Problem 1. The external angles of a triangle are $20\%, 35\%$ and $45\%$ of the sum of its external angles. Determine the angle between the bisector of the smallest angle and the shortest side.
Solution. Let $\alpha, \beta$ and $\gamma$ be the angles in the triangle. Then the exterior angles are $180^{\circ}-\alpha$,

![](https://cdn.mathpix.com/cropped/2024_06_05_1cb7beb611b796adc950g-08.jpg?height=348&width=397&top_left_y=1249&top_left_x=1085)
$180^{\circ}-\beta$ and $180^{\circ}-\gamma$, and their sum is

$$
\left(180^{\circ}-\alpha\right)+\left(180^{\circ}-\beta\right)+\left(180^{\circ}-\gamma\right)=540^{\circ}-(\alpha+\beta+\gamma)=540^{\circ}-180^{\circ}=360^{\circ}
$$

However, the exterior angles of the triangle are $72^{\circ}, 126^{\circ}$ and $162^{\circ}$. Therefore, its interior angles are $108^{\circ}, 54^{\circ}$ and $18^{\circ}$.

Let $ABC$ be a triangle with the calculated angles. Without loss of generality, we can assume that $\measuredangle ABC=108^{\circ}$ and $\measuredangle ACB=18^{\circ}$. If $CD$ is the angle bisector of the smallest angle ( $D \in AB$ ), then $\measuredangle DCB=9^{\circ}$ and

$$
\measuredangle CDB=180^{\circ}-\left(9^{\circ}+108^{\circ}\right)=180^{\circ}-127^{\circ}=53^{\circ}
$$