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.

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?

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

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

Attachments:

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 $Z$

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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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 $$\frac{FA}{FS} \div \frac{GA}{GS}=\frac{CE}{BD} \div \frac{SE}{SD}=\frac{LC}{KB} \div \frac{SL}{SK}=\frac{LA}{KA} \div \frac{SL}{SK}=-1.$$

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 $.

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.

They want two points on segment $BC$ rather than line $BC$, I guess. oops typoed last time

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!!!!

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

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.

Attachments:

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

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. Anyway, nice solution!

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

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.

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.

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.$\blacksquare$