orl wrote: Let $ ABC$ be a triangle with $ G$ as its centroid. Let $ P$ be a variable point on segment $ BC$. Points $ Q$ and $ R$ lie on sides $ AC$ and $ AB$ respectively, such that $ PQ \parallel AB$ and $ PR \parallel AC$. Prove that, as $ P$ varies along segment $ BC$, the circumcircle of triangle $ AQR$ passes through a fixed point $ X$ such that $ \angle BAG = \angle CAX$. An ugly solution . The first I will display two lemmas . Lemma 1 : A,B is two fixed point on $ O_x,O_y$ Let M,N varies on $ O_x,O_y$ and satisfy : $ \frac{OA}{OM}+\frac{OB}{ON}=1$ Then the line MN pass through a fixed point X such that $ OAXB$ is a parallelogram. Proof : Have some solutions for this problem . A simple way is draw $ A_t||O_y$ ,it cuts $ PQ$ at the point X . Apply Thales law we have : $ BX||O_x$ ,therefore X is a fixed point . Problem claim . Lemma 2 : A,B is fix points on $ O_x,O_y$ Let M,N varies on $ O_x,O_y$ and satisfy $ \frac{OM}{OA}+\frac{ON}{OB}=1$ then the circles $ (OMN)$ pass through a fixed point T such that $ OT,OX$ conjugate with $ \angle{xOy}$ Proof : Invert pole O with $ k=OA.OB$ .Then we have : $ A\to A',B\to B' ,M\to M',N\to N'$ Then $ (OMN)\to M'N'$ It becomes lemma 1 . The lines $ M'N'$ pass through the point X' which $ OA'X'B'$ is a parallelogram . Therefore $ (OMN)$ pass through X that is the imagine of X' . Because $ OA=OA',OB=OB'$ so that $ OT,OX$ conjugate with $ \angle{xOy}$ Now apply to our problem . Easy to check that $ \frac{AR}{AB}+\frac{AP}{AB}=1$ therefore the circle pass through fixed point X . Morethan $ OX,OG$ conjugate with $ \angle{BAC}$ so we have : $ \angle{BAG}=\angle{CAX}$

Dear Tung, your proof is not ugly at all. It is really nice. In addtion, the fixed point X lies on segment HK where H is the orthocenter and K is the midpoint of median AM.

let $ \alpha = \angle BAG$, and let $ \frac {BP}{PC} = \frac {x}{y}$... let $ l$ be the line through $ A$ such that $ l$ and line $ AG$ are isogonal lines, and let it intersect the circumcircle of triangle $ AQR$ at $ T$... from ptolemy's theorem we have that $ AQ\dfrac{\sin(A - \alpha)}{\sin A} + AR\dfrac{\sin\alpha}{\sin A} = AT$... it's easy to obtain that $ \dfrac{\sin(A - \alpha)}{\sin A} = \dfrac{c}{2m_a}$ and $ AQ = \dfrac{bx}{x + y}$... from these it's easy to conclude that $ \dfrac{bc}{2m_a} = AT$, which is fixed, and we're done

orl wrote: Let $ ABC$ be a triangle with $ G$ as its centroid. Let $ P$ be a variable point on segment $ BC$. Points $ Q$ and $ R$ lie on sides $ AC$ and $ AB$ respectively, such that $ PQ \parallel AB$ and $ PR \parallel AC$. Prove that, as $ P$ varies along segment $ BC$, the circumcircle of triangle $ AQR$ passes through a fixed point $ X$ such that $ \angle BAG = \angle CAX$.

Just an outline.First,consider the midpoints of $AB$ and $AC$,$M$ and $N$ and let $X$ be the intersection of the circumcircle of $ABC$ with the symmedian of $ABC$.Now,it is easy to prove $XM/XN=AB/AC$ and now from this we just need to prove that for any two points $Q$ and $R$ with the point $P$ that $XMQ$ is similar to $XNR$,but this is an easy lenght chase,so we are finished.

