Sorry for the question, but what's the Olympic Revenge?

Orestis_Lignos wrote: Sorry for the question, but what's the Olympic Revenge? It is a brazilian olympiad made by the olympic students for the teachers.

Ok thanks pablock. When you say teachers, you mean Brazilian ones, right?

Orestis_Lignos wrote: Ok thanks pablock. When you say teachers, you mean Brazilian ones, right? Yes, the brazilian teachers.

Let $T=BB_0\cap CC_0$ and $S$ the center of the circumcircle of $BAC$. It is well known that $E$ is actually the $A$-Humpty point, so for the sake of the proof, let us renote $E=H_A$. Obviously, $\angle B_0H_AC_0=180-\angle A$, and $\angle BH_AC=180-\angle A$. Now $\angle C_0B_0H_A=AB_0H_A-AB_0C_0=AH_AB_0-(90-A)=A-\angle H_AAC=\angle H_AAB=\angle H_ABC$. This gives that $H_AB_0C_0$ and $H_ABC$ are similar, so by spiral similarity $T$ will be the intersection of the circle $(BH_AC)$ and $B_0H_AC_0$. Since their centers are $S$, respectively $A$, we have $AT=AH_A$ and $ST=SH_A$. Let us suppose that $A-T-D$ are collinear. We have that $AS$ is the angle bisector of $TAH_A$. Since $AT$ is the symmedian and $AH_A$ is the median, they are isogonal so $AS$ will be the angle bisector of $\angle BAC$. As $S$ is on the perpendicular bisector of $BC$, we have that $S$, the center of the circumcircle of $BHH_AC$, is the midpoint of the arc $BC$ of the circumcircle of $BAC$. As $\angle BSC=2A$ and $\angle BSC+A=180$, we have that $\angle A=60$. If $A=60$, similar angle chasing gives us that $DB_0C_0$ is equilateral, so $AD,BB_0,CC_0$ will be the altitudes of this equilateral triangle so indeed they are concurrent. Done!

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Post a diagram should you have the time. Thanks

Overkill In a \(60^{\circ}\) triangle, the Dumpty point coincides with the First Fermat point and since neuberg is isogonal, so it also passes through the \(HM -\) point i.e, \(E\). \(\boxed{Q.E.D}\)