Concurrent Lines Problem from a TST

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Concurrent Lines Problem from a TST

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Source: 2018 China TST 3 Day 2 Problem 5

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#1 h Mar 27, 2018, 1:57 PM • 3 Y

Y by anantmudgal09, Adventure10, Mango247

Let $ABC$ be a triangle with $\angle BAC > 90 ^{\circ}$ , and let $O$ be its circumcenter and $\omega$ be its circumcircle. The tangent line of $\omega$ at $A$ intersects the tangent line of $\omega$ at $B$ and $C$ respectively at point $P$ and $Q$ . Let $D,E$ be the feet of the altitudes from $P,Q$ onto $BC$ , respectively. $F,G$ are two points on $\overline{PQ}$ different from $A$ , so that $A,F,B,E$ and $A,G,C,D$ are both concyclic. Let M be the midpoint of $\overline{DE}$ . Prove that $DF,OM,EG$ are concurrent.

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#2 h Mar 27, 2018, 4:43 PM • 6 Y

Y by den_thewhitelion, fastlikearabbit, Math-Ninja, Muradjl, Adventure10, Mango247

Here's How to Crack It
Let $OQ$ intersect $AC$ at $S$ and $BC$ at $L$ and let $R$ be the midpoint of $AB$ and let $OP$ intersect $BC$ at $Z$ . It is well-known that $PL \perp QL$ and respectively that $ALOBP$ is cyclic, and respectively that $QLZP$ lie on a circle with diameter $PQ$ . Let $G_1$ be the foot of the perpendicular from $L$ to $CQ$ , and let $ES \cap PQ= G'$ , note that the pairs of line $\{\overline{ES} ,\overline {G_1S}\}$ and $\{\overline{QG_1},\overline{QG'}\}$ are reflections over $\overline{OQ}$ , thus $G'$ is the reflection of $G_1$ over $OQ$ , respectively $LG' \perp PQ$ , now by Reims on the circles $(LG'PD)$ and $(LAPB)$ we have that $DG' || AB$ , now by Reims on $(CAAB)$ and $(CG'AD)$ and $DG' || AB$ we have that $A,G',C,D$ are concyclic, thus $G' \equiv G$ , note that now suffices to show that $\triangle{OED}$ and $\triangle{MSR}$ are perspective, thus it is enough to show that $OM$ bisects $RS$ , thus it is enough to show that $M$ is the midpoint of $LZ$ , but this follows directly from the facts that $P,Z,L,Q$ lie on a circle with diameter $PQ$ and that the perpendicular from $M$ to $BC$ bisects $PQ$ , the later one is true because $QPDE$ is a trapezoid.

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#3 h Mar 28, 2018, 10:18 AM • 2 Y

Y by Adventure10, Mango247

rmtf1111 wrote:

by Reims on $(CAAB)$ and $(CG'AD)$ and $DG' || AB$ we have that $A,G',C,D$ are concyclic.

Can you please explain this step?

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#4 h Mar 28, 2018, 10:25 AM • 2 Y

Y by Adventure10, Mango247

"Reim's" Is a theorem.
You can find it Here

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#5 h Mar 29, 2018, 3:56 AM • 3 Y

Y by YUANYNU, Adventure10, Mango247

My English is not good, so I can't take part in the discussion of many problems. Please look at the picture.

Attachments:

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#6 h Mar 30, 2018, 8:09 PM • 2 Y