We can rewrite the problem as following: Let $ P_1, P_2, P_3 $ be three points on $ BC $ . Let $ Q_1, Q_2, Q_3 $ be three points on $ AC $ satisfy $ P_1Q_1\parallel P_2Q_2\parallel P_3Q_3\parallel AB $ . Let $ R_1, R_2, R_3 $ be three points on $ AB $ satisfy $ P_1R_1\parallel P_2R_2\parallel P_3R_3\parallel AC $ . Then $ \odot (AQ_1R_1), \odot (AQ_2R_2), \odot (AQ_3R_3) $ are coaxial with common radical axis is A -symmedian . Proof: Let $X \equiv \odot (AQ_1R_1) \cap \odot (AQ_3R_3) $ and $ Y \equiv AX \cap \odot (ABC) $ . (i.e. $ X $ is the Miquel point which maps $ Q_1Q_3 $ to $ R_1R_3 $ ) Since $ \frac {Q_1Q_2}{Q_2Q_3}= \frac {P_1P_2}{P_2P_3}=\frac {R_1R_2}{R_2R_3} $ , so $ X $ is also the Miquel point maps $ Q_1Q_2 $ to $ R_1R_2 $ . ie. $ X\equiv \odot (AQ_1R_1) \cap \odot (AQ_2R_2) \cap \odot (AQ_3R_3) $ Consider the case when $ P \equiv B $ and $ P \equiv C $ , then we get $ X $ is the Miquel point which map $ CA $ to $ AB $ . i.e. $ X $ is a point satisfy $ \triangle XCA \sim \triangle XAB $ Since $ \angle XAC=\angle XBA=\angle YBC , \angle XCA=\angle XAB=\angle YCB $ , so we get $ \triangle XAC \sim \triangle XBA \sim \triangle YBC $ , hence $ AB\cdot CY=AX\cdot BC=AC\cdot BY $ . i.e. $ ABYC $ is a harmonic quadrilateral and $ AX $ is A-symmedian Q.E.D ____________________________________________________________ Remark: (1) Since $ XY $ bisect $ \angle BXC $ , so from the property of harmonic quadrilateral we get $ X $ is the midpoint of $ AY $ . (2) This problem has something to do with Arzt parabola Since $ QR $ is tangent to $ A $ -Arzt parabola when $ P $ varies on $ BC $ , so $ \odot (APQ) $ passes through the focus $ X $ of A -Arzt parabola .

Yes, it does Telv. Note that $QR$ is simply tangent to a fixed parabola (affine homography, so tangent at line at infinity). Hence, $APQ$ passes through a fixed point. Taking $P = B, C$, we have the point is the intersection of the circles through $B$ and $A$ tangent to $AC$ and $C$ and $A$ tangent to $AB$. We know this intersection lies on $A$-Symmedian.

Let the $A$-symmedian cut $\odot{AQR}$ at $X$ and $BC$ at $M'$ and $AM$ be the median.Then $\frac{RX}{XQ}=\frac{sin\angle{RAX}}{sin\angle{XAQ}}=\frac{sin\angle{MAC}}{sin\angle{MAB}}=\frac{AB}{AC}=\frac{BR}{AQ}$ where the last line follows from parallelity.This along with $\angle{BRX}=\angle{AQX} \implies \triangle{BRX} \sim \triangle{AQX} \implies \angle{AYB}=180-A$. The equality may also be written as $\frac{RX}{XQ}=\frac{AR}{CQ}$ and this in combination with $\angle{ARX}=\angle{XQC}$ implies that $\triangle{ARX} \sim \triangle{CQX}$ so once again $\angle{AYC}=180-A$.Thus we get $\angle{BXC}=2A$ or $X$ is actually the point of intersection of $\odot{BOC}$ with the $A$-symmedian,where $O$ is the circumcenter of $\triangle{ABC}$.This is indeed fixed,as desired.

