74th Romanian Mathematical Olympiad

We consider an orthonormal coordinate system centred in $O$, where $O$ is the centre of the circumcircle $C$ of the pentagon $ABCDE$, with the unit length equal to the radius of the circle $C$ and with the real axis being the perpendicular bisector of the segment $BC$. Let $z_X$ denote the complex number representing the position of point $X$ in this coordinate system.

Since $AB=BC=CD$, we have $\angle AOB=\angle BOC=\angle COD=2\alpha \in (0,\frac{2\pi }{3})$. It follows that the positions of the vertices of the pentagon are $z_A=\cos 3\alpha +i\sin 3\alpha$, $z_B=\cos \alpha +i\sin \alpha$, $z_C=\cos \alpha -i\sin \alpha$, $z_D=\cos 3\alpha -i\sin 3\alpha$, and $z_E=\cos \beta +i\sin \beta$, with $\beta \in (3\alpha ,2\pi -3\alpha )$.

If the centroid of the pentagon coincides with $O$, then $z_A+z_B+z_C+z_D+z_E=0$, from which we obtain $\cos \beta +2\cos 3\alpha +2\cos \alpha =0$ and $\sin \beta =0$. We get $\beta =\pi$, along with $-\sin \alpha +2\sin \alpha \cos 3\alpha +2\sin \alpha \cos \alpha =0$ (by multiplying by $\sin \alpha$) and, by transforming sums into products, $\sin 4\alpha =\sin \alpha$.

Since $0<\alpha <4\alpha <\frac{4\pi }{3}$, we deduce $4\alpha =\pi -\alpha$, so $\alpha =\frac{\pi }{5}$, from which $\angle AOB=\angle BOC=\angle COD=\angle DOE=\angle EOA=\frac{2\pi }{5}$, meaning $ABCDE$ is a regular pentagon.

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Alternative solution.

We consider an orthonormal coordinate system centred in $O$, where $O$ is the centre of the circumcircle $C$ of the pentagon $ABCDE$, with the unit length equal to the radius of the circle $C$ and with the real axis being the line $OA$. Let $z_X$ denote the complex number representing the position of the point $X$ in this coordinate system.

Since $AB=BC=CD$, we have $\angle AOB=\angle BOC=\angle COD$. It follows that the positions of the vertices of the pentagon are $z_A=1$, $z_B=z$, $z_C=z^2$, $z_D=z^3$, and $z_E=w$, with $z=\cos \alpha +i\sin \alpha$, $\alpha \in (0,\frac{2\pi }{3})$, and $w=\cos \beta +i\sin \beta$, $\beta \in (3\alpha ,2\pi )$.

If the centroid of the pentagon coincides with $O$, then $z_A+z_B+z_C+z_D+z_E=0$, from which we obtain $1+z+z^2+z^3+w=0$. We get $w=\frac{1-z^4}{z-1}$, from which $\overline{w}=\frac{1-z^4}{z-1}$, and using $w\overline{w}=z\overline{z}=1$, we deduce the relation $\frac{1-z^4}{z-1}\cdot \frac{z^4-1}{z^3(1-z)}=1$.

Performing the calculations, we obtain $(z^5-1)(z^3-1)=0$. The only solution of this equation with an argument in $(0,\frac{2\pi }{3})$ is $\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}$, so $\angle AOB=\angle BOC=\angle COD=\angle DOE=\angle EOA=\frac{2\pi }{5}$, meaning $ABCDE$ is a regular pentagon.