Y by e_plus_pi, Adventure10

We use areals. The equation of the $A$ tangent is $b^2 z+c^2 y=0$ and similarly for $B,C$ so:
$P=(a^2,b^2,-c^2) \, , \, Q=(a^2,-b^2,c^2)$ We then find $D$ :
$\begin{vmatrix} 0 & y & z \\ -a^2 & S_C & S_B \\ a^2 & b^2 & -c^2 \end{vmatrix}=0 \Leftrightarrow y S_A=-z (b^2+S_C)$ $D=(0,b^2+S_C,-S_A) \, , \, E=(0,-S_A,S_B+c^2)$ We now observe that:
$\measuredangle DEF=\measuredangle BEF=\measuredangle BAP=\measuredangle BCA \implies EF \parallel AC$ This saves us having to use circle formulae when finding $F,G$ (which is still doable even without the synthetic observation just messy). So to find $F$ :
$\begin{vmatrix} k & -b^2 & c^2 \\ 1 & 0 & -1 \\ 0 & -S_A & S_B+c^2 \end{vmatrix}=0 \Leftrightarrow k S_A=-(c^2 S_C+b^2S_B)$ $F=(b^2 S_B+c^2 S_C,-b^2 S_A,c^2 S_A) \, , \, G=(b^2 S_B+c^2 S_C,b^2 S_A,-c^2 S_A)$ We know let $Z=(b^2 S_B+c^2 S_C,S_{AC},S_{AB})$ and claim all the lines pass through $Z$ ( $Z$ can be found by intersecting two lines or guessing the concurency point is on the $A-$ altitude from a diagram):
$\text{ZFD colinear} \Leftrightarrow 0=\begin{vmatrix} b^2 S_B+c^2 S_C & S_{AC} & S_{AB} \\ b^2 S_B+c^2 S_C & -b^2 S_A & c^2 S_A \\ 0 & b^2+S_C & -S_A \end{vmatrix}= \begin{vmatrix} 0 & S_A (S_C+b^2) & -S_{A^2} \\ b^2 S_B+c^2 S_C & -b^2 S_A & c^2 S_A \\ 0 & b^2+S_C & -S_A \end{vmatrix}= \begin{vmatrix} 0 & 0 & 0 \\ b^2 S_B+c^2 S_C & -b^2 S_A & c^2 S_A \\ 0 & b^2+S_C & -S_A \end{vmatrix}$ So $Z$ lies on $DF$ and similarly on $EG$ . It remains to check it is on $OM$ :
$M=S_B (0,b^2+S_C,-S_A) +S_C(0,-S_A,S_B+c^2)=(0,a^2 S_C+b^2(c^2-b^2),a^2 S_B+c^2(b^2-c^2))$ $\text{OMZ colinear} \Leftrightarrow 0=\begin{vmatrix} b^2 S_B+c^2 S_C & S_{AC} & S_{AB} \\ a^2 S_A & b^2 S_B & c^2 S_C \\ 0 & b^2 S_B+S_{BC}-S_{AC} & c^2 S_C+S_{BC}-S_{AB} \end{vmatrix}= \begin{vmatrix} b^2 S_B+c^2 S_C & S_{AC} & S_{AB} \\ a^2 S_A & b^2 S_B & c^2 S_C \\ 2S_{BC} & S_{BC} & S_{BC} \end{vmatrix}$ $\cdots \Leftrightarrow 0=\begin{vmatrix} b^2 S_B+c^2 S_C & S_{AC} & S_{AB} \\ a^2 S_A & b^2 S_B & c^2 S_C \\ 2 & 1 & 1 \end{vmatrix}=a^2 b^2(S_{B^2}-S_{A^2})+b^2 c^2(S_{C^2}-S_{B^2})+a^2c^2(S_{A^2}-S_{C^2})=a^2b^2c^2(a^2-b^2+b^2-c^2+c^2)$ Hence $OM$ also passes through $Z$

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#7 h Mar 30, 2018, 10:41 PM • 4 Y

Y by xdiegolazarox, k12byda5h, Adventure10, Mango247

Let $OP,OQ \cap BC = S,T, BP\cap CQ=R$ . Then from the perspective of $\triangle PQR$ , we know $S$ is the intersection of the $P$ -external angle bisector and the $R$ -external touch chord, hence it's well-known that $S\in (PQ)$ , as with $T$ . Furthermore, by Reim, since $AABC, ABFE$ are cyclic, we have $EF||AC$ and similarly $DG||AB$ . Then by Reim again we have $DEFG$ cyclic. Furthermore, from earlier we get $QS||DG$ , so by Reim we have $QSEF$ cyclic, so $SF\perp PQ$ and similarly $TG\perp PQ$ .

