1.1. The lines containing the bisectors of the exterior angles of a triangle with angles of 42 and 59 degrees intersected pairwise and formed a new triangle. Find the degree measure of its largest angle.
Answer: 69

Solution. Let the vertices of the given triangle be $A, B, C$, and its angles be $\alpha, \beta$, and $\gamma$, respectively. Additionally, let $A_{1}, B_{1}$, and $C_{1}$ be the points of intersection of the external angle bisectors of angles $B$ and $C$, $A$ and $C$, and $A$ and $B$, respectively. Then, $\angle B_{1} A C = \angle C_{1} A B = \left(180^{\circ} - \alpha\right) / 2 = 90^{\circ} - \alpha / 2$. Similarly, $\angle B_{1} C A = 90^{\circ} - \gamma / 2$. From the sum of the angles in triangle $B_{1} A C$, we find $\angle A B_{1} C = 180^{\circ} - \left(90^{\circ} - \alpha / 2\right) - \left(90^{\circ} - \gamma / 2\right) = \alpha / 2 + \gamma / 2$. Therefore, the angles of triangle $A_{1} B_{1} C_{1}$ are the half-sums of the angles of triangle $A B C$. Clearly, to obtain the largest angle of triangle $A_{1} B_{1} C_{1}$, we need to take the half-sum of the two largest angles of triangle $A B C$. The angles of $A B C$ are $42^{\circ}, 59^{\circ}$, and $79^{\circ}$, so the largest half-sum is $69^{\circ}$.

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