4. The side $\overline{B C}=a$ of the regular pentagon $A B C D E$ is extended through vertex $C$ to point $F$ such that $\overline{C F}=\frac{a(1+\sqrt{5})}{2}$. Express the length of the segment $A F$ in terms of the side length $a$.
Solution. Let $d=\overline{C F}=\frac{a(1+\sqrt{5})}{2}$, the intersection of $A F$ and $C D$ at $G$, $\overline{C G}=x$, and $\overline{A D}=e$. Then $\overline{G D}=a-x$. From $\triangle D A G \sim \triangle C F G$ it follows that $e:(a-x)=d: x$, so

$$
x=\frac{a d}{e+d} .
$$

Let $H$ be a point on the diagonal $A D$ such that $\overline{D H}=a$. Then $\overline{A H}=e-a$. Further, from $\triangle A D E \sim \triangle E A H$ it follows that $e: a=a:(e-a)$, so

![](https://cdn.mathpix.com/cropped/2024_06_05_45d932ac9ba548f86db1g-03.jpg?height=450&width=490&top_left_y=777&top_left_x=990)

$$
e(e-a)=a^{2}
$$

We will prove that the segment $A F$ bisects the side $C D$, for which it is sufficient to prove that $e=d$. From $d=\frac{a(1+\sqrt{5})}{2}$ it follows that

$$
d(d-a)=\frac{a(1+\sqrt{5})}{2}\left(\frac{a(1+\sqrt{5})}{2}-a\right)=\frac{a(1+\sqrt{5})}{2} \frac{a(\sqrt{5}-1)}{2}=a^{2} \frac{\sqrt{5}^{2}-1}{4}=a^{2}
$$

Now, from the last equality and (2) it follows that $e(e-a)=d(d-a)$, from which we get $(d-e)(d+e-a)=0$. Since $d>a$ and $e>a$, from the last equality it follows that $d=e$. Now, from $d=e$ and the similarity $\triangle D A G \sim \triangle C F G$ it follows that $\triangle D A G \cong \triangle C F G$. From this, we conclude that the segment $A F$ bisects the side $C D$. Furthermore, since $A F$ bisects $C D$ and $\triangle D G A \cong \triangle C G A\left(\overline{D A}=\overline{C A}=e, \measuredangle A D G=\measuredangle A C G=72^{\circ}\right.$ and $\overline{D G}=\overline{C G}=\frac{a}{2}$ ), it follows that $A F \perp C D$. From the Pythagorean theorem applied to triangle $A C G$ it follows that

$$
\overline{A F}=\overline{A G}=\sqrt{\overline{A D}^{2}-\overline{D G}^{2}}=\sqrt{\frac{a^{2}(1+\sqrt{5})^{2}}{4}-\frac{a^{2}}{4}}=\frac{a}{2} \sqrt{5+2 \sqrt{5}}
$$

Finally, from $\triangle D A G \cong \triangle C F G$ it follows that $\overline{A G}=\overline{G F}$, so

$$
\overline{A F}=2 \overline{A G}=a \sqrt{5+2 \sqrt{5}}
$$

## II year