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{\bf Question}

\begin{description}
\item[(a)] Four non-coplanar points $A,B,C,D$ lie at equal
distances from the point $O$.  A point $P$ is defined by
$$2\vec{OP} = \vec{OA} + \vec{OB} + \vec{OC} + \vec{OD}.$$  show
that the line joining $P$ to the mid point of any edge of the
tetrahedron $ABCD$ is at right angles to the opposite edge.
\item[(b)] Suppose that $O$ is the centre of a circle
$A-1A_2A_3$ of unit radius in a plane $\pi$.  If the points
$B_1,,B_2,B_3$ are defined by $\vec{OB_1} = \vec{OA_2} +
\vec{OA_3},$

$ \vec{OB_2} = \vec{OA_3} + \vec{OA_1}, \, \vec{OB_3} = \vec{OA_1}
+ \vec{OA_2}$ show that $B_1,\,B_2,\,B_3$ are the centres of other
unit circles in $\pi$ which pass through $A_2$ and $A_3,$ $A_3$
and $A_1,$ $A_1$ and $A_1$ respectively.

Prove that the three circles, centers $B_i$ meet at $C$ where
$\vec{OC} = \vec{)A_1} + \vec{OA_2} + \vec{OA_3}.$
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] The midpoint $M$ of $CD$ is $\frac{1}{2}(\vec{OC} +
\vec{OD})$.

The point $P$ is $\frac{1}{2}(\vec{OA} + \vec{OB} + \vec{OC} +
\vec{OD}$

So $ \vec{MP} = \frac{1}{2}(\vec{OA}+ \vec{OB})$

The direction of the opposite edge $\vec{AB} = \vec{OB} -
\vec{OA}$

So $\vec{MP} \cdot \vec{AB} = \frac{1}{2}(|OB|^2 - |OA|^2 ) = 0$

${}$
\item[(b)]

\begin{center}
\epsfig{file=g77-4-1.eps, width=60mm}
\end{center}

${}$

first part os obvoius from parallelograms.

${}$

$\vec{B_1C} = \vec{OC} - \vec{OB_1} = \vec{OC} - (\vec{OA_2} +
\vec{OA_3}) = \vec{OA_1}$

So $|B_1C|=1$

Similarly $|B_2C|=1$ and $|B_3C|=1$

So $C$ lies on all three circles.
\end{description}


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