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Source: Vietnam Mo 2018 1st day 2nd problem

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#1 h Jan 11, 2018, 6:53 AM • 2 Y

Y by Adventure10, Mango247

We have a scalene acute triangle $ABC$ (triangle with no two equal sides) and a point $D$ on side $BC$ . Pick a point $E$ on side $AB$ and a point $F$ on side $AC$ such that $\angle DEB=\angle DFC$ . Lines $DF,\, DE$ intersect $AB,\, AC$ at points $M,\, N$ , respectively. Denote $(I_1),\, (I_2)$ by the circumcircles of triangles $DEM,\, DFN$ in that order. The circle $(J_1)$ touches $(I_1)$ internally at $D$ and touches $AB$ at $K$ , circle $(J_2)$ touches $(I_2)$ internally at $D$ and touches $AC$ at $H$ . $P$ is the intersection of $(I_1),\, (I_2)$ different from $D$ . $Q$ is the intersection of $(J_1),\, (J_2)$ different from $D$ .
a. Prove that all points $D,\, P,\, Q$ lie on the same line.
b. The circumcircles of triangles $AEF,\, AHK$ intersect at $A,\, G$ . $(AEF)$ also cut $AQ$ at $A,\, L$ . Prove that the tangent at $D$ of $(DQG)$ cuts $EF$ at a point on $(DLG)$ .

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#3 h Jan 12, 2018, 5:58 PM • 2 Y

Y by gausskarl, Adventure10

Let $(I_1)\cap AB=Z,W$ and $(I_2)\cap AC=X,Y$ where $Z,E,W$ and $X,F,Y$ collinear in this order.
We've $\angle{NEM}=\angle{NFM}\implies NEFM$ concyclic.
This gives $AX\times AY=AE\times AM=AF\times AN=AZ\times AW$ .
So, $WXZY$ is cyclic.
Since $\angle{DEZ}=180^{\circ}-\angle{DEB}=\angle{DZN}\implies DZ^2=DE\times DN$ and $\angle{DWN}=\angle{DFC}=\angle{DEW}\implies DW^2=DE\times DN$ .
We get that $DW=DZ$ . Hence $D$ lie on perpendicular bisector of $WZ$ .
Similarly, $D$ lie on perpendicular bisector of $XY$ . This gives us $D$ must be the centre of $(WXZY)$ .
Let $M_1$ and $M_2$ denote the midpoint of minor arc $FN$ and $EM$ of $(I_1)$ and $(I_2)$ , respectively.
It's well-known that $D,K,M_1$ and $D,H,M_2$ are collinear.
It's also easy to see that $\angle{M_1DN}+\angle{M_2DM}=\frac{\angle{EDF}}{2}+\frac{\angle{EDF}}{2}=\angle{EDF}$ , so $D,M_1,M_2$ collinear.
Hence, $D,M_1,M_2,K,H$ are all collinear. We've
It's well-known that $DK\times DM_2=DX^2=DY^2$ (since $D$ is the midpoint of arc $XY$ of $(I_2)$ ).
Similarly, $DM_1\times DH=DZ^2$ . Comparing two equations gives us $DM_1\times DH =DK\times DM_2$ .
Also, note that $\triangle{DJ_1K}\sim \triangle{DI_1M_1}$ and $\triangle{DJ_2H}\sim \triangle{DI_2M_2}$ .
We get $\frac{DJ_1}{DI_1}=\frac{DK}{DM_1}=\frac{DH}{DM_2}=\frac{DJ_2}{DI_2}\implies J_1J_2\parallel I_1I_2$ .
Hence, $DQ$ which is radical axis of $(J_1)$ and $(J_2)$ , perpendicular to $J_1J_2$ , is the same line with $DP$ which is radical axis of $(I_1)$ and $(I_2)$ , perpendicular to $I_1I_2$ . This finishes part a.

Let $R=HK\cap EF$ . Note that by Miquel's point of quadrilateral, we get that $ERGH$ concyclic.
Since $A$ lie on radical axis of $(I_1)$ and $(I_2)$ , we get that $D,A,P,Q$ are all collinear.
Let $EF$ intersect the tangent line at $D$ of $(DQG)$ at $T$ .
We've $\angle{GRD}=180^{\circ}-\angle{GHK}=180^{\circ}-\angle{GEA}=180^{\circ}-\angle{GLA}=\angle{GLD}$ .
This means $R\in (DLG)$ .
And since $\angle{GRT}=\angle{GHE}=\angle{GQL}=\angle{GDT}$ , $GRDT$ is cyclic.
In other words, $T$ lies on $(DLG)$ . This finishes part b.

