Problem 4. Let the non-adjacent angles $A O C$ and $A O B$ be such that $\frac{m(\angle A O B)}{m(\angle A O C)}=1.5$. It is known that the measure of the angle formed by their bisectors is $30^{\circ}$.

a) Determine the measures of the two angles.

b) Show that $(O A$ and $(O B$ are opposite rays.

Nicolae Bivol, Corabia

NOTE. All subjects are mandatory. Time for work: 2 hours and 30 minutes.

## MATHEMATICS OLYMPIAD

## LOCAL STAGE - FEBRUARY 16, 2014  Grade 6

Solutions and grading criteria
## Problem 4.

a) We have: $\frac{m(\angle A O B)}{m(\angle A O C)}=\frac{3}{2} \Rightarrow \frac{m(\angle A O B)}{3}=\frac{m(\angle A O C)}{2}=k \Rightarrow m(\angle A O B)=3k$ and $m(\angle A O C)=2k$, Therefore (OC is in the interior of $\angle A O B)(2 \text{ points})$

If (OM is the bisector of $\angle A O C \Rightarrow m(\angle A O M)=m(\angle M O C)=k$

If (ON is the bisector of $\angle A O B \Rightarrow m(\angle A O N)=m(\angle N O B)=\frac{3k}{2}$

Then $m(\angle M O N)=\frac{3k}{2}-k=\frac{k}{2} \Rightarrow \frac{k}{2}=30^{\circ} \Rightarrow k=60^{\circ}$

We have $m(\angle A O C)=120^{\circ}$ and $m(\angle A O B)=180^{\circ}$ (1 point)

b) From $m(\angle A O B)=180^{\circ} \Rightarrow(O A$ and (OB are opposite rays. (1 point)