# Centroids and Circumcenters

### Problem

Sohail Farhangi
Submitted 9 October, 2015

Given $\triangle ABC$ with centroid $G$ and circumcenter $O$, construct the circumcenters of triangles $AGB, BGC,$ and $CGA$, which we will denote by $O_c, O_a,$ and $O_b$ respectively. Let the centroid of $\triangle O_aO_bO_c$ be $G'$. Show that $G'$ and $O$ coincide.

The proof depends on the following lemma.

### Lemma

Take a point $P$ in $\triangle ABC$ with circumcenter $O$. Let $O_c$ be the circumcenter of $\triangle APB$, and define $O_b$ and $O_a$ similarly. $P$ and $O$ are isogonal conjugates with respect to $\triangle O_aO_bO_c$.

### Proof of Lemma

Let $M_c$ be the midpoint of $AB$, and let $E$ be the midpoint of $BP$.

It can be seen that $O_cO_a \perp BP$ and $O_cO \perp AB$, so it follows that $O_cM_cEB$ is a cyclic quadrilateral, and that $\angle PBA = \angle EBM_c = \angle EO_cM_c = \angle O_aO_cO$. Furthermore, we have that $\displaystyle\angle PBA = \frac{\angle PO_cA}{2} = \angle PO_cO_b = O_aO_cO$. The argument can be repeated for vertices $O_b$ and $O_a$ to attain the desired result.

### Proof of the main result

It is well known that the isogonal conjugate of the centroid, is the Lemoine point, hence using Lemma we only need to show that $G$ is the Lemoine point of $\triangle O_aO_bO_c$.

Since the trilinear coordinates of the Lemoine point are proportional to the sides of the triangle, it only remains to verify that $GD:GE:GF = O_bO_c:O_cO_a:O_aO_b$.

To see that this is so, let $A',$ $B',$ $C'$ be the projections of $G$ on $BC,$ $CA,$ $AB,$ respectively; $D,$ $E,$ $F$ are the projections on $O_bO_c,$ $O_cO_a,$ $O_aO_b.$ For, say $GD:GE,$ we have

\displaystyle\begin{align} \frac{GD}{GE} &= \frac{GA}{GB}\\ &= \frac{GA}{GB}\cdot\frac{\sin(\angle GCB)}{\sin(\angle GCA)}\cdot \frac{\sin(\angle GCA)}{\sin(\angle GCB)}\\ &= \frac{GA}{\sin(\angle GCA)}\cdot \frac{\sin(\angle GCB)}{GB}\cdot \frac{\sin(\angle GCA)}{\sin(\angle GCB)}\\ &= \frac{AC}{\sin(\angle AGC)}\cdot \frac{\sin(\angle BGC)}{BC}\cdot \frac{\sin(\angle GCA)}{\sin(\angle GCB)}\\ &= \frac{AC}{BC}\cdot \frac{GC\cdot \sin(\angle GCA)}{GC\cdot \sin(\angle GCB)}\cdot \frac{\sin(\angle BGC)}{\sin(\angle AGC)}\\ &= \frac{AC}{BC}\cdot \frac{GB'}{GA'}\cdot \frac{\sin(\angle BGC)}{\sin(\angle AGC)}= \frac{[\Delta AGC]}{[\Delta BGC]}\cdot \frac{\sin(\angle BGC)}{\sin(\angle AGC)}\\ &= \frac{\sin(\angle BGC)}{\sin(\angle AGC)}= \frac{\sin(\angle EO_aF)}{\sin(\angle DO_bF)}= \frac{\sin(\angle O_cO_aO_b)}{\sin(\angle O_aO_bO_c)}\\ &= \frac{O_bO_c}{O_cO_a}. \end{align}

Repeating the argument to evaluate $\displaystyle\frac{GE}{GF}=\frac{O_cO_a}{O_aO_b}$ attains the desired result.