Median IPQ bisects arc BAC

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Median IPQ bisects arc BAC

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Source: S.A booklet 2021 P8 in problems without solution

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#1 h Sep 20, 2021, 9:04 PM

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Let $ABC$ be an non-isosceles triangle with incenter $I$ , circumcenter $O$ and a point $D$ on segment $BC$ such that $(BID)$ cut segments $AB$ at $E$ and $(CID)$ cuts segment $AC$ at $F$ Circle $(DEF)$ cuts segments $AB$ , $AC$ again at $M,N$ . Let $P$ The intersection of $IB$ and $DE$ , $Q$ The intersection of $IC$ and $DF$ . Prove that $EN,FM,PQ$ are parallel and the median of vertex $I$ in triangle $IPQ$ bisects the arc $BAC$ of $(O)$ .

This post has been edited 1 time. Last edited by Adam_oly, Sep 20, 2021, 9:09 PM

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#2 h Sep 21, 2021, 12:18 AM

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It's easy to see that $ID = IE = IF$ . Hence $IM = IN$ . But $AI$ is bisector of $\angle{BAC}$ then $I$ $\in$ $(AMN)$ . So $\angle{MIN} = 180^{\circ} - \angle{BAC} = \angle{EIF}$ . Hence $\triangle IEF = \triangle IMN$ or $EF = MN,$ which means $MENF$ is isosceles trapezoid. Therefore $EN$ $\parallel$ $FM$ . Note that $\angle{NME} = \angle{EFN}$ then $\triangle AEF = \triangle ANM$ or $AE = AN, AM = AF$ . Hence $AI$ $\perp$ $EN$ . Since $IP \cdot IB = ID^2 = IQ \cdot IC$ then $B, P, Q, C$ lie on a circle. So $AI$ $\perp$ $PQ$ or $EN$ $\parallel$ $FM$ $\parallel$ $AI$ . We also see that the $I$ - median of $\triangle IPQ$ is $I$ - symmedian of $\triangle IBC,$ which passes through midpoint of arc $BAC$

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#3 h Sep 21, 2021, 11:07 PM

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Claim 1 : $I$ is the circumcenter of $(DEF)$
Proof : easy to notice by simple angle chasing that $ID=IE=IF$
Claim 2 : $IQDP$ is cyclic
Proof : $\angle{IQF}=\angle{IFC}$ $(\triangle{IQF}\sim \triangle{IFC})$ $=\angle{IDB}=\pi-\angle{IEB}=\angle{IPD}$ $(\triangle{IEB}\sim \triangle{IPE})$
Claim 3 : $EN , FM , PQ$ are parallel
Proof : $\angle{QPD}=\angle{DIQ}=\angle{DFN}=\angle{NED}$ $\implies$ $PQ \parallel EN$
And $\angle{PQD}=\angle{PID}=\angle{DEB}=\angle{MFD}$ $\implies$ $PQ \parallel MF$
Claim 4 : The $I$ -median of $\triangle{IPQ}$ bisects $\overarc{BAC}$
Proof : Let $K$ be the midpoint of segment $\overline{PQ}$ , and $L$ the midpoint of segment $\overline{BC}$
By simple angle chasing $PQCB$ is cyclic , hence $\triangle{IKQ} \sim \triangle{ILB}$ , wich implies that the $I$ -median of $\triangle{IPQ}$ is the $I$ -symmedian of $\triangle{IBC}$ . $(1)$
Now let $X$ be the intersection of the tangents to $(BIC)$ at $B$ and $C$ , angle chasing gives that $X$ $\in$ $(ABC)$ , thus it suffices to show that points $X , I , K$ are collinear , wich is obviously true by $(1)$ .

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#4 h Sep 22, 2021, 8:23 AM

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Nice angle chasing exercise.

$\angle PIQ+\angle PDQ=90+\angle A/2+ \angle B/2+\angle C/2=180 \implies IPDQ$ is a cyclic quadrilateral. $\angle MFD=\angle MED=\angle BID= \angle PQD \implies PQ \parallel MF$ and by symmetry $PQ \parallel EN$ . $\angle IPQ=\angle IDQ=\angle ICF=\angle ICD \implies PQCD$ is a cyclic quadrilateral so $I$ -median of $IPQ$ is the same as $I$ -symmedian of $IBC$ .

Proof of the fact that $I$ -symmedian of $IBC$ passes through the midpoint of arc $BAC$ :

$K$ is the midpoint of arc $BAC$ , $BI \cap (ABC)=U$ , $CI\cap (ABC)=V$ , $T=KI \cap BC, I\cap BC=X, AI \cap (ABC)=S$ . From angle chasing $KUIV$ is a parallelogram. We know that $IX$ and $IS$ are isogonal conjugates, $\angle TIX=\angle TKM=\angle SIM \implies IT$ and $IM$ are isogonal conjugates as desired.

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