Question 49
Let $ABCDA'B'C'D'$ be a rectangular parallelepiped, and $M$, $N$, $P$ the projections of the points $A$, $C$, respectively $B'$, on the diagonal $BD'$.
a)
Show that $BM + BN + BP = BD'$.
b) Show that $3(AM^2 + B'P^2 + CN^2) \ge 2D'B^2$ if and only if the rectangular parallelepiped $ABCDA'B'C'D'$ is a cube.
a)
Show that $BM + BN + BP = BD'$.
b) Show that $3(AM^2 + B'P^2 + CN^2) \ge 2D'B^2$ if and only if the rectangular parallelepiped $ABCDA'B'C'D'$ is a cube.