## SUBJECT IV (7p)

Given the collinear points $A, O, B$, in this order, and points $C$ and $D$ on opposite sides of line $\mathrm{AB}$, such that $m(\overrightarrow{C O D})=90^{\circ}$ and $m(\overrightarrow{A O C})=4 m(\overrightarrow{A O D})$. If $[O M$ is the bisector of angle $\overparen{B O C}$ and [ON is the bisector of angle $\overparen{B O D}$, the following are required:

a) Determine the measure of angle $\widetilde{D O M}$;

b) Determine the measure of angle $\widehat{M O N}$.

Note:

- All questions are mandatory
- Time allowed: 2 hours


## MATHEMATICS OLYMPIAD  LOCAL STAGE  February 22, 2014

## GRADE 6

## Grading Rubric
## Subject IV

![](https://cdn.mathpix.com/cropped/2024_06_07_3c25989a4f125ee3747ag-3.jpg?height=540&width=884&top_left_y=1392&top_left_x=172)

$$
\begin{aligned}
& m(\widetilde{A O C})=4 m(\widehat{A O D}) \\
& m(\widetilde{C O D})=90^{\circ} \quad \Rightarrow m(\widetilde{A O D})=90^{\circ}: 5=18^{\circ}, m(\widetilde{A O C})=4 m(\widetilde{A O D})=72^{\circ} \quad 3 \text{p} \\
& m(\widetilde{B O C})=180^{\circ}-m(\widetilde{A O C})=108^{\circ} \Rightarrow m(\widetilde{C O M})=54^{\circ} \Rightarrow m(\widetilde{D O M})=90^{\circ}+54^{\circ}=144^{\circ}
\end{aligned}
$$

$$
m(\widetilde{B O D})=180^{\circ}-m(\widetilde{A O D})=162^{\circ} \Rightarrow m(\widetilde{B O N})=81^{\circ} \Rightarrow m(\widetilde{M O N})=54^{\circ}+81^{\circ}=135^{\circ}
$$