# An ellipse determined by circumcenters

Can you prove or disprove the following claim:

Claim. A convex hexagon $$ABCDEF$$ is circumscribed about an ellipse. Let $$G$$ be the point of concurrency of hexagon's principal diagonals , and let the points $$O_1$$ , $$O_2$$ , $$O_3$$ , $$O_4$$ , $$O_5$$ , $$O_6$$ be the circumcenters of $$\triangle ABG$$ , $$\triangle BCG$$ , $$\triangle CDG$$ , $$\triangle DEG$$ , $$\triangle EFG$$ and $$\triangle FAG$$ , respectively . Then, the points $$O_1$$ , $$O_2$$ , $$O_3$$ , $$O_4$$ , $$O_5$$ , $$O_6$$ lie on a new common ellipse. GeoGebra applet that demonstrates this claim can be found here.

Consider a hexagon $$O_1O_2O_3O_4O_5O_6$$ . My idea is to apply Braikenridge–Maclaurin theorem on it in order to prove that points $$O_1$$ , $$O_2$$ , $$O_3$$ , $$O_4$$ , $$O_5$$ , $$O_6$$ are conelliptic. But, to do that I need to show that three intersection points of the three pairs of lines through opposite sides of a hexagon $$O_1O_2O_3O_4O_5O_6$$ lie on a line l. How this can be done?

• Because the circumcenter lies on the perpendicular bisector of the sides, $O_{1}O_{6}\vert\vert O_{4}O_{3}$, etc. (The opposite sides of the hexagon are parallel). Maybe the Braikenridge-Maclaurin theorem can be applied here with the "line at infinity", but the Wikipedia article didn't say more about that. Perhaps another theorem can be applied now that this property is known? Jan 18, 2021 at 6:11
• A side remark: the (non evident) fact that the 3 main diagonals of a circuscribed haxagon to a conic are concurrent is a consequence of Brianchon's theorem. Jan 18, 2021 at 16:50

Since the circumcenter lies on the perpendicular bisector of the sides of each triangle, $$O_{1}O_{6}\perp AD$$ and $$O_{3}O_{4}\perp AD$$. This means that $$O_{1}O_{6}\ \vert\vert\ O_{3}O_{4}$$, and similarly for the other pairs of opposite sides. Thus, the hexagon is a parallelogon, which means that the lines through its opposite vertices concur and furthermore the hexagon is centrally symmetric through this point of concurrency (moved to the origin for then next part.)
Consider the parallelogram formed by $$2$$ pairs of opposite points, and note that any conic through them will have a center at the origin, which means that that conic will also be centrally symmetric. Next, we add in a $$5$$th point. Since a conic can go through any $$5$$ points, we can construct a conic through these $$5$$ points. This conic must still be centrally symmetric, which means that the $$6$$th point must also lie on this conic.
It remains to prove that this conic is an ellipse, which is trivial since each of the interior angles of the hexagon are between $$0^{\circ}$$ and $$180^{\circ}$$.
Thus, $$\boxed{O_{1}O_{2}O_{3}O_{4}O_{5}O_{6}\text{ has a circumscribed ellipse.}}$$