I've changed the notations a bit- $D$ is a variable point on $BC$ and $P, Q$ lie on $AB, AC$ respectively such that $DP \parallel AC, DQ \parallel AB$. Let the tangents to $B$ and $C$ intersect at $U$. By letting $D$ be the midpoint of $BC$, we see that $X$ must be the other point of intersection of the $A$-symmedian and the circle with diameter $AO$, where $O$ is the circumcenter of $\triangle ABC$. Let $T$ be the said point. Note that $T$ also lies on the circumcircle of $\triangle BOC$ as $OT \perp TU$. Now, \[\angle TCB = \angle TUB = 180^{\circ} - \angle ABU - \angle BAU \\ = 180^{\circ} - \angle BAC - \angle ABC - \angle BAU = \angle ACB - \angle BAU,\] so $\angle BAT = \angle TCA$. Similarly, $\angle TBA = \angle TAC,$ so $\triangle TBA \sim \triangle TAC \implies \dfrac{AB}{AC} = \dfrac{AT}{CT}$. Now, we have that \[\dfrac{BD}{BC} = \dfrac{BP}{AB} \quad \dfrac{CD}{BC} = \dfrac{CQ}{AC} \\ \implies \dfrac{BP}{AB} + \dfrac{CQ}{AC}= 1 \implies \dfrac{AP}{CQ} = \dfrac{AB}{BC} = \dfrac{AT}{TC} \\ \implies \triangle APT \sim \triangle CTQ \implies \angle APT = \angle TQC = 180^{\circ} - \angle AQT,\] so $T$ lies on $(APQ)$, as desired.

Here's a solution with a combination of synthetic geometry and complex numbers. Note that $AQ \parallel PR$ and $AR \parallel PQ \implies AQPR$ is a parallelogram. Furthermore, from $AC \parallel RP$, it follows that \[\triangle BAC \sim \triangle BRP \implies \frac{AB}{AC} = \frac{RB}{RP} = \frac{RB}{AQ} = \frac{AB - AR}{AQ} \implies AB \cdot AQ + AC \cdot AR = AB \cdot AC.\] Now, let $A$ be the origin of the complex plane, and let us denote in lowercase letters the complex numbers corresponding to points. The above relation can then be rewritten as $|b||q| + |c||r| = |b||c|.$ But note that all three terms in this equation have equal arguments; hence, we may remove the magnitudes to obtain $bq + cr = bc.$ Now, we will prove that $x = \tfrac{bc}{b + c}$ is the fixed point. First, we show that $\angle GAB = \angle CAX.$ This is equivalent to $\tfrac{g}{b} / \tfrac{c}{x} \in \mathbb{R}.$ Note that $g = \tfrac{b + c}{3}$, so that \[\frac{g}{b} \bigg/ \frac{c}{x} = \frac{\frac{b + c}{3}}{b} \bigg/ \frac{c}{\frac{bc}{b + c}} = \frac{b + c}{3b} \bigg/ \frac{b + c}{b} \in \mathbb{R},\] as desired. Now, in order to prove that $A, Q, R, X$ are concyclic, we need only prove that $\tfrac{q}{r} / \tfrac{q - x}{r - x} \in \mathbb{R}.$ Using the condition $bq + cr = bc$, we find that \[\frac{q - x}{r - x} = \frac{q - \frac{bc}{b + c}}{r - \frac{bc}{b + c}} = \frac{(b + c)q - bc}{(b + c)r - bc} = \frac{cq + (bq - bc)}{br + (cr - bc)} = \frac{cq - cr}{br - bq} = -\frac{c}{b}.\] Then since $A, R, B$ and $A, Q, C$ are collinear, it follows that \[\frac{q}{r} \bigg/ \frac{q - x}{r - x} = -\frac{q / b}{r / c} \in\mathbb{R},\] as desired. $\square$

