IMO Mock-1 problem 2

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Tahmid Hasan
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IMO Mock-1 problem 2

Unread post by Tahmid Hasan » Tue Sep 18, 2012 1:02 pm

Let $ABCD$ be a convex quadrilateral with no pair of parallel sides, such that $\angle ABC=\angle CDA$. Assume that the intersections of the pairs of neighboring angle bisectors of $ABCD$ form a convex quadrilateral $EFGH$. Let $K$ be the intersection of the diagonals of $EFGH$. Prove that the lines $AB$ and $CD$ intersect on the circumcircle of the triangle $BKD$.
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User avatar
Tahmid Hasan
Posts:665
Joined:Thu Dec 09, 2010 5:34 pm
Location:Khulna,Bangladesh.

Re: IMO Mock-1 problem 2

Unread post by Tahmid Hasan » Tue Sep 18, 2012 1:12 pm

My solution
In my figure, $AB<DC,AD<BC$.
Let $\ell_A,\ell_B,\ell_C,\ell_D$ denote the internal bisectors of $\angle A,\angle B,\angle C,\angle D$ respectively.
Let $\ell_ A \cap \ell_B=E,\ell_B \cap \ell_C=H,\ell_C \cap \ell_D=G,\ell_D \cap \ell_A=F$.
Let $BA \cap CD=I,DA \cap CB=J$.Let $\ell_I,\ell_J$ be the internal bisectors of $\angle BIC,\angle DJC$ respectively.
In $\triangle IBC,H=\ell_B \cap \ell_C$.So $H \in \ell_I$
In $\triangle IAD,F$ is the intersection of external bisectors of $\angle IAD,\angle IDA$.So $F \in \ell_I$.
Hence $I,H,K,F$ are collinear.
Similarly $J,E,K,G$ are collinear.From here we conclude $\ell_I \cap \ell_J=K$
Now $\angle ABC=\angle ADC \rightarrow IBJ=\angle IDJ \Rightarrow I,D,B,J$ are cyclic.Let the circle be $\alpha$.
Let $K'$ be the midpoint of arc $BD$ of $\alpha$ which does not contain $I,J$.
So $K' \in \ell_I,K' \in \ell_I \Rightarrow K=K'$.
Hene $I$ lies on the circumcircle of $BKD$.
Note:I also found an interesting property:$EFGH$ is a cyclic trapezoid which follows from direct angle chasing.
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