## PROBLEM 2

a) The square ABCD and the rectangle ABEF are located in perpendicular planes. If $A B=40 \text{ cm}, B E=30 \text{ cm}$, find $d(D, E F)$ and $d(E, A C)$.

b) The cube OLEMPICS has the edge length $O P=6 \text{ cm}$.

Calculate: i) the length of the diagonal of the rectangular parallelepiped formed by joining two such cubes;

ii) $d(M,(P L C))$;

iii) $m(<(O C,(O L E)))$.

(Math Gazette No.1/2011-Supplement with exercises)
Solution:
a) Figure:
0.50 points
$D A \perp(A B E)$
$\left.\begin{array}{c}\text { From } E F \subset(A B E) \\ A F \perp F E\end{array}\right\} \underset{\pi_{3} P}{\Rightarrow} D F \perp F E \Rightarrow d(D, E F)=D F=50 \mathrm{~cm}$ $.1.50$ points

## From

$$
\left.\begin{array}{l}
E B \perp(A B C) \\
A C \subset(A B C) \\
draw B O \perp A C
\end{array}\right\} \Rightarrow E O \perp A C \Rightarrow d(E, A C)=E O=10 \sqrt{17} \mathrm{~cm} \ldots \ldots \ldots . .1.50 \text { points }
$$

b) Figure: $.0.50$ points

i) $\mathrm{d}=6 \sqrt{6} \mathrm{~cm}$ 1 point

ii) d(M, (PLC)) is the height of the tetrahedron MPLC ... $.0.50$ points,

$\mathrm{d}(\mathrm{M},(\mathrm{PLC}))=4 \sqrt{3} \mathrm{~cm}$ $.0.50$ points

iii) In $\triangle O E C \text{, } \mathrm{m}(<\mathrm{E})=90^{\circ},(0.50 \mathrm{p}), \sin (<(\mathrm{OC},(\mathrm{OLE})))=\frac{\sqrt{3}}{3}$.... .. 1 point