Question
Consider the regular tetrahedron $ABCD$. The points $M$ and $P$ are the means of the segments $BC$ and $AM$, respectively. On the edge $BD$ take the point $K$ so that $BK = 3*KD$ Knowing that the distance between the lines $PK$ and $AC$ is $\sqrt{3}$ cm, calculate the distance between $AC$ and $BD$.
My idea
**Drawing **
I know that most of these types of problems are solved using the volumes expressed in two ways.
So we can easily show that $AC \perp BD$, which makes the angle between these two lines $90$ degrees.
We can use Chasles Formula(I explained it in the comments) and we get that the volume of the regular tetrahedron is $= \frac{AC*BD* dis(AC, BD)*sin(90)}{6}$.
We can say that because we have a regular tetrahedron $AC=BD$ and we also know that $sin(90)=1$
If we can express the volume again using the given distance we can find the distance between $AC$ and $BD$.
I don't know where to start! I hope one of you can help me! Thank you!
Note first of all that $$ volume(KPAC)={1\over4}volume(KBAC)={3\over16}volume(ABCD). $$ Hence, from the formula given in the question we get: $$ PK\cdot AC\cdot distance(PK,AC)\sin\alpha={3\over16}AB\cdot CD\cdot distance(AB,CD), $$ where $\alpha$ is the angle between lines $PK$ and $AC$, that is: $$ distance(AB,CD)= {16\over3}{PK\over CD}\cdot distance(PK,AC)\sin\alpha ={16\over\sqrt3}\ {PK\over CD}\sin\alpha. $$ To evaluate $PK/CD$ and $\alpha$, draw from $P$ and $K$ two lines parallel to $BC$, intersecting edges $AC$ and $CD$ respectively at $E$ and $F$ (see figure below). Observe that $PE={1\over4}BC=KF$, hence $PEFK$ is a parallelogram. As a consequence, $PK=EF$ and $\alpha=\angle CEF$.
From the sine rule in triangle $CEF$ we then get: $$ {PK\over CD}\sin\alpha={EF\over CD}\sin(\angle CEF)= {CF\over CD}\sin 60°={3\sqrt3\over8}. $$ Inserting that into the above formula we finally get: $$ distance(AB,CD)=6. $$