Now we consider from the perspective of triangle $OPQ$ , so $A,S,T$ are the feet of the altitudes. Define $Z=DF\cap EG$ . Since $DEFG, PSTQ$ are cyclic, we deduce $Z,O$ lie on the radical axis of $(PDSF), (GTEQ)$ . Furthermore, letting $N$ be the midpoint of $PQ$ , we have $NM\perp ST$ , hence $M$ is the midpoint of $ST$ as well as $DE$ , so $MD\cdot MS=MT\cdot ME$ , so $M$ lies on the radical axis too as desired.

Remark: We can also prove that $(DEFG)$ has diameter $DE$ using the above configuration.

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#8 h Apr 6, 2018, 1:58 PM • 1 Y

Y by Adventure10

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I'll alter the wording a bit because that's how I did it on scratch paper

Photaesthesia wrote:

Let $ABC$ be a triangle with $\angle BAC > 90 ^{\circ}$ , and let $O$ be its circumcenter and $\omega$ be its circumcircle. The tangent line of $\omega$ at $A$ intersects the tangent line of $\omega$ at $B$ and $C$ respectively at point $K$ and $L$ . Let $D,E$ be the feet of the altitudes from $K,L$ onto $BC$ , respectively. Suppose $F,G$ are two points on $\overline{KL}$ different from $A$ , so that $A,F,B,E$ and $A,G,C,D$ are both concyclic. Let $M$ be the midpoint of $\overline{DE}$ . Prove that the lines $\overline{DF}, \overline{OM}, \overline{EG}$ concur.

Let $P$ be the foot of $A$ -altitude in $\triangle ABC$ , let $X=\overline{DF} \cap \overline{EG}$ and $T=\overline{DG} \cap \overline{EF}$ . Note that $\overline{DG} \parallel \overline{AB}$ and $\overline{EF} \parallel \overline{AC}$ hence antiparallelism of $\overline{AB}, \overline{AC}$ in angle $\measuredangle (\overline{KL}, \overline{BC})$ gives that $DEGF$ is cyclic.

Lemma. $APTX$ is a line (and also a harmonic quadruple).

(Proof) Let $S=\overline{DE} \cap \overline{FG}$ and observe that $\overline{TX}$ is the polar of $S$ in $\odot(DEGF)$ . Note that $(F,G; A,S)=(K,L;A,S)=(D,E;P,S)=-1$ hence $A, P$ also lie on the polar so $S$ with respect to $\odot(DEGF)$ . $\blacksquare$

Lemma: $\overline{DE}$ is a diameter of $\odot(DEGF)$ .

(Proof) In $\triangle XDE$ we note that point $T$ lies on the $X$ -altitude and $\angle XDT=\angle XET$ hence $T$ is the orthocenter of this triangle (standard proof with isogonal conjugates). Hence $\angle DFE=\angle DGE=90^{\circ}$ . $\blacksquare$

Now let $J$ be the circumcenter of $\triangle XDE$ , $Y$ be the reflection of $X$ in $J$ , point $J'$ it's reflection in side $\overline{DE}$ and $I=\overline{XJ} \cap \overline{DE}$ . Define point $O'$ such that $XIO'A$ is a parallelogram.

Lemma: $O' \equiv O$ and $O$ lies on line $\overline{XM}$ .

(Proof) Note that $\overline{AO'} \perp \overline{FG}$ hence $A, O, O'$ are collinear. Note that $\triangle TDE$ and $\triangle ABC$ are homothetic at point $P$ . In particular, $P$ lies on the line $\overline{J'O}$ . Now $\frac{MJ'}{IO'}=\frac{MJ}{XA}=\frac{1}{2} \cdot \frac{XT}{XA}$ while $\frac{PM}{PO}=\frac{XJ}{XI}=\frac{1}{2} \cdot \frac{XY}{XI}.$ Since $XFPG$ and $XEYD$ are similar quadrilaterals, these two ratios are equal. Hence $P, J', O'$ are collinear. Thus, $O \equiv O'$ .

