Team selection tests for GMO 2018

Denote $\angle B=\beta$ then we have $$\angle XYZ=180^{\circ }-\angle AIC=180^{\circ }-\left(90^{\circ }+\frac{\beta }{2}\right)=90^{\circ }-\frac{\beta }{2}.$$ So if $O$ is the circumcenter of $XYZ$ then $$\angle OXZ=90^{\circ }-\left(90^{\circ }-\frac{\beta }{2}\right)=\frac{\beta }{2}=\angle IBC,$$ which implies that $OX\perp BC$. Similarly, we also have $OY\perp CA$, $OZ\perp AB$.

Denote $M,N$ as the midpoints of $BC,B{B}_1$ respectively. From the cyclic quadrilateral and parallel line, we have $\angle B{B}_{1}C=\angle BNM=180^{\circ }-\angle BXM$. Similarly, $\angle B{C}_{1}C=180^{\circ }-\angle CXM$. Since $MX\perp BC$, triangle $XBC$ is isosceles, then $\angle BXM=\angle CXM$, thus $\angle B{B}_{1}C=\angle B{C}_{1}C$, which implies that $BC{C}_1{B}_1$ is cyclic.

Angle chasing again, we have $\angle I{C}_1{B}_1=\angle CB{B}_1=\frac{\beta }{2}$, but $\angle AIC=90^{\circ }+\frac{\beta }{2}$, implies that $AI\perp B_1{C}_1$. By similar way, we get $BI\perp C_1{A}_1$ and $CI\perp A_1{B}_1$, hence $I$ is the orthocenter of triangle $A_1{B}_1{C}_1$.

It is easy to check that these conditions are equivalent, which finishes the proof.