3. (42nd IM() problem) In $\triangle A B C$, $A P$ bisects $\angle B A C$, intersecting $B C$ at $P$, and $B Q$ bisects $\angle A B C$, intersecting $C A$ at $Q$. It is known that $\angle B A C=60^{\circ}$, and $A B+B P=A Q+B Q$. What are the possible values of the angles of $\triangle A B C$?
3. Solution 1 Extend $A B$ to $B^{\prime}$ such that $B B^{\prime}=B P$, and take $C^{\prime}$ on the ray $Q C$ such that $Q C^{\prime}=Q B$, as shown in Figure 3 or Figure 4. If $C$ and $C^{\prime}$ do not coincide, then connect $B^{\prime} C^{\prime}$, $B^{\prime} P$, $C^{\prime} P$, and $B C^{\prime}$.
Since $A B + B P = A Q + Q B$,
then $A B + B B^{\prime} = A Q + Q C^{\{\prime}$, i.e., $A B^{\prime} = A C^{\prime}$.
Also, $\angle B^{\prime} A C^{\prime} = 60^{\circ}$, so $\triangle A B^{\prime} C^{\prime}$ is an equilateral triangle.
Since $A P$ bisects $\angle B^{\prime} A C^{\prime}$, then $A P$ is the axis of symmetry of $\triangle A B^{\prime} C^{\prime}$.
Thus, $\angle A B^{\prime} P = \angle A C^{\prime} P$, and $B^{\prime} P = C^{\prime} P$.
Also, $\angle A B C = \angle A B Q + \angle C B Q = \angle B B^{\prime} P + \angle B P B^{\prime}$, $\angle A B Q = \angle C B Q$, $\angle B B^{\prime} P = \angle B P B^{\prime}$, so $\angle Q B C = \angle B B^{\prime} P = \angle Q C^{\prime} P = \alpha$.
Since $B Q = Q C^{\prime}$, then $\angle Q B C^{\prime} = \angle Q C^{\prime} B = \beta$. Thus, $\angle P B C^{\prime} = \angle P C^{\prime} B = \alpha - \beta$ or $\beta - \alpha$.
Therefore, $B P = C^{\prime} P = B^{\prime} P = B B^{\prime}$.
Then $\angle B B^{\prime} P = 60^{\circ}$, $\angle A B C = 120^{\circ}$.
Also, $\angle B A C = 60^{\circ}$, which is a contradiction, so $C$ and $C^{\prime}$ coincide.
Similarly, we can get $B^{\prime} P = C P$, as shown in Figure 5.
Then $\angle P B^{\prime} C = \angle P C B^{\prime}$.
Since $\angle B P B^{\prime} = \angle B B^{\prime} P = \angle P B^{\prime} C + \angle P C B^{\prime} = \gamma$,
then $\angle P B^{\prime} C = \frac{\gamma}{2}$,
$\gamma + \frac{\gamma}{2} = 60^{\circ}$, $\gamma = 40^{\circ}$,
$\angle A B C = 2 \gamma = 80^{\circ}$.
Solution 2 In $\triangle A B Q$, by the Law of Sines,
$\frac{A Q}{\sin \frac{B}{2}} = \frac{A B}{\sin \left(\frac{B}{2} + 60^{\circ}\right)} = \frac{B Q}{\sin 60^{\circ}}$.
Thus, $A Q = \frac{A B \cdot \sin \frac{B}{2}}{\sin \left(\frac{B}{2} + 60^{\circ}\right)}$, $B Q = \frac{A B \cdot \sin 60^{\circ}}{\sin \left(\frac{B}{2} + 60^{\circ}\right)}$.
Similarly, $B P = \frac{A B \cdot \sin 30^{\circ}}{\sin \left(30^{\circ} + B\right)}$.
Since $A B + B P = A Q + B Q$, $\angle B + \angle C = 120^{\circ}$, and $0^{\circ} < \angle B < 120^{\circ}$,
then $1 + \frac{\sin 30^{\circ}}{\sin \left(B + 30^{\circ}\right)} = \frac{\sin \frac{B}{2}}{\sin \left(\frac{B}{2} + 60^{\circ}\right)} + \frac{\sin 60^{\circ}}{\sin \left(\frac{B}{2} + 60^{\circ}\right)}$.
Thus, $1 + \frac{\sin 30^{\circ}}{\sin \left(B + 30^{\circ}\right)} = \frac{\cos \left(\frac{B}{4} - 30^{\circ}\right)}{\cos \left(\frac{B}{4} + 30^{\circ}\right)}$,
then $\frac{\sin 30^{\circ}}{\sin \left(B + 30^{\circ}\right)} = \frac{\sin \frac{B}{4}}{\cos \left(\frac{B}{4} + 30^{\circ}\right)}$.
Thus, $\sin 30^{\circ} \cdot \cos \left(\frac{B}{4} + 30^{\circ}\right) = \sin \frac{B}{4} \cdot \sin \left(B + 30^{\circ}\right) = \frac{1}{2} \left[\cos \left(\frac{3 B}{4} + 30^{\circ}\right) - \cos \left(\frac{5 B}{4} + 30^{\circ}\right)\right]$.
Then $\cos \left(\frac{B}{4} + 30^{\circ}\right) + \cos \left(\frac{5 B}{4} + 30^{\circ}\right) = \cos \left(\frac{3 B}{4} + 30^{\circ}\right)$.
Thus, $\cos \left(\frac{3 B}{4} + 30^{\circ}\right) \cos \frac{B}{2} = \cos \left(\frac{3 B}{4} + 30^{\circ}\right)$.
When $\cos \left(\frac{3 B}{4} + 30^{\circ}\right) = 0$, equation (1) holds. Then $\frac{3 \angle B}{4} + 30^{\circ} = 90^{\circ}$.
Thus, $\angle B = 80^{\circ}$.
When $\cos \left(\frac{3 B}{4} + 30^{\circ}\right) \neq 0$, from equation (1), we know $\cos \frac{B}{2} = \frac{1}{2}$. Then $\angle B = 120^{\circ}$, which contradicts the problem statement.
In summary, the internal angles of $\triangle A B C$ are
$\angle B A C = 60^{\circ}$, $\angle A B C = 80^{\circ}$, $\angle A C B = 40^{\circ}$.