3. [5] Pentagon $A B C D E$ is circumscribed around a circle. The angles at its vertices $A, C$ and $E$ are $100^{\circ}$. Find the angle $A C E$.
Answer: $40^{\circ}$. It is not hard to understand that such a pentagon exists.

Solution 1. The lines connecting the vertices with the center $O$ of the inscribed circle $\omega$ are the angle bisectors of the pentagon. Therefore, $\angle O A E=\angle O E A=50^{\circ}, \angle A O E=80^{\circ}$. Let $\omega$ touch the sides $B C$ and $A E$ at points $K$ and $M$ respectively. Then $\angle O C K=50^{\circ}$, and the right triangles $O M A, O K C$ and $O M E$ are equal by the leg and the opposite angle. Thus, $O A=O C=O E$, i.e., points $A, C$ and $E$ lie on a circle with center $O$. Therefore, $\angle A C E=\frac{1}{2} \angle A O E=40^{\circ}$. Solution 2. Since the angles $A, C, E$ are equal, all tangents to the circle from these vertices are equal. Since the tangents from vertex $B$ are also equal, triangle $A B C$ is isosceles, and triangle $C D E$ is also (analogously). The sum of angles $B$ and $D$ is $540^{\circ}-3 \cdot 100^{\circ}=240^{\circ}$, so $\angle A C B+\angle E C D=\left(2 \cdot 180^{\circ}-240^{\circ}\right): 2=$ $=60^{\circ}, \mathrm{a} \angle A C E=100^{\circ}-60^{\circ}=40^{\circ}$.

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