Now $AXIO$ is a parallelogram. Now we show that $\overline{XM}$ bisects segment $\overline{IA}$ . Note that $DTEY$ is a parallelogram and $\overline{XM}$ bisects $\overline{XY}$ . Now similarity of $XGTF$ and $XDYE$ establishes that $\frac{XA}{XT}=\frac{XI}{XY} \implies \overline{XY} \parallel \overline{AI}$ hence the lemma follows. $\blacksquare$

Evidently, $\overline{DF}, \overline{OM}, \overline{EG}$ concur at point $X$ ; thus, completing the proof!

Remark: It is worth mentioning how we obtain $(F,G;A,S)=-1$ ; the other two cross-ratios are standard. Observe that $\frac{FA}{FS}=\frac{CE}{SE}$ hence $\frac{FA}{FS} \div \frac{GA}{GS}=\frac{CE}{BD} \div \frac{SE}{SD}=\frac{LC}{KB} \div \frac{SL}{SK}=\frac{LA}{KA} \div \frac{SL}{SK}=-1.$

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1000 posts

#9 h Apr 12, 2018, 3:34 PM • 5 Y

Y by Anar24, huricane, Adventure10, Mango247, Anancibedih

Wow why such hard solutions for that easy problem?
Mine is basically same with @tastymath 's.Since it took me long to solve it I am going to post my solution.
Let $OQ\cap BC=X$ , $OP\cap BC=Y$ .An easy angle chasing yields that $PQXY$ and $DGFE$ are cyclic and by PoP $OP\cdot OY=OQ\cdot OX$ ... $(*)$
Again angle-chase to get $QFYE$ is cyclic so $\angle PDY=90^{\circ}=\angle QEY=\angle PFY$ $\implies$ $PDYF$ is cyclic.Similarly $QGXE$ is cyclic.
Claim: $DX=EY$ (i.e $M$ is also midpoint of $XY$ )
Proof:Easy Sine Bash in triangles $PDX,PQX,PEY,PQY$ .
Radical axis on $DGFE,PDYF,QGXE$ yields that $DF\cap EG$ lies on radical axis of $PDYF$ and $QGXE$ .Because of $(*)$ $O$ lies on radical axis of the two circles.So we need to prove that $M$ also lies on radical axis which is equivalent with $MD\cdot MY=ME\cdot MX$ which is clear since $M$ is midpoint of $XY$ and $DE$ .

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#10 h Apr 13, 2018, 3:53 PM • 2 Y

Y by Adventure10, Mango247

Photaesthesia wrote:

Let $ABC$ be a triangle with $\angle BAC > 90 ^{\circ}$ , and let $O$ be its circumcenter and $\omega$ be its circumcircle. The tangent line of $\omega$ at $A$ intersects the tangent line of $\omega$ at $B$ and $C$ respectively at point $P$ and $Q$ . Let $D,E$ be the feet of the altitudes from $P,Q$ onto $BC$ , respectively. $F,G$ are two points on $\overline{PQ}$ different from $A$ , so that $A,F,B,E$ and $A,G,C,D$ are both concyclic. Let M be the midpoint of $\overline{DE}$ . Prove that $DF,OM,EG$ are concurrent.

Where can we use $\angle BAC>90?$
If $\angle BAC<90,$ the problem is also true.

This post has been edited 1 time. Last edited by falantrng, Apr 13, 2018, 3:53 PM

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#11 h Apr 13, 2018, 4:12 PM • 1 Y

Y by Adventure10

They want two points on segment $BC$ rather than line $BC$ , I guess.

oops typoed last time

This post has been edited 1 time. Last edited by wu2481632, Apr 13, 2018, 4:15 PM

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Anar24

#12 h Jun 7, 2018, 11:16 AM • 2 Y

Y by Adventure10, Mango247

tenplusten wrote:

