Problem 4. Consider the rhombus $ABCD$ where $AB=6 \, \text{cm}$ and $m(\angle BAD)=60^\circ$. From the same side of the plane $(ABC)$, perpendiculars $AM$ and $CQ$ are raised to the plane $(ABC)$ such that $AM=9 \, \text{cm}$ and $CQ=3 \, \text{cm}$.

a) Prove that the planes $(MBD)$ and $(QBD)$ are perpendicular.

b) Calculate the distance between the lines $BD$ and $MQ$.

c) Determine the cosine of the angle formed by the planes $(MBQ)$ and $(ABC)$.

Dorin Popa, Slatina
## Solution and marking scheme

a) $m(\Varangle M O A)=60^{\circ}, m(\Varangle Q O C)=30^{\circ}$ ..... (1p)
The measure of the dihedral angle is $m(\Varangle M O Q)=90^{\circ}$, thus $(M B D) \perp(Q B D)$ ..... (1p)
b) Let $\{O\}=A C \cap B D$; by constructing $O P \perp M Q$ it is shown that $d(B D, M Q)=O P$ ..... (1p)
$O P=3 \sqrt{3} \text{ cm}$ ..... (1p)
c) The projection of triangle $M B Q$ onto the plane $A B C$ is triangle $A B C$ ..... (1p)
The area of triangle $A B C$ is $9 \sqrt{3} \text{ cm}^2$, and the area of triangle $B M Q$ is $36 \text{ cm}^2$ ..... (1p)
Let $\alpha=(M B Q),(A B C)$, then $S_{A B C}=S_{M B Q} \cdot \cos \alpha$, from which $\cos \alpha=\frac{\sqrt{3}}{4}$ ..... (1p)