3-3. The lateral surface development of a cone represents a sector with an angle of $120^{\circ}$; a triangular pyramid is inscribed in the cone, the angles of the base of which form an arithmetic progression with a difference of $15^{\circ}$. Determine the angle of inclination to the base plane of the smallest of the lateral faces.
Solution 3. Let $l$ be the length of the generatrix of the cone. The length of the circumference of the base is equal to the length of the arc of the development, so the radius of the base circle is $l / 3$. Considering that the sum of the angles of any triangle is $180^{\circ}$, we get that the base of the pyramid is a triangle $A B C$ with angles $45^{\circ}, 60^{\circ}, 75^{\circ}$. Let $B C$ be the shorter side, $O$ be the center of the circumscribed circle of triangle $A B C$. Then $\angle B O C=2 \cdot 45^{\circ}=90^{\circ}$, so $B O C$ is an isosceles right triangle with legs $l / 3$. Therefore, $B C=l \sqrt{2} / 3$. Let $D$ be the midpoint of segment $B C, S-$ be the vertex of the cone. Then $O D=\frac{l}{3 \sqrt{2}}$ and $S D=\sqrt{S B^{2}-D B^{2}}=\sqrt{l^{2}-\frac{l^{2}}{18}}=\frac{l \sqrt{17}}{3 \sqrt{2}}$. If $\varphi$ is the desired angle, then $\cos \varphi=\frac{O D}{S D}=\frac{1}{\sqrt{17}}$.

## Second Round

## Series A