Let $D$ and $T$ be the intersections of the $A$-symmedian with $(ABC)$ and $(ARQ)$. I claim that $T$ is the midpoint of $AD$. Consider the inversion $\mathcal{I}(A,1)$ and denote by $X^\prime$ the inverse of $X$. Obviously, $A-D^\prime-T^\prime,\ R^\prime-T^\prime-Q^\prime$ and $B^\prime-D^\prime-C^\prime$ are collinear. Note that $\dfrac{D^\prime B^\prime}{D^\prime C^\prime}=\dfrac{\frac{DB}{AD\cdot AB} }{\frac{DC}{AD\cdot AC}}=\dfrac{DB}{DC}\cdot \dfrac{AC}{AB}=1$, so $D^\prime$ is the midpoint of $B^\prime C^\prime$. From $\dfrac{AR}{RB}=\dfrac{QC}{QA}$ it is easy to infer that $\dfrac{AB^\prime}{B^\prime R^\prime}=\dfrac{C^\prime Q^\prime}{C^\prime A}$. Let $S$ be a point on $Q^\prime R^\prime$ such that $B^\prime S\parallel AQ^\prime$. We have then $\dfrac{SR^\prime}{SQ^\prime}=\dfrac{B^\prime R^\prime}{B^\prime A}=\dfrac{C^\prime A}{C^\prime Q^\prime}$, so $SC^\prime \parallel AR^\prime$. Therefore, $AB^\prime SC^\prime$ is a parallelogram, which implies $A-D^\prime-S$ collinear, so $S=T^\prime$. This implies $$AT^\prime=2AD^\prime\Leftrightarrow AD=2AT\Leftrightarrow (ARQ)\ \mathrm{passes\ through\ the\ midpoint\ of}\ AD$$

My solution: Let $\omega_1$ be a circle passes $A, B$ and tangents to $AC$. Let $\omega_2$ be a circle passes $A, C$ and tangents to $AB$. Let $X$ be intersection of $\omega_1$ and $\omega_2$, so $X$ is fixed. Since two triangle $ABX$ and $AXC$ are similar to each other, hence $\dfrac{AX}{BX}=\dfrac{AC}{AB}$. On the other hand, $\dfrac{AC}{AB}=\dfrac{PR}{BR}=\dfrac{AQ}{BR}$ so $\dfrac{AX}{BX}=\dfrac{AQ}{BR}$, also $\widehat{RBX}=\widehat{QAX}$ hence two triangle $BRX$ and $AQX$ are tangent to each other, therefore $\widehat{BRX}=\widehat{AQX}$, following that $A, Q, X, R$ is cyclic. Moreover, we have: $$\dfrac{\sin BAX}{\sin CAX}=\dfrac{\sin BAX}{\sin ABX}=\dfrac{BX}{AX}=\dfrac{AB}{AC}$$So $AX$ is the symmedian line of triangle $ABC$, which means $\widehat{BAG}=\widehat{CAX}$. We have done.

@fvlfollower why $d$ has to cancel out?

Reynan wrote: @fvlfollower why $d$ has to cancel out? $P= (0:d:a-d)$, so locus of point $P$ on side $BC$ depends on $d$ and this makes points $Q, R$, and the equation of $(AQR)$ contain $d$ as well. $(AQR)$ passes through a fixed point implies that there is a point on $(AQR)$ which its coordinates do not contain $d$.

oke thank you

Hmm...Let $E$, $F$, and $M$ by the midpoints of $AB$, $AC$, and $BC$, respectively. Let the $A$-symmedian intersect the circumcircle again at $D$. Let $(AEF)$ intersect the $A$-symmedian at point $X$. It is easy to see that $X$ is the midpoint of $AD$ due to the homothety $AEF$ to $ABC$. It suffices to show that $ARXQ$ is cyclic. Also note that $ARPQ$ is a parallelogram. Lemma 1: $X$ is the center of the spiral similarity that takes $BA$ to $AC$ Proof: It is easy to see that $\bigtriangleup ABD$ and $\bigtriangleup AMC$ are similar. Since $X$ and $F$ are midpoints of $AD$ and $AC$, respectively, $\bigtriangleup ABX \sim \bigtriangleup AMF$. Then, $\angle ABX = \angle AMF = \angle BAM = \angle CAX$. Similarly, $\angle ACX = \angle BAX$. This means $\bigtriangleup XBA \sim \bigtriangleup XAC$, and the lemma is proven. Lemma 2: $X$ is also the center of the spiral similarity that takes $ER$ to $FQ$ Proof: It suffices to show $\frac{AR}{AB} = \frac{CQ}{CA}$. So, $\frac{AR}{AB} = \frac{QP}{AB} = \frac{CP}{CB}$. On the other hand, $\frac{CQ}{CA} = 1 - \frac{QA}{CA} = 1 - \frac{PR}{CA} = 1 - \frac{BP}{BC} = \frac{CP}{CB}$, and the lemma is proven. Lemma 2 implies that $X$ is also the center of spiral similarity that takes $EF$ to $RQ$. Since $ER$ and $FQ$ intersect at $A$, $(AEF)$ and $(ARQ)$ concur at $X$. This implies $ARXQ$ is cyclic.

