3. From point $O$, rays $O A, O B, O C$ and $O D$ were laid out, and $\angle A O B=\angle B O C=\angle C O D=3 \angle A O D$. What can $\angle A O D$ be equal to?
Answer: $36^{\circ}, 45^{\circ}$.

## Solution:

Notice that the angles $A O B, B O C$ and $C O D$ follow each other in one direction (otherwise, some rays would coincide).

Let $\angle A O D=X$. Then $\angle A O B=\angle B O C=\angle \underline{\mathrm{COD}}=3 X$.

If the sum of the angles $A O B, B O C$ and $C O D$ is less than $360^{\circ}$, we get $9 X+X=360^{\circ}$, from which $X=36^{\circ}$. If the sum is greater than $360^{\circ}$, we get $9 X-X=360^{\circ}$, from which $X=45^{\circ}$.

## Criteria:

1 point for each correct answer (without correct justification).

Case when the sum of the angles AOB, BOC, and COD is less than $360^{\circ}$ is explained (justified) - plus 2 points.

Case when the sum of the angles AOB, BOC, and COD is greater than $360^{\circ}$ is explained (justified) - plus 3 points.