I think this problem is the best problem in this test.. I solved it with 4steps. Let $Z$ be the intersection of $BE, CD$, $P,Q$ be the midpoint of $arc{ABC}$ and $arc{ACB}$. Let $I_a$ be the intersection of $BP, CQ$. Trivially, $Z \in (ABC)$ and $Z$ is on the perpendicular bisecter of $DE$. step1) $(D,I_a,E)$colinear and $Z$ is the orthocenter of $\triangle{BI_aC}$. step2) Let $OI_a$ and $BC$ meet at $F$. then $(P,X,F,Y,Q)$ colinear. step3)$(O,X,I_a,Y)$ cyclic and $DE$ is tangent to the circle. step4) $\angle{XAY} \ge 60$.

KPBY0507 wrote: I think this problem is the best problem in this test.. I solved it with 4steps. Let $Z$ be the intersection of $BE, CD$, $P,Q$ be the midpoint of $arc{ABC}$ and $arc{ACB}$. Let $I_a$ be the intersection of $BP, CQ$. Trivially, $Z \in (ABC)$ and $Z$ is on the perpendicular bisecter of $DE$. step1) $(D,I_a,E)$colinear and $Z$ is the orthocenter of $\triangle{BI_aC}$. step2) Let $OI_a$ and $BC$ meet at $F$. then $(P,X,F,Y,Q)$ colinear. step3)$(O,X,I_a,Y)$ cyclic and $DE$ is tangent to the circle. step4) $\angle{XAY} \ge 60$. Thanks for the nice solution! I was sad that almost everyone tried to solve this problem by calculating.

KPBY0507 wrote: I think this problem is the best problem in this test.. I solved it with 4steps. Let $Z$ be the intersection of $BE, CD$, $P,Q$ be the midpoint of $arc{ABC}$ and $arc{ACB}$. Let $I_a$ be the intersection of $BP, CQ$. Trivially, $Z \in (ABC)$ and $Z$ is on the perpendicular bisecter of $DE$. step1) $(D,I_a,E)$colinear and $Z$ is the orthocenter of $\triangle{BI_aC}$. step2) Let $OI_a$ and $BC$ meet at $F$. then $(P,X,F,Y,Q)$ colinear. step3)$(O,X,I_a,Y)$ cyclic and $DE$ is tangent to the circle. step4) $\angle{XAY} \ge 60$. Could you please explain the steps more detailed?( I have no idea how to prove the second step