BdMO Online Forum

Let $\triangle ABC$ be a triangle and let $A_1, B_1$, and $C_1$ be points of contact of the excircles with the sides $BC, AC$, and $AB$, respectively. Prove that if the circumcenter of $\triangle A_1B_1C_1$ lies on the circumcircle of $\triangle ABC$, then $\triangle ABC$ is a right triangle.

I have used traditional notations here.
Assume that $O\in \widehat{BAC}$. Then $\angle B_{1}A_{1}C_{1}$ is obtuse. Let us assume that $P$ is the midpoint of $\widehat{BAC}$. Then, $\angle C_{1}BP=\angle ABP=\angle ACP=\angle B_{1}CP$ and $PB=PC$.
Also,$BC_{1}=s-a=B_{1}C$. Thus $\triangle PBC_{1}\cong \triangle PCB_{1}$. So, $PC_{1}=PB_{1}$,that is $P$ is on the perpendicular bisector of $B_{1}C_{1}$. Again $P\in \omega$ where $\omega=\bigodot A_{1}B_{1}C_{1}$.
Thus $P\equiv O$.
Now Let $\omega \cap C_{1}A$ (extended)$=K$. Now simple calculations with angles implies that $AK=AB_{1}=s-c$.
Using power of point the power of $B$ WRT $\omega$ is $BC_{1}\times BK=OB^{2}-OC_{1}^{2}\Rightarrow (s-a)s=OB^{2}-OC_{1}^{2}$
Next use Stewart's theorem in $\triangle OBC$ for cevian $OA_{1}$ to get $OB^{2}-OA_{1}^{2}=(s-b)(s-c)$.
Thus, $s(s-a)=(s-b)(s-c)\Rightarrow a^{2}=b^{2}+c^{2}$. That is $\angle A=90^{\circ}$.

Should the converse necessarily been true? That is, given ABC right-angled, should necessarily the center of A1B1C1 lie on the circumcircle of ABC?

King Jakaria wrote:Should the converse necessarily been true? That is, given ABC right-angled, should necessarily the center of A1B1C1 lie on the circumcircle of ABC?

Yes.