We use barycentric coordinates on $\triangle{ABC}$. We let $BP=m$ and $PC=m$. Noting that $\triangle{BRP} \sim \triangle{BAC}$ and $\triangle{CQP} \sim \triangle{CAB}$. It follows that $P=(0:n:m)$, $Q=(bn:0:ab-bn)$ and $R=(cm:ac-cm:0)$. We know the circle $(AQR)$ is of the form $$-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0$$, for some real numbers $u$, $v$, and $w$. Plugging in $A$, $Q$ and $R$, we find that the equation is $$-a^2yz-b^2xz-c^2xy+\frac{1}{a}(c^2my+b^2nz)(x+y+z)$$. The angle condition is easily translated to the fact that our magical point $X$, should it exist, lies on the $A$ symmedian of $\triangle{ABC}$, and as a result, we let $X=(t:b^2:c^2)$. Plugging $X$ into out circle formula and simplifying, we have $$-a^2b^2c^2-tb^2c^2-tb^2c^2+\frac{1}{a}(b^2c^2(m+n))(t+b^2+c^2)$$At this point, the crucial observation is that $m+n=a$, which deletes $m$ and $n$ from the equation entirely, implying that $t$, and in turn $X$ is independent of our choice of $P$. $\square$

It`s dumpty point

complex! Let A=0, B=b, C=c. Let the intersection of the symmedian of ABC and (ABC) be D. Since ABDC is a harmonic quad, BA * CD / BC *CA = -1. Therefore, (b-a)(c-d) / (b-d)(c-a) = -1. Simplyfying, D= 2bc/(b+c). Let X the midpoint of AD. X=bc/ (b+c). Additionally, since we know CQ/QA= AR/RB, let, R= b(1-k) and Q=ck for some constant k. Now we wish to show if AQRX is cyclic and we know all of their coordinates which simplifies out nicely!

yunseo wrote: complex! Let $A=0$, $B=b$, $C=c$. Let the intersection of the symmedian of $ABC$ and $(ABC)$ be $D$. Since $ABDC$ is a harmonic quad, $BA \cdot CD / BC \cdot CA = -1$. Therefore, $(b-a)(c-d) / (b-d)(c-a) = -1$. Simplyfying, $D= 2bc/(b+c)$. Let $X$ be the midpoint of $AD$. We see that $X=bc/ (b+c)$. Additionally, since we know $CQ/QA= AR/RB$, let, $R= b(1-k)$ and $Q=ck$ for some constant k. Now we wish to show if $AQRX$ is cyclic and we know all of their coordinates which simplifies out nicely! FTFY

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Here is my solution for this problem Solution Let $AD$ be $A$ - symmedian of $\triangle$ $ABC$ ($D$ $\in$ ($ABC$)); $X$ be second intersection of ($APQ$) with $AD$; $S$, $M$, $N$, $T$ be midpoint of $BC$, $AB$, $AC$, $AP$ It's easy to prove that: $\triangle$ $BDC$ $\sim$ $\triangle$ $RXQ$ But: $S$, $T$ are midpoint of $BC$, $RQ$, respectively then $\triangle$ $BDS$ $\sim$ $\triangle$ $RXT$ So: $\widehat{RXT}$ = $\widehat{BDS}$ = $\widehat{ABC}$ = $\widehat{AMN}$ or $X$ $\in$ ($TMR$) Hence: $X$ is Miquel point of completed quadrilateral $ARTN.MQ$ or $X$ $\in$ ($AMN$) Therefore: $X$ is midpoint of $AD$ or $X$ is fixed point which satisfies $\widehat{GAB}$ = $\widehat{CAX}$

