Team Selection Test for IMO 2019

Let $\angle ABC=\alpha$ and $\angle BAC=90-\alpha =\beta$. $\Rightarrow \angle BCD=\angle EAC=\beta$ and $\angle ACD=\angle ACE=\alpha$. Let $K$ be the second intersection point of the circumcircle of the triangle $ECB$ and the line $AB$. $\Rightarrow \angle AKE=\angle KEA=\beta$ and hence $AKE$ is an isosceles triangle.



Note that $O_1$ is the intersection point of the perpendicular bisectors of the line segments $[EK]$ and $[BC]$. Let $O$ and $L$ be the midpoints of $[AB]$ and $[BC]$, respectively. Then, $L$, $O$

and $O_1$ are collinear. By angle chasing we see that $A{O}_{1}LC$ is a rectangle. Moreover, since $OLB\cong O{O}_{1}A$ we have $|OL|=|O{O}_1|$. Consequently, we obtain $O{O}_1\parallel AC$ and $|O{O}_1|=|AC|/2$. Similarly, one can get $O{O}_2\parallel BC$ and $|O{O}_2|=|BC|/2$. Therefore, we see that $ABC\sim O_1{O}_{2}O$ and hence $|O_1{O}_2|=|AB|/2$.