## Task A-1.7.

Let $A B C$ be a triangle in which the longest side is $\overline{B C}$, and the angle $\varangle B C A$ is three times the angle $\varangle A B C$. The external angle bisector at vertex $A$ intersects the line $B C$ at point $A_{0}$, and the external angle bisector at vertex $B$ intersects the line $A C$ at point $B_{0}$. If $\left|A A_{0}\right|=\left|B B_{0}\right|$, determine the angles of the given triangle.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_acf83088931d085246bdg-06.jpg?height=659&width=1288&top_left_y=590&top_left_x=310)

Let $\varangle A B C=\beta$. Then $\varangle B C A=3 \beta$.

The angle $\varangle B A B_{0}$ is the exterior angle at vertex $A$, so $\varangle B A B_{0}=4 \beta$. Therefore, $\varangle A_{0} A C=2 \beta$ (the line $A A_{0}$ is the angle bisector of $\varangle B A B_{0}$).

The angle $\varangle B C A$ is the exterior angle of triangle $A C A_{0}$, so we have

$$
\varangle A A_{0} C=\varangle B C A-\varangle A_{0} A C=3 \beta-2 \beta=\beta.
$$

We observe that triangle $A B A_{0}$ is isosceles because $\varangle A A_{0} B=\varangle A B A_{0}=\beta$. 1 point

Therefore, $|A B|=\left|A A_{0}\right|$.

According to the problem, $\left|A A_{0}\right|=\left|B B_{0}\right|$, so it follows that $|A B|=\left|B B_{0}\right|$. 1 point

Therefore, triangle $A B B_{0}$ is isosceles and $\varangle B A B_{0}=\varangle B B_{0} A$.

Since $\varangle B A B_{0}=4 \beta$, the third angle of this triangle is $\varangle A B B_{0}=180^{\circ}-2 \varangle B A B_{0}=180^{\circ}-8 \beta$. But $\varangle A B B_{0}$ is also half of the exterior angle at vertex $B$, so $\varangle A B B_{0}=90^{\circ}-\frac{\beta}{2}$.

Thus, we have

$$
180^{\circ}-8 \beta=90^{\circ}-\frac{\beta}{2}
$$

from which we get $\beta=12^{\circ}$. 1 point

The angles of the given triangle are $\varangle A B C=12^{\circ}, \varangle B C A=36^{\circ} \text{ and } \varangle C A B=132^{\circ}$.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

2nd grade - high school - A variant
January 3, 2014.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.