## PROBLEM 4

In the plane of square $\mathrm{ABCD}$, a perpendicular $\mathrm{BM}$ is raised with $\mathrm{BM}=2 \sqrt{3} \mathrm{~cm}$. Knowing that triangle $\mathrm{MAC}$ is equilateral, calculate:

a) the side of the square $\mathrm{ABCD}$;

b) the distance between the lines $\mathrm{AC}$ and $\mathrm{MD}$.
## Solution:

a) Let $l$ be the length of the side of the square, $\triangle M A C \equiv \triangle M B C$ (S.S.) $\Rightarrow[M C] \equiv[M A]$. Since $\triangle M A C$ is equilateral $=>\mathrm{AC}=\mathrm{MC}=\mathrm{MA}=\mathrm{l} \sqrt{2}$. 2p

Using the Pythagorean Theorem in $\triangle M B C$ we get $\mathrm{l}=2 \sqrt{3}$ 1p

Therefore, $\mathrm{AB}=2 \sqrt{3} \mathrm{~cm}$ 1p

b) Let $\{\mathrm{O}\}=\mathrm{AC} \cap \mathrm{BD}$. Construct $\mathrm{OP} \perp M D(1)$

From $A C \perp B D, A C \perp M B \Rightarrow A C \perp(M B D) \Rightarrow A C \perp O P(2)$ 1p

From relations (1) and (2) we obtain that OP is the common perpendicular of lines MD and AC, so $\mathrm{d}(\mathrm{MD}, \mathrm{AC})=\mathrm{OP} \Rightarrow$ From $\triangle D P O \sim \triangle D B M \Rightarrow O P=\frac{3 \sqrt{2}}{2}=\mathrm{d}(\mathrm{MD}, \mathrm{AC}) \ldots \ldots \ldots 1 \mathrm{p}$