A self made geo problem

[Thanic Nur Samin]

Let $\Delta ABC$ be a triangle with orthocentre $H$. Let $X$ be an arbitrary point in the circumcircle of $\Delta ABC$. Let $Y$ be a point such that $HAXY$ is a parallelogram. Let $P=AB \cap HY$ and $Q=AH \cap BY$. Prove that, $APQY$ is a cyclic quadrilateral.

[tanmoy]

Assume $AH$ and $XY$ intersect the circumcircle of $\Delta ABC$ at $R$ and $S$,respectively.Then $HYSR$ is an isosceles trapezoid.And as $BC$ is the perpendicular bisector of $HR$,it is also the perpendicular bisector of $YS$.So,$Y$ is the orthocenter of $\Delta BXC$.So,$\angle QYH=\angle BYH=\angle BCH=\angle BAH=\angle PAH$.So,$APQY$ is cyclic.

[rah4927]

Note that the reflection of $H$ across $AB$ lies on the circumcircle. Now by $2012$ G2, we are done.

Note: The 2012 G2 :

Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D,H,F,G$ are concyclic.

The key idea used in this problem recurs throughout olympiads.

[Thanic Nur Samin]

Nice solutions.

We use complex numbers with the circumcircle of $\Delta ABC$ as the unit circle. We denote the coordinate of a point by the lowercase letters of the letter denoting them. Now, $h=a+b+c$ and $x+h=a+y$. So $y=x+b+c$.

Now, $\dfrac{\frac{b-a}{h-a}}{\frac{h-y}{b-y}}=\dfrac{\frac{b-a}{b+c}}{\frac{a-x}{-x-c}}=-\dfrac{\frac{a-b}{-c-b}}{\frac{a-x}{-c-x}}$ which is a real number as $A, B, X$ and the diametrically opposite point of $C$ are cyclic.

So, $\angle BAH=\angle HYB$, and so $APQY$ is cyclic.