IMOC 2019 G5 (DH bisects <EDF, circumcircle, orthocenter related)

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IMOC 2019 G5 (DH bisects <EDF, circumcircle, orthocenter related)

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Source: https://artofproblemsolving.com/community/c6h1958463p13536764

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#1 h Mar 22, 2020, 11:01 AM

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Given a scalene triangle $\vartriangle ABC$ with orthocenter $H$ and circumcenter $O$ . The exterior angle bisector of $\angle BAC$ intersects circumcircle of $\vartriangle ABC$ at $N \ne A$ . Let $D$ be another intersection of $HN$ and the circumcircle of $\vartriangle ABC$ . The line passing through $O$ , which is parallel to $AN$ , intersects $AB,AC$ at $E, F$ , respectively. Prove that $DH$ bisects the angle $\angle EDF$ .

https://3.bp.blogspot.com/-F1mFwojG_I0/XnYNR8ofqSI/AAAAAAAALeo/zge24WF0EO8umPAaXprKAeXJHAj7pr6tQCK4BGAYYCw/s1600/imoc2019g5.png

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#3 h Mar 24, 2020, 4:11 PM • 1 Y

Y by Kanep

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Let us define some points. Let $M$ be the midpoint of arc $BC$ , $D'=(EFM)\cap (ABC)$ , $G=(EFM)\cap AM$ , $L=(AEF)\cap AM$ , $X=(AEF)\cap (ABC)$ . $J$ , $K$ are on $AB$ , $AC$ respectively such that $JN\perp AC$ and $KN\perp AB$ .

By symmetry, $AEMF$ , $AELF$ and $MFGE$ are kites, and $AL$ and $GM$ are diameters.
In particular, $AEMF$ is a rhombus (by letting $Y=EF\cap AM$ , $NO=OM\rightarrow AY=YM$ )
We have $XA\perp XL, XN\perp XM, AN\perp LM$ , so $X$ spirals $AN$ to $LM$ . Since $KJAN$ is clearly congruent to $FELM$ , $X$ also spirals $KJ$ to $FE$ , and $XKAJ$ is cyclic.
Hence, $X$ spirals $JEB$ to $KFC$ , so $\frac{JE}{EB}=\frac{KF}{FC}$ . Since $JN\parallel EG\parallel BH$ (all are $\perp AC$ ) and $KN\parallel GF\parallel HC$ (all $\perp AB$ ), this means that $NGH$ is collinear.
Since $\angle NDM=90^\circ$ , $\angle GDM=90^\circ$ , so $GEFDM$ cyclic. Then, since $EG=GF$ by symmetry, we have $\angle EDH=\angle HDF$ as angles subtending equal arcs.

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#4 h Aug 23, 2020, 5:11 PM • 1 Y

Y by GuvercinciHoca

My solution:

https://scontent.fdad3-1.fna.fbcdn.net/v/t1.15752-9/117770228_633255090650839_4614696949313122246_n.png?_nc_cat=110&_nc_sid=b96e70&_nc_ohc=N0CN3RGUOeoAX-TkhEU&_nc_ht=scontent.fdad3-1.fna&oh=ed7c421b8a6f768f22c3e68a5fb7528b&oe=5F69C468

Let $K$ be the midpoint of minor arc $BC$ , $A'$ be the antipode of $A$ , $CL$ be the altitude of triangle $ABC$ . The line passing through $H$ and parallel $AN$ meets $AB, AC$ at $I$ and $J$ respectively. We can easily get $E, F, I, J$ are concyclic. Since, $EF \parallel AN \parallel AK$ , $K$ is the reflection of $A$ in $EF$ . Then, $\angle EAF = \angle EKF$ . We have $\triangle AHL \sim \triangle AA'C$ . Then, $\dfrac{AH}{AO} = 2 \cdot \dfrac{AH}{AA'} = 2 \cdot \dfrac{AL}{AC} = 2 \cdot \cos \angle A$ Also, we have $\triangle AIH \sim \triangle AOF$ . Then, $\dfrac{AI}{AF} = \dfrac{AH}{AO} = 2 \cos \angle A$ Then, $FIA$ is an isosceles triangle. Then, $\angle AIF = \angle IAF = \angle EKF$ . Therefore, $E, F, K, I$ are concyclic. According to the Reim's Theory, since $IJ \parallel AN$ , we have $I, H, B, D$ are concyclic and $J, H, D, C$ are concyclic. Then, $\angle IDJ = \angle IDH + \angle JDH = \angle HBA + \angle HCA = 2(90^o-\angle BAC) = \angle IFJ$ Hence, $I, F, J, D$ are concyclic. Since $EI = FJ$ and $DH$ bisects angle $\angle IDJ$ , $DH$ bisects angle $\angle EDF$ .