This post has been edited 1 time. Last edited by ThE-dArK-lOrD, May 23, 2019, 12:49 PM

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#4 h Jan 12, 2018, 11:28 PM • 1 Y

Y by Adventure10

gausskarl wrote:

We have a scalene acute triangle $ABC$ (triangle with no two equal sides) and a point $D$ on side $BC$ . Pick a point $E$ on side $AB$ and a point $F$ on side $AC$ such that $\angle DEB=\angle DFC$ . Lines $DF,\, DE$ intersect $AB,\, AC$ at points $M,\, N$ , respectively. Denote $(I_1),\, (I_2)$ by the circumcircles of triangles $DEM,\, DFN$ in that order. The circle $(J_1)$ touches $(I_1)$ internally at $D$ and touches $AB$ at $K$ , circle $(J_2)$ touches $(I_2)$ internally at $D$ and touches $AC$ at $H$ . $P$ is the intersection of $(I_1),\, (I_2)$ different from $D$ . $Q$ is the intersection of $(J_1),\, (J_2)$ different from $D$ .
a. Prove that all points $D,\, P,\, Q$ lie on the same line.
b. The circumcircles of triangles $AEF,\, AHK$ intersect at $A,\, G$ . $(AEF)$ also cut $AQ$ at $A,\, L$ . Prove that the tangent at $D$ of $(DQG)$ cuts $EF$ at a point on $(DLG)$ .

$a)$ first remark that $EFMN$ is cyclic so $A$ is on the radical axis of $(I_1),(I_2)$ ;let the tangents of $(I_1),(I_2)$ cut $AB,AC$ at $C' ,B'$ respectively.it s easy to show that $AB'DC'$ is parallelogram thus $AK=AH$ then $A$ is also on the radical axis of $(J_1),(J_2)$ hence $P,Q,D,A$ are collinear.
$b)$ $\angle C'KD=\angle KDC', \angle DHB' =\angle B'DH ,\angle C'DB'= \angle C'AB'$ thus $K,H,D$ are collinear ,consider $T$ the intersection of $EF$ and $KH$ complete quadrilateral property leads $TGFH$ is cyclic so $\angle GTD =\angle GFA =\angle GLA$ thus $T \in (DLG)$ . $\angle STD =\angle SLA =\angle GLA -\angle GLS=\angle GFA -\angle GDS= \angle GFA- \angle GQD =\angle GFA-\angle GQA=\angle GFA-\angle GHA=\angle FGH$ since $Q \in (AKH) [\because \angle AQH=\angle DQH=\angle DHA=\angle AKH]$ so $\angle STD=\angle FGH = \angle FTH$ which means that $T,S,F$ are collinear hence the result follows.

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#5 h May 23, 2019, 8:11 AM • 2 Y

