# Construct the intersection of a cube by a plane through $3$ points on its edges, no pair of which is on the same face

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.

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.

• Thank you! The answer, indeed, very cozy and elegant = ) – hayk Sep 11 '15 at 19:01
• Sorry, what software did you use to draw this? – Who Save Me Save Entire World Jun 15 at 23:13
• @TheShortestMustacheTheorem I used GeoGebra 5. – Intelligenti pauca Jun 16 at 6:38
• Thank you very much! – Who Save Me Save Entire World Jun 16 at 7:00

Another way to solve the problem is the following. Given are the points $$A$$, $$B$$ and $$C$$. Consider the following image:

• 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.