11.4. The lengths of the sides of the pentagon $A B C D E$ are equal to 1. Let points $P, Q, R, S$ be the midpoints of sides $A B, B C, C D, D E$ respectively, and points $K$ and $L$ be the midpoints of segments $P R$ and $Q S$ respectively. Find the length of segment $K L$.
Answer: $\frac{1}{4}$

Solution 1. Under $X Y$ in this solution, we will understand the vector with the origin at point $X$ and the end at point $Y$. Then $A K=\frac{1}{2}(A P+A R)=\frac{1}{2}\left(\frac{1}{2} A B+A B+B C+\frac{1}{2} C D\right)=\frac{3}{4} A B+\frac{1}{2} B C+\frac{1}{4} C D$. Similarly, $A L=\frac{1}{2}(A Q+A S)=\frac{1}{2}\left(A B+\frac{1}{2} B C+A B+B C+C D+\frac{1}{2} D E\right)=A B+\frac{3}{4} B C+\frac{1}{2} C D+\frac{1}{4} D E$. Then $K L=A L-A K=\frac{1}{4}(A B+B C+C D+D E)=\frac{1}{4} A E$. Therefore, the segment $K L$ is parallel to the segment $A E$ and its length is $\frac{1}{4}$ of the length of $A E$.

Solution 2. Mark the point $M$, which is the midpoint of the diagonal $B E$. Then the points $Q, R, S$, and $M$ are the midpoints of the sides of the quadrilateral $B C D E$ and, by a well-known fact, are the vertices of a parallelogram (Varignon's theorem for $B C D E$). The point $L$ is the midpoint of its diagonal $Q S$, and therefore also the midpoint of its diagonal $R M$. Consequently, the segment $K L$ is the midline in the triangle $P R M$. On the other hand, the segment $R M$ is the midline in the triangle $A B E$. Therefore, the length of $K L$ is half the length of $R M$, and a quarter of the length of $A E$. Note that the convexity of the pentagon and the lengths of its other sides are almost irrelevant here.

Evaluation criteria. The idea of the parallelogram $Q R S M: 3$ points.

Noted that the segment $K L$ is the midline in the triangle $P R M: 3$ points.

Noted that the segment $R M$ is the midline in the triangle $A B E: 1$ point.