Wow why such hard solutions for that easy problem?
Mine is basically same with @tastymath 's.Since it took me long to solve it I am going to post my solution.
Let $OQ\cap BC=X$ , $OP\cap BC=Y$ .An easy angle chasing yields that $PQXY$ and $DGFE$ are cyclic and by PoP $OP\cdot OY=OQ\cdot OX$ ... $(*)$
Again angle-chase to get $QFYE$ is cyclic so $\angle PDY=90^{\circ}=\angle QEY=\angle PFY$ $\implies$ $PDYF$ is cyclic.Similarly $QGXE$ is cyclic.
Claim: $DX=EY$ (i.e $M$ is also midpoint of $XY$ )
Proof:Easy Sine Bash in triangles $PDX,PQX,PEY,PQY$ .
Radical axis on $DGFE,PDYF,QGXE$ yields that $DF\cap EG$ lies on radical axis of $PDYF$ and $QGXE$ .Because of $(*)$ $O$ lies on radical axis of the two circles.So we need to prove that $M$ also lies on radical axis which is equivalent with $MD\cdot MY=ME\cdot MX$ which is clear since $M$ is midpoint of $XY$ and $DE$ .

This is indeed one of the most beautiful solution that I HAVE EVER SEEN!!!!!CONGRATULATIONS!!!!

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#13 h Jun 7, 2018, 12:43 PM • 2 Y

Y by Adventure10, Mango247

sbealing wrote:

$....0=a^2 b^2(S_{B^2}-S_{A^2})+b^2 c^2(S_{C^2}-S_{B^2})+a^2c^2(S_{A^2}-S_{C^2})=a^2b^2c^2(a^2-b^2+b^2-c^2+c^2)$ Hence $OM$ also passes through $Z$

Here is a minute typo, in the last bracket it should be $a^2b^2c^2(a^2-b^2+b^2-c^2+c^2-a^2)$
Ignoring this, it is a brilliant solution in my opinion

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#14 h Jun 7, 2018, 11:45 PM • 2 Y

Y by Adventure10, Mango247

Dear MLs

Extend DF and EG to intersect the given circles
in Y and Z collinear with O. Circle XYZ with center U
is congruent to circle ABC. XW is radical axis of
O1 and O2 and X-altitude in XYZ as well. Center line
O1O2 is mediatrix of OU. H is orthocenter and A is Euler
point in XYZ.MO is X-median. 9pc has AO as diameter.

Friendly,
M.T.

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111 posts

#15 h Jul 9, 2019, 9:47 AM • 3 Y

Y by top1csp2020, Adventure10, Mango247

DF-EG=K, PQ-BC=H, I,J-midpoint AC,AB so HE/HD=PD/EQ=BD/EC (PDB~QEC). By Reim EF//AC, let DJ-AP=F' by Mene EC/EH=DB/DH=F'A/F'H so F=F' so DFE=180-ACB-JPB=90 so F,G on (DE). OJ//IK, OI//JK so KO cuts S-midpoint IJ but IJ//ED so KM also cuts S so q.e.d

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290 posts

jred

#16 h Sep 22, 2019, 3:56 AM • 2 Y

Y by Adventure10, Mango247

ylmath123 wrote:

My English is not good, so I can't take part in the discussion of many problems. Please look at the picture.

It‘s weird that one can find two $'Q’$ s in the diagram. :blush:

:blush:

Anyway, nice solution!

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65 posts

Arefe

#17 h May 15, 2020, 11:26 AM • 1 Y

Y by Mango247

It's easy to see that $GD||AB$ , $FE||AC$ $(1)$ . We can see that $DFGE$ is cyclic .
Get the intersection of $DG$ and $FE$ , $H$ .
We can show that $\triangle{FHD}{\sim}\triangle{DBP}$ (it's easy to check that $\frac{BD}{FG}=\frac{AP}{DE}$ ) , so ${EF}\perp{DF}$ , ${DG}\perp{GE}$ .
If $I$ is the intersection of $DF$ , $EG$ . Name $K$ intersection of $AH$ with $BC$ .With use of $K$ and $(1)$ it's easy to see that ${AH}\perp{BC}$ so we can see $I$ , $A$ , $H$ are collinear .
If we get $R$ and $N$ be the intersection of ${FD},{PO}$ and ${EG},{OQ}$ , it's easy to check that $RN||DE$ and $IRON$ is parallelogram and the problem is finished easily

