Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.2 National Olympiad Prove it Romania

A regular hexagonal prism $ABCDEF A'B'C'D'E'F'$ has the edge $AB = 12$ and the height $AA' = 12\sqrt{3}$ . Let $N$ be the midpoint of the edge $CC'$ .

a.
Prove that the lines $BF'$ and $ND$ are perpendicular.

b.
Find the distance between the lines $BF'$ and $ND$ .

Solution

a.
Let $M$ be the midpoint of the segment $BB'$ . Since $MN \parallel AD$ , the points $A$ , $D$ , $M$ and $N$ are coplanar. Let $Q$ be the midpoint of the segment $BF$ . The intersection of the planes $(BFF')$ and $(ADN)$ is the line $MQ$ . Notice that $BFF'B'$ is a square, hence $BF' \perp MQ$ and then $AD \perp (BFF')$ , implying $BF' \perp AD$ . Therefore $BF' \perp (ADN)$ and then $BF' \perp ND$ , as claimed.

b.
Let $S$ be the intersection point of the lines $MQ$ and $BF'$ . Let $P$ be the projection of $S$ onto line $ND$ . From $BF' \perp (ADN)$ we derive that $BF' \perp SP$ , hence line $SP$ is the common perpendicular of the lines $BF'$ and $ND$ .

In order to find the length of the line segment $[SP]$ , we evaluate the area of the triangle $SND$ in two ways: $S[SND] = S[MNDQ] - S[SDQ] - S[SMN] = \frac{1}{2}SP \cdot ND$ . As $MN = 12$ , $QD = 18$ , $ND = 6\sqrt{7}$ , $MQ = 6\sqrt{6}$ , we get $SP = \frac{15\sqrt{42}}{7}$ .

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