143. Three consecutive sides of a flat convex pentagon are equal to 1, 2, and $a$. Find the remaining two sides of this pentagon, given that it is an orthogonal projection onto a plane of a regular pentagon. For what values of $a$ does the problem have a solution?
143. Let the pentagon $A B C D E$ be the projection of a regular pentagon,

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-084.jpg?height=249&width=409&top_left_y=1146&top_left_x=681)

Fig. 26. where $|A B|=1,|B C|=2,|C D|=$ $=a, A B C D$ is a trapezoid in which $\frac{|A D|}{|B C|}=\lambda=\frac{1+\sqrt{5}}{2}, F$ is the intersection point of its diagonals, and $A F D E$ is a parallelogram. Draw $C K$ parallel to $A B$ (Fig. 26). In $\triangle C K D$, we have $|C K|=1$, $|K D|=2(\lambda-1),|C D|=a$. Let $\widehat{C D K}=\varphi$. Write the law of cosines for triangles $C K D$ and $A C D$:

$$
\begin{aligned}
1= & a^{2}+4(\lambda-1)^{2}-4(\lambda-1) a \cos \varphi \\
& |A C|^{2}=a^{2}+4 \lambda^{2}-4 a \lambda \cos \varphi
\end{aligned}
$$

From these two equations, we get

$$
\begin{gathered}
|A \epsilon|=\sqrt{\frac{4 \lambda^{2}-3 \lambda-a^{2}}{\lambda-1}} \\
|E D|=|A F|=\frac{\lambda}{\lambda+1} \sqrt{\frac{4 \lambda^{2}-3 \lambda-a^{2}}{\lambda-1}}
\end{gathered}
$$

Similarly, we find

$$
|A E|=|F D|=\frac{\lambda}{\lambda+1} \sqrt{\frac{a^{2} \lambda-1+4 \lambda^{2}-4 \lambda}{\lambda-1}}
$$

Answer. The other two sides are equal to:

$$
\frac{\sqrt{5}-1}{4} \sqrt{14+10 \sqrt{5}-2(\sqrt{5}+1) a^{2}}
$$

and

$$
\frac{\sqrt{5}-1}{4} \sqrt{a^{2}(6+2 \sqrt{5})+6(\sqrt{5}+1)} .
$$

The problem has a solution for $\sqrt{5}-2<a<\sqrt{5}$.