This post has been edited 2 times. Last edited by Arefe, May 15, 2020, 11:57 AM

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308 posts

#18 h Jul 21, 2020, 2:11 AM • 1 Y

Y by Rounak_iitr

Let $FG$ and $DE$ meet at $Y$ . $DF$ and $EG$ meet at $X$ .
CLAIM 1. $D,E,G,F$ are concyclic.
Proof. Using power of a point
$YB\times YE=YF\times YA$ $YD\times YC=YA\times YG$ $YB\times YC=YA^2$ multiplying them we have $YD\times YE=YF\times YG$ . Therefore by converse of power of a point we have $D,E,G,F$ are conyclic. This proves CLAIM 1.

CLAIM 2. $AB\|DG$ and $AC\|FE$
Proof. This is just angle chasing.
$\angle FAB=\angle FEB=\angle FED=\angle FGD$ which implies that $AB\|DG$ . Similarly $FE\|AC$ .

CLAIM 3.
Proof. $\angle DFE=\angle DGE=90^{\circ}$ .
Let $N$ and $R$ be the midpoint of $AB$ and $AC$ . Since $\angle PNB=\angle PDB=90^{\circ}$ , $P,N,D,B$ are concyclic. Now
$\angle GAR=\angle QAC=\angle ABC=\angle NBD$ $\frac{AG}{BD}=\frac{YA}{YB}=\frac{AC}{AB}=\frac{AR}{AN}=\frac{AR}{BN}$ hence $\triangle AGR\sim\triangle BDN$ Therefore
$\angle ARG=\angle BND=\angle BPD=90^{\circ}-\angle PBD=90^{\circ}-\angle QCE=\angle EQC=\angle ERC$ hence $G,R,E$ are collinear, as a result
$\angle QEG=\angle QER=\angle QCR=\angle ABC$ Since $QE\perp BC$ and $AB\|DG$ we have $DG\perp GE$ . Similarly $DF\perp FE$ .

CLAIM 4. Let $H_1$ be the projection of $A$ on $BC$ . Then $AH_1$ , $EF$ and $DG$ are concurrent.
Proof.Notice that $\triangle PBD\sim \triangle QCE$ . Hence
$\frac{BD}{EC}=\frac{PB}{QC}=\frac{PB}{BA}\cdot\frac{CA}{QC}\cdot\frac{BA}{CA}=\frac{2\cos B}{2\cos C}\cdot\frac{\sin C}{\sin B}=\frac{\tan C}{\tan B}=\frac{H_1 B}{H_1C}$ Therefore if $DG$ meet $AH_1$ at $H$ then
$\frac{HH_1}{HA}=\frac{H_1D}{DB}=\frac{H_1E}{EC}$ . Hence $E,H,F$ are collinear.

Now notice that $H_1$ is the homothetic center of $\triangle DHE$ and $\triangle BAC$ . Hence the same homothety sends $J$ ,the circumcenter of $\triangle DHE$ to $O$ .

CLAIM 5. $\frac{OJ}{IH_1}=\frac{JM}{XH_1}$
Proof. From the above homothety,
$\frac{OJ}{IH_1}=\frac{AH}{AH_1}=\frac{AH}{AF}\cdot \frac{AH_1}{AF}=\frac{\sin\angle XDE}{\cos\angle DEX}\cdot\frac{\sin 2\angle XDE}{\cos\angle DXE}=\frac{\cos\angle DEX}{2\cos\angle XDE\cos\angle DXE}$ Now notice that $(DHE)$ and $(XDE)$ has the same circumradius, denote it by $r$ . Then
$JM=R\cos\angle DXE$ $XH_1=2R\cos\angle XDE\cos \angle DXE$ Comparing the above identites we have proved claim 5.

Now since $JM$ and $XH_1$ are both perpendicular to $BC$ , they are parallel. Therefore a homothety at $O$ sends $MJ$ to $H_1J$ which shows that $I,M,X$ are collinear.