orl wrote: Let $ ABC$ be a triangle with $ G$ as its centroid. Let $ P$ be a variable point on segment $ BC$. Points $ Q$ and $ R$ lie on sides $ AC$ and $ AB$ respectively, such that $ PQ \parallel AB$ and $ PR \parallel AC$. Prove that, as $ P$ varies along segment $ BC$, the circumcircle of triangle $ AQR$ passes through a fixed point $ X$ such that $ \angle BAG = \angle CAX$. solution(with amar_04 and RAMUGAUSS): I claim that $X$ is the midpoint of the $A$ symmedian chord in $\triangle ABC$. It is very well known that $X$ is the center of spiral similarity sending $BA \mapsto AC$. Now $$\frac{AR}{BR}=\frac{PC}{PB}=\frac{QC}{AQ}$$. Hence $X: R\mapsto Q \implies \triangle XRA \sim \triangle XQC \implies \angle XQC=\angle XRA \implies A,R,X,Q$ are conclyc

It suffices to show that there is a fixed point on the symmedian, or a point with coordinates $(t:b^2:c^2)$. Let $D = (0,m,n)$ where $m+n=1$. Note that $R = (n,m,0)$and $Q = (m,0,n)$. The equation of $(AQR)$ is \[ -a^2 yz - b^2xz -c^2xy+(x+y+z)(vy+wz)=0 \]where $v = c^2n$ and $w = b^2 m$. If we plug in a point $X = (t:b^2:c^2)$, we obtain \[ -a^2b^2c^2 - b^2c^2t - b^2c^2t + (b^2+c^2+t)(b^2c^2)(m+n) = 0 \]\[ -a^2 - 2t + b^2 + c^2 + t = 0 \]Therefore, $t = b^2 +c^2-a^2$, and there exists a fixed point on the symmedian, as desired.

It doesn't look like anyone posted a linearity solution yet, so here it is: Define a function $F(P)=pow_{(ARQ)}P-pow_{(ABC)}P,$ which we know is linear. Now let $D$ be the foot of the $A$-symmedian on $BC$, it suffices to show $F(D)$ is fixed. But $$F(D)= \frac{b^2}{b^2+c^2}F(B)+\frac{c^2}{b^2+c^2}F(C) = \frac{b^2}{b^2+c^2}(BR \cdot c)+\frac{c^2}{b^2+c^2}(CQ\cdot b )$$and by similar triangles, $BR = \frac{c}{a} BP$, $CQ = \frac{b}{a}CP$, so hence $$F(D)= \frac{b^2c^2}{a(b^2+c^2)} (BP + CP) = \frac{b^2c^2}{(b^2+c^2)} $$and we are done.

[asy][asy]defaultpen(fontsize(12pt));size(8cm);pair A=Drawing("A",dir(120),dir(120));pair B=Drawing("B",dir(210),dir(210));pair C=Drawing("C",dir(330),dir(330));pair P=Drawing("P",(-1,-2.5)/5,S);pair G=extension(A,(B+C)/2,B,(C+A)/2);pair Q=Drawing("Q",(0.13,1.7)/5,E);pair R=Drawing("R",(-3.63,0.13)/5,W);pair X=Drawing("X",(-2.5,-0.33)/5,dir(320));draw(A--B--C--cycle,red);draw(Q--P--R,heavygreen);draw(circumcircle(A,Q,R),royalblue);[/asy][/asy] As usual, set $A=(1:0:0), B=(0:1:0), C=(0:0:1)$, and let $P=(0:m:n)$, and $Q=(x:0:z)$ Now, $\overline{PQ}$ and $\overline{AB}$ intersect at the point at infinity along $\overline{AB}$. One can easily check that the point at infinity along $\overline{AB}$ is just $(-t:t:0)$. This means that $(x:0:z)$, $(0:m:n)$, and $(-t:t:0)$ are collinear. Hence $\begin{vmatrix} -t & t & 0 \\ 0 & m & n \\ x & 0 & z \\ \end{vmatrix}=0$.$\implies Q=(m:0:n)$. Similarly, $R=(n:m:0)$. Now, we can compute the equation of $(AQR)$. It is just $-a^2yz-b^2zx-c^2xy+(x+y+z)\left(\frac{c^2ny+b^2mz}{m+n}\right)=0.$ Since $\angle BAG=\angle CAX$, it follows that $X$ lies on the $A-symmedian$, so $X=(k:b^2:c^2)$ for some $k$. Putting this into the equation of $(AQR)$, we see that $k=b^2+c^2-a^2$. Consequently, $X=(b^2+c^2-a^2:b^2:c^2)$, which does not depend on $m$ and $n$ and hence does not depend on $P$ and so we are done.