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#5 h Aug 25, 2020, 7:57 PM

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Let $a,b,c$ be complex numbers such that $a^2,b^2,c^2$ are vertices of the triangle and $n=bc$ . Then
$h=a^2+b^2+c^2,\ d=\frac{h-n}{1-n\cdot\overline{h}}=\frac{a^2bc(a^2+b^2+c^2-bc)}{a^2bc-(a^2b^2+b^2c^2+c^2a^2)}.$ $E\in AB\iff \overline{e}=\frac{a^2+b^2-e}{a^2b^2}$ Hence $EO\parallel AN\iff e=\frac{c(a^2+b^2)}{c-b}.$ Let $P$ be reflection of $C$ wrt the circumcircle. Then $p=-c^2$ and
$e+c^2=\frac{c(a^2+b^2+c^2-bc)}{c-b},\ e-d=\frac{c^2(a^2c+b^3)(a^2+bc)}{(c-b)(a^2b^2+b^2c^2+c^2a^2-a^2bc)}.$ Hence $\frac{e-d}{e+c^2}=\frac{c(a^2c+b^3)(a^2+bc)}{(a^2+b^2+c^2-bc)(a^2b^2+b^2c^2+c^2a^2-a^2bc)}\in\mathbb{R}\implies P\in DE.$ Analogously we can prove that if $Q$ is reflection of $B$ wrt $O$ then $Q\in DF$ . Because $BN=BC$ and $PQ\parallel BC$ there holds $NP=NQ$ and
$\measuredangle EDN=\measuredangle PDN=\measuredangle NDQ= \measuredangle NDF$ that is line $DN$ bisects angle $EDF.\square$
#1713

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#6 h Feb 18, 2025, 12:04 PM

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Lemma: Let $O,H,N$ be the circumcenter, orthocenter and midpoint of arc $BAC$ on a triangle $ABC$ . The parallel from $O$ to $AN$ , $(BOC),NH$ concur.
Proof: Work on the complex plane. Let $PO\parallel AN$ and $P\in (BOC)$ . Let $a=x^2,b=y^2,c=z^2$ and $n=yz$ .We have $\overline{p}=-\frac{p}{x^2yz}$ and
$\frac{y^2}{z^2}.\frac{p-z^2}{p-y^2}=\frac{z^2}{y^2}.\frac{-\frac{p}{x^2yz}-\frac{1}{z^2}}{-\frac{p}{x^2yz}-\frac{1}{y^2}}=\frac{pz^2+x^2yz}{py^2+x^2yz}\iff -py^2-x^2yz=pz^2-y^2z^2\iff p=\frac{yz(yz-x^2)}{y^2+z^2}$ $\frac{yz-x^2-y^2-z^2}{yz-\frac{yz(yz-x^2)}{y^2+z^2}}=\frac{(y^2+z^2)(yz-x^2-y^2-z^2)}{yz(x^2+y^2+z^2-yz)}=\frac{-(y^2+z^2)}{yz}\in R$ Which proves the lemma.

Lemma: $ABC$ is a triangle with circumcenter, orthocenter $O,H$ . Let $N$ be the midpoint of arc $BAC$ , $B'$ be the antipode of $B$ on $(ABC)$ . Let $NH$ meet $(ABC)$ at $D$ and $DB'$ intersects $AC$ at $K$ . Then, $OK\parallel NA$ .
Proof: Let $A'$ be the antipode of $A$ on $(ABC)$ . Note that $P,O,A',D$ are concyclic since $PO\parallel AN$ . Radical axises of $(BOCP),(ABCA'D),(DA'PO)$ hence $PO,DA',BC$ concur. Let $V$ be the concurrency point. Pascal at $B'DA'ACB$ gives $K,V,O$ are collinear thus, $K$ lies on $OP$ which is equavilent to $OK\parallel NA$ .

Let $B',C'$ be the antipode of $B,C$ on $(ABC)$ . By the latter lemma, we see that $D,E,C'$ and $D,F,B'$ are collinear. We have $\measuredangle EDH=\measuredangle C'DN=\frac{\measuredangle A}{2}=\measuredangle NDL=\measuredangle HDF$ as desired. $\blacksquare$

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#7 h Feb 18, 2025, 1:46 PM

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Simple one,

$EF$ meets $(BOC)$ again at $K$ . $AH$ intersects $OK$ at $J$ and $AM$ intersects $OK$ at $I$ . Let $S$ be the antipode of $O$ in $(BOC)$ , $L$ be the midpoint of $BC$ and $M$ the midpoint of minor arc $BC$ .
Claim 1. $IL \parallel KM$ .
Notice that: $OL.OS = OC^2 = OM^2$ , thus $\cfrac{OL}{OM} = \cfrac{OM}{OS} = \cfrac{OI}{OK}$ . So $IL \parallel KM$ .
Claim 2. $K, H, N$ are collinear.
Observe how $ANOJ$ is a parallelogram, thus $AJ = ON = OM$ . Now:
$\cfrac{JH}{AH} = \cfrac{AJ}{AH} - 1 = \cfrac{OM}{2OL} - 1 = \cfrac{OK}{2OI} - 1 = \cfrac{OK}{OJ} - 1 = \cfrac{KJ}{NA}$ So, $K, H, N$ are collinear.
Claim 3. $BEKD$ and $CFKD$ are cyclic
Since $\angle EBD = 180^{\circ} - \angle AND = 180^{\circ} - \angle NKO = 180^{\circ} - \angle EKD$ , we have $BEKD$ is cyclic. Similar for $CFKD$ .
Claim 4. $DK$ bisects $\angle EDF$ .
Now, because $KO$ is the external bisector of $\angle BKC$ , we have the angle chase:
$\angle EDB = \angle EKB = \angle FKC = \angle FDC$ But then, $DN$ is the internal bisector of $\angle BDC$ , so we must have that $DN$ or $DK$ bisects $\angle EDF$ , as desired.

This post has been edited 4 times. Last edited by Retemoeg, Feb 18, 2025, 1:59 PM

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