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#19 h Mar 23, 2021, 11:06 AM

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Let $DE$ meet $PQ$ at $T$ . By Reim's theorem, it follows that $EF\parallel AC$ , and $DG\parallel BA$ . This again by Reim's theorem shows that $D,E,F,G$ are concyclic. We now claim that $(T,A;F,G)=-1$ . This is because
$\frac{AF}{FT}\div \frac{AG}{GT}=\frac{CE}{ET}\div \frac{BD}{DT}=\frac{CE}{BD}\cdot \frac{DT}{ET}=\frac{QE}{PD}\cdot \frac{PD}{QE}=-1$ because $A$ lies inside segment $FG$ which in turn lies inside segment $BC$ . Now projecting through $E$ shows that
$-1=(T,A;F,G)=(C,A;P_\infty,EG\cap CA)$ so $EG$ cuts $CA$ at the midpoint of $CA$ , say $I$ . Similarly, $FD$ cuts $AB$ at the midpoint of $AB$ , say $J$ . Now let $H$ be the foot of perpendicular from $A$ to $H$ . Since $(T,A;P,Q)=(T,H;D,E)=-1$ , it follows that $EG$ , $FD$ and $AH$ are concurrent at a point $K$ . Moreover, as $DEFG$ is cyclic, it follows that $\angle EFD=\angle EGD=90^\circ$ . In particular, this shows that $OP\parallel EG$ and $OQ\parallel DF$ . Let $L$ be the midpoint of $IJ$ . Since $OIKJ$ is a parallelogram, $O,K,L$ are collinear. Since $IJ\parallel ED$ , by homothety at $K$ one can also deduce that $K,L,M$ are collinear. Hence $OM$ passes through $K$ as desired.

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#21 h Nov 8, 2024, 9:48 PM

Y by

Lemma: $ABC$ is a triangle with orthocenter $H$ and altitudes $AD,BE,CF$ . $M$ is the midpoint of $BC$ and $S$ is the intersection of $EF$ and $AD$ . Perpendicular from $S$ to $AC,AB$ intersect $BC$ at $X,Y$ . Prove that circumcenter of $(XYS)$ lies on $AM$ .
Proof: Let $N$ be the midpoint of $AH$ and $W,V$ be the circumcenters of $(BHC)$ and $(XYS)$ . Let $M'$ be the image of $M$ under the homothety centered at $D$ with radius $\frac{DH}{DS}$ . $N,A$ swap under this homothety. Let $O$ be the circumcenter of $(ABC)$ .
$\frac{DM'}{MM'}=\frac{DN}{NA}=\frac{DN}{OM}=\frac{DN}{WM}$ Thus, $N,M',W$ are collinear. This yields the collinearity of $A,M,V$ . $\square$

Let $DF\cap EG=K,DG\cap EF=L,PQ\cap BC=T$ .
$\measuredangle BCA=\measuredangle BAP=\measuredangle BAF=\measuredangle BEF$ Hence $EF\parallel AC$ . Similarily, we conclude that $DG\parallel AB$ .
$\frac{TF}{FA}=\frac{TE}{EC}=\frac{TD}{DB}=\frac{TG}{GA}$ Which implies $(T,A;F,G)=-1$ . So $K,A,L$ are collinear. Let $\overline{KAL}\cap BC=S$ . If $S'$ is the altitude from $A$ to $BC$ , then $(T,S;D,E)=-1=(T,A;P,Q)=(T,S;D,E)$ or $S=S'$ . Hence $AS\perp BC$ . Let the feet of the altitudes from $D,E$ to $EK,DK$ be $G',F'$ .
$(G'F'\cap BC,S;D,E)=-1=(T,S;D,E)$ Thus, $G'F'$ passes through $T$ . Also $TF.TG=\frac{TE.TA}{TC}.\frac{TD.TA}{TB}=TD.TE$ hence $D,E,F,G$ are concyclic. Since $F'G'$ and $FG$ are antiparallel to $DE$ and these lines intersect, $F'G'\equiv FG$ or $F'=F,G'=G$ . This gives that $(K,L,D,E)$ is an orthogonal system. Problem follows from Lemma as desired. $\blacksquare$

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