AoPS - Problem

Source: Romanian National Olympiad 2024 - Grade 10 - Problem 2

Tags: geometry, circumcircle, pentagon, inscriptible, Centroid

Filipjack

06.04.2024 12:49

We consider the inscriptible pentagon $ABCDE$ in which $AB=BC=CD$ and the centroid of the pentagon coincides with the circumcenter. Prove that the pentagon $ABCDE$ is regular. The centroid of a pentagon is the point in the plane of the pentagon whose position vector is equal to the average of the position vectors of the vertices.

Pintea_Andrei wrote: Bump this! Sketch of my in contest solution: Consider the complex plane where $(ABCDE)$ is unit circle and $A(1)$. Let $B(z)$ so because $AB=BC=CD$ it follows that $C(z^2), D(z^3)$. Let $E(\epsilon)$. Because centroid is circumcircle we get $1+z+z^2+z^3+\epsilon=0$. By taking the conjugate we get $1+z+z^2+z^3+z^3/\epsilon=0$, so $\epsilon^2=z^3$. So if $z=e^{2\theta i}$ then $\epsilon = -w^3$ where $w=e^{i\theta}$. Hence $\frac{w^8-1}{w^2-1}=w^3$, which implies $1-w^3=w^8-w^5=w^5(w^3-1)$, so $w^5=-1$. This implies $z^5=1$ and also $\epsilon = w^8=z^4$, hence $ABCDE$ regular.