Y by Adventure10, Mango247

Here is my solution for this problem
Solution
a) Let $I$ , $J$ be midpoint of $\stackrel\frown{ME}$ , $\stackrel\frown{NF}$ then it's well - known that: $D$ , $H$ , $J$ are collinear and $D$ , $K$ , $I$ are collinear
So: $\dfrac{J_2D}{I_2D} = \dfrac{DH}{DJ}$ and $\dfrac{J_1D}{I_1D} = \dfrac{DK}{ID}$
Hence: $\dfrac{DH}{HJ} = \dfrac{DN}{NJ} . \dfrac{DF}{FJ} = \dfrac{DN . DF}{NJ^2}$
Similarly: $\dfrac{DK}{KI} = \dfrac{ED . DM}{MI^2}$
Then: $\dfrac{DH}{HJ} : \dfrac{DK}{KI} = \dfrac{DN . DF}{NJ^2} : \dfrac{ED . DM}{MI^2} = \dfrac{DN . DF}{ED . DM} . \dfrac{MI^2}{NJ^2}$
Since: $\angle{DEB} = \angle{DFC}$ so: $M$ , $N$ , $E$ , $F$ lie on a circle
Then: $\triangle DFN$ $\sim$ $\triangle DEM$ or $\dfrac{DF}{ED} = \dfrac{DN}{DM} = \dfrac{FN}{EM} = \dfrac{NJ}{MI}$
So: $\dfrac{DH}{HJ} = \dfrac{DK}{KI}$ or $\dfrac{DH}{DJ} = \dfrac{DK}{ID}$
Hence: $\dfrac{J_2D}{I_2D} = \dfrac{J_1D}{I_1D}$ or $J_1J_2$ $\parallel$ $I_1I_2$
But: $J_1J_2$ $\perp$ $QD$ then: $QD$ $\perp$ $I_1I_2$
Combine with: $DP$ $\perp$ $I_1I_2$ so: $D$ , $P$ , $Q$ are collinear
b) Let point $S$ on $EF$ which satisfies $\dfrac{\overline{SE}}{\overline{SF}} = \dfrac{\overline{DK}}{\overline{DH}}$
We have: $\angle{GKH} = \angle{GAH} = \angle{GAF} = \angle{GEF}$ and $\angle{GHK} = \angle{GAK} = \angle{GFE}$
So: $\triangle GKH$ $\sim$ $\triangle GEF$
Then: $\dfrac{GK}{KH} = \dfrac{GE}{EF}$ or $\dfrac{GK}{GE} = \dfrac{KH}{EF}$
Since: $\dfrac{SE}{SF} = \dfrac{DK}{DH}$ , we have: $\dfrac{SE}{DK} = \dfrac{SF}{DH} = \dfrac{SE - SF}{DK - DH} = \dfrac{EF}{HK}$
Hence: $\dfrac{EG}{KG} = \dfrac{EF}{KH} = \dfrac{SE}{DK}$
But: $\angle{GKH} = \angle{GEF}$ so: $\triangle DGK$ $\sim$ $\triangle SGE$
Then: $\angle{DGK} = \angle{SGE}$ and $\dfrac{KG}{DG} = \dfrac{EG}{SG}$
Hence: $\angle{SGD} = \angle{EGK}$ or $\triangle GKE$ $\sim$ $\triangle GDS$
So: $\angle{GKE} = \angle{GDS} = \angle{GQD}$ or $DS$ tangents $(DQG)$ at $D$
Since: $\triangle GKE$ $\sim$ $\triangle GDS$ then: $\angle{DSG} = \angle{KEG}$
But: $\angle{KEG} = \angle{AEG} = \angle{ALG}$ so: $\angle{DSG} = \angle{ALG}$ or $D$ , $L$ , $G$ , $S$ lie on a circle
Hence: tangent at $D$ of $(DQG)$ intersects $EF$ at a point $S$ lies on $(DLG)$

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#6 h Jul 2, 2023, 6:35 AM • 1 Y

Y by GeoKing

https://i.ibb.co/7WQzqhN/2018-VMO.png

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ohiorizzler1434

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ohiorizzler1434

#7 h Mar 31, 2025, 8:35 AM

Y by

Solved in 25 minutes.

We start by observing that by angle conditions $(MEFN)$ is cyclic. Then, note that by death star lemma, $DK$ and $DH$ are angle bisectors giving $KDH$ collinear. We observe that by radical axis on $(EDPM)$ , $(FDPN)$ , and $(EFMN)$ that $A$ , $D$ , $P$ are collinear. Now, note that by similarity of $EDM$ and $FDN$ , that $\angle EKD = \angle FHD$ , implying $AK=AH$ by isosceles, and now radical axis on the circle tangent to $AM$ , $AN$ at $K$ , $H$ (exists by ice cream cone), $(KDQ)$ , and $(HDQ)$ implies that $A,D,Q$ collinear. Putting this together gives $A,D,Q,P$ collinear.

Now, we see that $A,K,Q,H$ is concyclic as $\measuredangle KQH = \measuredangle AKD + \measuredangle DHA = -\measuredangle HAK$ by tangencies. Construct the point $T$ such that the spiral symmetry at $G$ sends $ETF$ to $KDH$ . We observe that $\measuredangle GLD = \measuredangle GLA = \measuredangle GFA = \measuredangle GFH = \measuredangle GTD$ giving $GTLD$ cyclic. Now $\measuredangle GQD = \measuredangle GQA = \measuredangle GHA = \measuredangle GHF = \measuredangle GDT$ , giving the tangency.

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