## SUBJECT IV

Let $V A$ be a line perpendicular to the plane of a triangle $A B C$, with $m(\hat{A})=90^{\circ}$, $m(\hat{B})=75^{\circ}$, $\mathrm{BC}=4(\sqrt{3}+1) \mathrm{cm}$, and $V A=25 \% \cdot B C$. Find:

a) The distance from point $V$ to the line $B C$.

b) The distance between the lines $A C$ and $VB$.

(We consider it known that $\sin 15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}$ )

Note: All subjects are mandatory;

Each subject is graded with 7 points;

No points are awarded ex officio;

The actual working time is 3 hours from the moment the subject is received.

## MATHEMATICS OLYMPIAD  LOCAL STAGE
## Problem 4.

| Problem4 | Stage of solving ...................................................................................................................... | Score |
| :---: | :---: | :---: |
| a) | $\mathrm{AT} \perp \mathrm{BC}, \quad \mathrm{AT}=\sqrt{3}+1$ | 2 |
|  | $\mathrm{d}(\mathrm{V}, \mathrm{BC})=\mathrm{VT}=(\sqrt{3}+1) \sqrt{2}$ | 1 |
| b) | $\mathrm{AB}=\mathrm{BC} \sin 15=2 \sqrt{2}$ | 1 |
|  | $\mathrm{AU} \perp \mathrm{VB}$ so $\mathrm{d}(\mathrm{AC}, \mathrm{VB})=\mathrm{AU}$ | 2 |
|  | $\mathrm{AU}=\frac{V A \cdot A B}{V B}$, algebraic calculation | 1 |