IMO-2011-G3

[Tahmid Hasan]

Let $ABCD$ be a convex quadrilateral whose sides $AD$ and $BC$ are not parallel. Suppose that the circles with diameters $AB$ and $CD$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_E$ be the circle through the feet of the perpendiculars from $E$ to the lines $AB,BC$ and $CD$. Let $\omega_F$ be the circle through the feet of the perpendiculars from $F$ to the lines $CD,DA$ and $AB$. Prove that the midpoint of the segment $EF$ lies on the line through of $\omega_E$ and $\omega_F$.

[Tahmid Hasan]

My solution

I have some changes in the notations.Let the circles with diameters $AD,BC$ meet at $X,Y$ inside the quadrilateral $ABCD$.
So $\angle AXD=\angle AYD=\angle BXD=\angle BYD=90^{\circ}$.
Let $X_1,X_2,X_3,X_4$ be the foot of perpendiculars from $X$ to $BC,CD,DA,AB$ respectivley.
Let $Y_1,Y_2,Y_3,Y_4$ be the foot of perpendiculars from $Y$ to $BC,CD,DA,AB$ respectively.
Observation:From the perpendiculars we get $(X,X_3,A,X_4);(X,X_4,B,X_1);(X,X_1C,X_2);(X,D,X_2,X_3)$ are cyclic.The similar cases with $Y$.
Note that these properties are independent of the positions of the foot of perpendiculars meaning the properties are valid even if some the foots lie on the extensions of sides $AB,BC,CD,DA$.
We'll be using these properties from now on.
Claim1:$X_1X_2X_3X_4,Y_1Y_2Y_3Y_4$ are cyclic.
Proof:$\angle X_4X_1X_2+\angle X_4X_3X_2=\angle X_4X_1X+\angle XX_1X_2+\angle X_4X_3X+\angle XX_3X_2$
$=\angle ABX+\angle XCD+\angle BAX+\angle CDX$
$=360^{\circ}-(\angle XBC+\angle XCB)-(\angle XAD+\angle XDA)$
$=360^{\circ}-90^{\circ}-90^{\circ}=180^{\circ}$.
So $X_1X_2X_3X_4$ is cyclic.Let the circle be $\omega_X$.
Similarly $Y_1Y_2Y_3Y_4$ is cyclic.Let the circle be $\omega_Y$.
Let $P,Q=\omega_X \cap \omega_Y$.
Let $X_2'=XX_2 \cap AB,Y_2'=YY_2 \cap AB,X_4'=XX_4 \cap CD,Y_4'=YY_4 \cap CD$
Claim:$X_2',X_4' \in \omega_X,Y_2',Y_4' \in \omega_Y$.
$\angle X_4X_2'X_2=\angle X_2'BX+\angle X_2'XB=\angle X_4BX+90^{\circ}-\angle CXX_2$
$=\angle X_4X_1X+\angle XCX_2=\angle X_4X_1X+\angle XX_1X_2=X_4X_1X_2$.
So,$X_2' \in \omega_X$.The others can be proved similarly.
Further observation:$\angle X_2'Y_4Y_4'=\angle X_2'X_2Y_4'=90^{\circ}$,so $X_2',Y_4,X_2,Y_4'$ are cyclic.
Similarly $X_4,Y_2',X_4',Y_2$ are cyclic.
Let $M=Y_4Y_4' \cap X_2X_2',N=X_4X_4' \cap Y_2Y_2'$
So from intersecting chord theorem $MY_4.MY_4'=MX_2.MX_2'$
But $LHS=$ power of $M$ wrt $\omega_Y$ and $RHS=$ power of $M$ wrt $\omega_X$
$\Rightarrow M \in PQ$.
Similarly $N \in PQ$.
$MY \bot AB,NX \bot AB \rightarrow MY \parallel NX;MX \bot CD,NY \bot CD \rightarrow MX \parallel NY$
So $MXNY$ is a parallelogram of which $MN,XY$ are diagonals.
Hence $MN$ bisects $XY$ meaning the midpoint of $XY$ lies on $PQ$.