$ABCDE$ is a regular pentagon with $\overrightarrow{AB}=a$ and $\overrightarrow{BC}=b$. Find $\overrightarrow{EA}$

Question

$ABCDE$ is a regular pentagon with $\overrightarrow{AB}=a$ and $\overrightarrow{BC}=b$.

Show that $$\overrightarrow{EA}=\frac{\left(\sqrt{5}-1\right)a-2b}{2}$$

I tried using dot product using the angle $36^0$.

Any help appreciated

Answer 1

Observe that $$ \overrightarrow EA+\overrightarrow a+\overrightarrow b=\overrightarrow EC $$ and $\overrightarrow EC$ is parallel to $\overrightarrow a$, $\overrightarrow EC= k\overrightarrow a$ with some positive $k$. We just need to find $k$ which is the ratio of lengths $EC/a$. Call $O$ the center of the pentagon, and draw a line through $O$ perpendicular to $\overrightarrow a$, thus going through $D$ given the symmetry. Call $P$ the point of intersection of this line with $EC$. Then $$ EC=2PC=2OC\sin 72=2OB\sin 72=2\frac{a}{2\sin 36}\sin 72=2a\cos 36 $$ Thus $k=2\cos 36=(1+\sqrt{5})/2$. Plugging in this value for $k$ in the above vector relation you get your result.

The fact that $\cos 36=(1+\sqrt{5})/4$ follows from the fact that $\alpha=36$ satisfies $\sin 2\alpha=\sin 3\alpha$.

Hope it is clear.