Task A-1.4.

A regular nonagon with side length $a$ is given. What is the difference in length between its longest and shortest diagonals?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_a7570febbe914cc2404eg-08.jpg?height=692&width=679&top_left_y=432&top_left_x=577)

Without loss of generality, we consider the lengths from vertex $A$. The length of the longest diagonal is $|A E|$, and the shortest is $|A C|$, and we need to find $|A E| - |A C|$.

Notice that triangles $A D G$ and $B E H$ are equilateral. Therefore, and due to the equality of angles subtended by the same arc of the circumscribed circle of the nonagon, we have:

$$
\varangle A E D = \varangle A G D = 60^{\circ} \quad \text{and} \quad \varangle E A B = \varangle E H B = 60^{\circ}
$$

Therefore, quadrilateral $A B D E$ is an isosceles trapezoid.

Draw a line through vertex $B$ parallel to $\overline{D E}$. Let $T$ be its intersection with $\overline{A E}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_a7570febbe914cc2404eg-08.jpg?height=264&width=598&top_left_y=1737&top_left_x=620)

Quadrilateral $T E D B$ is a parallelogram, and we have

$$
|A E| - |A C| = |A E| - |B D| = |A E| - |T E| = |A T|
$$

Notice that triangle $A B T$ is equilateral, so we have:

$$
|A E| - |A C| = |A T| = |A B| = a
$$