Let $ABCDS$ be a regular pyramid ($AB=BC=CD=AD$ and $SA=SB=SC=SD$). $E$ is the midpoint of $CD$ and $O$ is the centre of the base. Let $d$ be the distance between $AC$ and $SE$, and $h$ be the distance of $O$ from $SD$.
Show that $d\gt \frac{h}{2}$
I came across this problem in a textbook. In that textbook this problem is a little bit different, the line segments have lengths and there is a numerical answer. In the answers the lengths of $d$ and $\frac{h}{2}$ are the same. I checked them with a 3d constructor and found that $d\gt \frac{h}{2}$. Can you help me find which of us is right?
A hint would be much more appreciated.
Thanks in advance.
The distance from line $AC$ to line $SE$ is the same as the distance between line $AC$ and the plane parallel to $AC$ and passing through $SE$. That plane is easy to construct: if $M$ is the midpoint of $AD$, line $ME$ is parallel to $AC$ and $SEM$ is then the requested plane.
To compute the distance $d$ between $AC$ and plane $SEM$ we can conveniently choose point $O$ on $AC$ and observe that the altitude $OG$ of triangle $OFS$ is perpendicular to the plane, where $F$ is the midpoint of $EM$. Hence: $$ d=OG={OF\cdot OS\over FS}= {OF\cdot OS\over \sqrt{OF^2+OS^2}}. $$ On the other hand, the distance $h$ of $O$ from line $SD$ is the altitude of triangle $OSD$, hence: $$ h={OD\cdot OS\over DS}= {2OF\cdot OS\over \sqrt{4OF^2+OS^2}} $$ and: $$ {d\over h}={1\over2}{\sqrt{4OF^2+OS^2}\over\sqrt{OF^2+OS^2}}= \sqrt{1+2r^2\over1+8r^2}>{1\over2}, \quad\text{where}\quad r={OS\over CD}. $$