, $B$ and $C$. Consider the following image:

enter image description here

  • Draw the line AC.
  • Project the point $A$ on the $xy$-plane. Because the point $A$ lies in the $yz$-Plane, the projected point will lie on the $x$-Axis. Mark the projected point as $A'$.
  • Draw the line $A'C$. This line lies on the $xy$-plane.
  • Call the plane parallel to the $yz$ plane, which contains the other side of the cube, $\tau_1$. This plane contains the point $B$, which is also contained in the plane given by the three points $A,B,C$, call this last plane $\epsilon ABC$. If you manage to find another point on $\tau_1$, which also lies on the plane $\epsilon ABC$, then you get a line, which lies on both planes, in particular in $\tau_1$.
  • Find this point by drawing $M$ and its projection on the line $AC$. Call this point $D$.
  • The line $BD$ lies on $\tau_1$, therefore the segment $BE$ is the first segment of the intersection of the cube with your plane.
  • Connect $E$ with $C$ to find the second segment
  • The last drawn segment is goint to be parallel to the segment going trough $A$. (The $xy$-plane an the upper plane containing $A$ are parallel, therefore the intersection lines of your plane with those planes must be parallel). Draw the third segment and find $F$.
  • Connect $F$ and $B$ to find the fourth segment and with the same reasoning as in the last point complete the intersection.
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