Wow, pretty easy for a problem 6... Suppose that $BD>CD$. (i) follows immediatly by easy angle chasing, and also if we denote $G$ the midpoint of the arc$BC$ that doesn't contain $A$, we'' get easily again that $\angle IGD=\angle IQD$. (ii) Easy angle chasing give us $\angle BIC=180-\angle BDA=90+\frac{A}{2}$, so $BDFI$ is cyclic. Analogeously, $EIDC$ is cyclic. So $\angle BXC=\angle EFX+\angle FEX=\angle BFD+\angle DEC=\angle BID+\angle DIC=\angle BIC$ So $BIXC$, $FXDC$ and $EXDB$ are cyclic. Now $\angle IXD=\angle IXB+\angle BXD=\angle ICB+\angle FCD=\angle ACF+\angle FCD=\angle ACD$. But $\angle ACD=180-\angle AGD$, so $\angle AGD+\angle IXG=180$. So $I,D,G,X$ are concyclic. So $G,D,P,Q,I,X$ are concyclic. Done.

could you post other problems from itamo?

@2 above You probably mistyped something, since $M,D,P$ are collinear.

@GGPiku wrong sol $\angle AMD = \angle IXD = \angle ACD$ $I,M,D,X$ are not concyclic

$(a)$ Assume that $\angle ABI=\angle IBC=\angle BCI=\angle ACI=\alpha$. Hence, $\angle BIC=180^\circ- 2\alpha$. Other hand, $\angle QDP=\angle NDM=2\alpha$. So, $\angle QIP+\angle QDP=180^\circ$. Hence, $I,Q,D,P-$concyclic. $(b)$ We know $\angle BIC=180^\circ-2\alpha, \angle BIF=2\alpha$. Other hand, $\angle EDC=\angle ABC=2\alpha$. So, quadrilateral $DEIC$ is cyxlic. Similarly, quadrilateral $DIFB$ is also cyclic. Now, assume that $X=CE \cap BF$, $\angle BXC=\angle XFE+\angle XEF=\angle BID+\angle DIC=\angle BIC$ hence, quadrilateral $BXIC$ is cyclic.From quadrilaterals $BXIC$ and $BFID$ are cyclic implies $FE \cdot ED=BE \cdot EI=XE \cdot EC$ so, quadrilateral $DXFC$ cyclic.If $\angle BAD=\phi$, then $\angle BMD=\angle BCD=\phi$.In triangle $DQM$, $\angle DQM=180^\circ-\phi-2\alpha$. Other hand, in triangle $DFC$, $\angle DFC=180^\circ-\phi-3\alpha$.From quadrilateral $DXFC$ is cyclic implies $\angle DXC=\angle DFC$ and quadrilateral $BXIC$ is cyclic implies $\angle FXI=\angle ICB=\alpha$, quadrilateral $DXFC$ is cyclic implies $\angle FXC=\angle FDC=2\alpha$ so,$\angle IXC=\alpha$. Hence, $\angle IXD=\angle IXC+\angle DXC=180^\circ-2\alpha-\phi$. So, $\angle DQI=\angle DXI$ implies quadrilateral $DQXI$ is cyclic. Hence points $D,Q,X,I,P$ are concyclic. Proved.

FedeX333X wrote: @2 above You probably mistyped something, since $M,D,P$ are collinear. Indeed, A typo. It’s G instead of M(on my configuration, I have M as the midpoint of the arc)

ITAMO has 6 problems like most olympiads but goes on for only one day (4h30), that's why the problems are easier than "expected".

Is there a way to solve this problem using the method of moving point or the projective geometry?

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/O.M.%20Italie%202018%20Probleme%206.pdf Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, here Sincerely Jean-Louis