Let $ABCD$ be a quadrilateral inscribed a circle $(O)$. Assume that $AB$ and $CD$ intersect at $E, AC$ and $BD$ intersect at $K$, and $O$ does not belong to the line $KE$. Let $G$ and $H$ be the midpoints of $AB$ and $CD$ respectively. Let $(I)$ be the circumcircle of the triangle $GKH$. Let $(I)$ and $(O)$ intersect at $M, N$ such that $MGHN$ is convex quadrilateral. Let $P$ be the intersection of $MG$ and $HN,Q$ be the intersection of $MN$ and $GH$. a) Prove that $IK$ and $OE$ are parallel. b) Prove that $PK$ is perpendicular to $IQ$.
Problem
Source: 2017 Saudi Arabia IMO TST III p2
Tags: geometry, cyclic quadrilateral, parallel, perpendicular
DNCT1
21.09.2020 20:39
a) First, we have that $\bigtriangleup KAB\sim\bigtriangleup KDC$ . Hence $G,H$ is the midpoint of segment $AB,CD$ respectively so we have
$\bigtriangleup KAG\sim\bigtriangleup KDH$ so $\angle KGA=\angle KHD$ and since $\angle OGA=\angle OHD=90^o$ so we have $\angle KGO=\angle KHO$ too.Now, since $OGEH$ is cyclic quadrilateral so by angle chasing we have
$$\angle (\overline{IK},\overline{GO})=180^o-\angle GKI-\angle KGO=180^o-(90^o-\angle GHK)-\angle KHO=90^o-(\angle KHO-\angle GHK)=90^o-\angle GHO=\angle GHE=\angle GOE=\angle (\overline{OE},\overline{GO})\Rightarrow \overline{IK}\parallel\overline{OE}$$(complete the proof).
Let $\overline{DA}\cap\overline{CB}=Z$.$\overline{ZC}\cap(O),\overline{AB},\overline{DC}=U,V,X,Y$ respectively
Then we easily get that $$(E,X,D,C)\overset{AB}{=}(E,V,A,B)=C(E,V,A,B)\overset{ZK}{=}(Z,K,V,X)=-1(1)$$So by Maclaurin we have $$\overline{EV}.\overline{EG}=\overline{EA}.\overline{EB}=\overline{ED}.\overline{EC}=\overline{EX}.\overline{EH}$$Implies that $V,G,H,X$ are concyclic.Let $\overline{HG}\cap\overline{ZX}=W$,by Gauss's line in quadrilateral $ABCD$ we conclude that $W$ is the midpoint of $ZX$
From $(1)$ and Newton's relation we have $$WK^2=\overline{WV}.\overline{WX}=\overline{WG}.\overline{WH}$$So $\overline{ZK}$ is tagent to $(GKH)$ at $K$.Hence $Z$ is the union point with $K$ WRT $(O)$ so we have that $(Z,K,U,Y)=-1$
$$\Rightarrow WK^2=\overline{WU}.\overline{WY}=\overline{WG}.\overline{WH}$$so we get that $W$ lies on the radical axis of $(GHK)$ and $(O)$ or $\overline{N,M,W}$ this implies that in the circumscribed $(GHK)$ we have $\overline{KK},\overline{NM},\overline{HG}$ are concurrent at $W$ or $W\equiv Q$.
Let $QS$ be the secend tagents to $(GHK)$ with $S$ lies on $(GHK)$
Let $\overline{KS}\cap \overline{GH},\overline{MN},\overline{GN}=X_1,Y_1,T$ respectively hence we have $GKHS,MKNS$ are harmonic so we get that $$K(K,S,G,H)\overset{QH}{=}(Q,X_1,G,H)=(Q,Y_1,M,N)=-1\Rightarrow \overline{SK}\cap\overline{MG}\cap\overline{NH}=P$$and hence $N(H,G,K,S)\overset{PK}{=}(P,K,T,S)=-1$ so we can easily get that $\overline{PK},\overline{GM},\overline{NH}$ are concurrent at $T$
By Brocard's theorem on cyclic quadrilateral $MGHN.PQ$ we get that $\overline{PK}\bot\overline{IQ}$ (complete the proof).
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