Circumscribed sphere about a tetrahedron

Question

$ABCD$ is a tetrahedron such that $AB=2\sqrt{3}$, $BC=\sqrt{7}$, $AC=1$ and $AD=BD$. Plane $ABD$ is perpendicular to plane $ABC$. Find the shortest distance between lines $BD$ and $AC$ if the radius of the circumscribed sphere about the tetrahedron is equal to $2\sqrt{2}$.

Answer 1

I'll assume that $\triangle ABD$ is acute. Let $P$ be the center of the circumscribed circle about $\triangle ABC$ and $Q$ be the center of the circumscribed circle about $\triangle ABD$. Consider a line $p$ that passes through point $P$ and is perpendicular to plane $ABC$. Similarly, let $q$ be a line that passes through $Q$ and is perpendicular to plane $ABD$. Denote $O=p\cap q$. We've thereby constructed a point which is equidistant from the vertices of the tetrahedron and is thus the center of the circumscribed sphere about the tetrahedron. Note that $\triangle ABC$ is obtuse and hence point $P$ lies outside it, so $O$ must lie outside the tetrahedron. So far so good.

Let $M$ be the midpoint of $AB$. $DA=DB$ yields $DM\perp AB$. On the other hand, $OM\perp AB$ as $OA=OB$. The dihedral angle between the two perpendicular planes $(ABC)$ and $(ABD)$ is therefore $\sphericalangle QMP=90^\circ$. Note that $MPOQ$ is a quadrilateral with three right angles so it must be a rectangle, meaning that $OP=MQ$. We'll come back to that in a bit. Let's calculate the radius of the circumscribed circle about $\triangle ABC$. We calculate $\sphericalangle BAC=30^\circ$ and now we can use law of sines to calculate the radius, which is simply equal to $\sqrt{7}$. Consider the right $\triangle BPO$ where $BO=2\sqrt{2}$ and $BP=1$. The Pythagorean theorem gives us $OP=1$ which, according to our previous observations, is also equal to $MQ$. From here we easily calculate $QB=2=QD$ so $MD=3$ and hence $AD=BD=2\sqrt{3}$.

To find the shortest distance between lines $BD$ and $AC$, I suggest that we choose a point $N\in (ABC)$ such that $ABNC$ is a parallelogram. Here's a sketch of what we've done for now:

Note that $AC\parallel (BND)$ and $BD\subset (BND)$ so the shortest distance between these lines would be equal to the distance between an arbitrary point from $AC$ to plane $(BND)$. Let that point be $C$. In order to calculate the required distance, we need to find the volume of $BNCD$ (which is equal to the volume of $ABCD$) and the area of $\triangle BND$. Let's calculate the volume first. The height of the tetrahedron is $DM=3$ and we easily calculate $S_{ABC}=\dfrac{\sqrt{3}}{2}$. We therefore obtain $V_{ABCD}=V_{BNCD}=\dfrac{\sqrt{3}}{2}$. To find the area of $\triangle BND$. We have $\cos\sphericalangle ABD\cos\sphericalangle ABN=\cos\sphericalangle DBN$ and so $\cos\sphericalangle DBN=-\dfrac{\sqrt{3}}{4}$ (because $\sphericalangle ABD=60^\circ$ and $\sphericalangle ABN=150^\circ$). Therefore, $\sin\sphericalangle DBN=\dfrac{\sqrt{13}}{4}$, which we find using the well known Pythagorean identity $\sin^2\alpha+\cos^2\alpha=1$. We thus have $S_{DBN}=\dfrac{\sqrt{39}}{4}$ and so we're finally ready to find the required distance, which I'm going to denote by $d$. We have $\dfrac{d\sqrt{39}}{12}=\dfrac{\sqrt{3}}{2}$ or $d=\dfrac{6\sqrt{13}}{13}$. The case when $\triangle ABD$ is obtuse is left as an exercise to the OP. A very interesting problem indeed.