11.1. Inside a convex pentagon, a point is marked and connected to all vertices. What is the maximum number of the ten segments drawn (five sides and five segments connecting the marked point to the vertices of the pentagon) that can have a length of 1? (A. Kuznetsov)
Answer: 9 segments.

Solution. First, we prove that all 10 segments cannot have a length of 1. Assume the opposite. Let $A B C D E$ be a pentagon, $O$ be a point inside it, and all 10 drawn segments have a length of 1 (see Fig. 6). Then triangles $O A B, O B C$, $O C D, O D E$, and $O E A$ are equilateral, so $\angle A O B = \angle B O C = \angle C O D = \angle D O E = \angle A O E = 60^{\circ}$. The sum of these angles should be $360^{\circ}$, but $5 \cdot 60^{\circ} = 300^{\circ}$, which is a contradiction.

![](https://cdn.mathpix.com/cropped/2024_05_06_5e804b2a1d2ef00abb53g-3.jpg?height=365&width=371&top_left_y=827&top_left_x=307)

Fig. 6

![](https://cdn.mathpix.com/cropped/2024_05_06_5e804b2a1d2ef00abb53g-3.jpg?height=345&width=380&top_left_y=837&top_left_x=803)

Fig. 7

It remains to provide an example where 9 segments have a length of 1 (see Fig. 7). Mark points $A$ and $O$ on the plane at a distance of 1. Choose points $B, C, D$, and $E$ sequentially so that triangles $A O B, B O C, C O D$, and $D O E$ are equilateral. Then point $O$ lies inside the pentagon $A B C D E$, and out of the 10 drawn segments, all except $A E$ have a length of 1.

Remark. In the given example, points $A, B, C, D$, and $E$ are five vertices of a regular hexagon with side length 1.

Comment. Only the proof that there are no more than 9 segments of unit length - 3 points.

Only the example showing that 9 segments of unit length can be found - 3 points.