5. The diagonals of a regular pentagon with a side length of 1 form a new, smaller regular pentagon. What is the length of the side of the smaller pentagon?

Solve the problems independently. You have 150 minutes to solve them.

The use of notes, literature, or a pocket calculator is not allowed.

## Problems for 4th Grade
III/5. Let the larger pentagon be denoted by $A B C D E$, and the smaller one by $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}$, where $A^{\prime}$ is the vertex farthest from $A$. Let $\left|A^{\prime} B^{\prime}\right|=a$ and $\left|B^{\prime} D\right|=b$. Triangles $A B D^{\prime}$ and $E C D$ are similar, as they are isosceles with angles of $36^{\circ}$ at the base and $108^{\circ}$ at the vertex. This gives the ratio $\frac{b}{1}=\frac{1}{2 b+a}$. We observe that triangle $C D B^{\prime}$ is isosceles with the vertex at $C$, so $a+b=1$. This results in the quadratic equation $b^{2}+b-1=0$, which has the only positive solution $b=\frac{-1+\sqrt{5}}{2}$, and from this, $a=1-b=\frac{3-\sqrt{5}}{2}$.

Similarity $\triangle A B D^{\prime} \sim \triangle E C D: 1$ point.

Relation $\frac{b}{1}=\frac{1}{2 b+a}: 2$ points. Relation $a+b=1: 2$ points.

![](https://cdn.mathpix.com/cropped/2024_06_07_af92d6d33b11d7748e87g-10.jpg?height=563&width=597&top_left_y=1089&top_left_x=1232)

Equation $b^{2}+b-1=0$: $\mathbf{1}$ point. Solution $a=\frac{3-\sqrt{5}}{2}: \mathbf{1}$ point.

If the contestant introduces different labels and writes the correct system of 2 equations with 2 unknowns: 5 points. If the contestant simplifies such a system: 1 point.