By the gliding principle, there is a fixed point $X$ lying on each of the circles $(AQR)$, which must lie on $(AAC)$ and $(AAB)$ but not be $A$, where $(AAC)$ is tangent to $AB$ and $(AAB)$ is tangent to $AC$. Then by $\sqrt{bc}$-inversion at $A$, $X$ must lie on the $A$-symmedian of $\triangle ABC$ implying the desired.

Sol:- Inverting with radius √AB*AC and reflecting across A- angle bisector. We get the following problem. P is an arbitrary point on (ABC) in arc BC not containing A. Let π be the circle through A and P tangent to AC & let ∆ be the circle through A and P tangent to AB. Let AB intersect π again at D & let AC intersect ∆ again at E. Prove that as P varies DE passes through a fixed point. Proof:- Let A' be the reflection of A across midpoint of BC. We find that P is the center of spiral sim. Sending DA to AE and also sends B to C. So we get DB*CE=AB*AC and hence DBA' is similar to A'CB which implies that D-A'-E are collinear which regardless of any P. Hence we are done.

Dumpty point! Let $S=PQ\cap (AQR)$ and $T=PR\cap (ARQ)$. Also, let $D=BS\cap (AQR)$. By Pascal on $ARTDSQ$ we get $T-D-C$ collinear. Now, $$ \measuredangle DBA=\measuredangle SBA=\measuredangle DSP=\measuredangle DAQ=\measuredangle DAC $$$$ \measuredangle ACD=\measuredangle ACT=\measuredangle PTD=\measuredangle RAD=\measuredangle BAD $$Hence $D$ is the $A-Dumpty$ point. Since $D\in (AQR)$ por every point $P$, we've finished.

The spiral similiarity $\psi$ sending $B\to A$, $A\to C$ also sends $R \to Q$. This is because $$\frac{BR}{RA}=\frac{BP}{PC}=\frac{AQ}{QC}$$ Now $\psi$ is centered at the Dumpty point $F$, so $AQRF$ is cyclic. Since $F$ is on the $A$-symmedian we are done. $\square$

Let $Q_a$ denote the $A$-Dumpty point. Because $Q_a$ lies on the $A$-Symmedian, we know $\angle BAG = \angle CAQ_a$, so it suffices to show $AQQ_aR$ is cyclic. It's well-known that $Q_aBA \overset{+}{\sim} Q_aAC$ and $(Q_aBA)$ is tangent to $AC$. In addition, the parallel lines imply $$\frac{BR}{RA} = \frac{BP}{PC} = \frac{AQ}{QC}$$so $Q_aBRA \overset{+}{\sim} Q_aAQC$ which means $Q_aBA \overset{+}{\sim} Q_aRQ \overset{+}{\sim} Q_aAC$. Thus, $$\angle QQ_aR = \angle AQ_aB = 180^{\circ} - \angle CAB = 180^{\circ} - \angle QAR$$as required. $\blacksquare$

Let the $A-$symmedian intersects $BC$ at $J$ and $(AQR)$ at $K$. We want to show $K$ is the fixed point. It is equivalent to show $P(J, (AQR))$ is fixed. Consider $f:R^2\to R$ such that $$f(\bullet)=P(\bullet, (AQR))-P(\bullet, (ABC))$$Or equivalently it is sufficient to prove $f(J)$ is fixed. We know that $J=(0:b^2:c^2)$. Let $W=b^2+c^2$. Now \begin{align*} f(J)&=\frac{c^2}{W}f(C)+\frac{b^2}{W} f(B)\\ &=\frac{1}{W}\left(c^2\cdot CQ\cdot b+b^2\cdot BR\cdot c\right)\\ &=\frac{bc}{W}\left(CQ\cdot c+BR\cdot b\right)\\ &=\frac{b^2c^2}{W} \end{align*}Where $CQ\cdot c+BR\cdot b=bc$ follows after adding $\frac{CQ}{CA}=\frac{CP}{CB}, \frac{BR}{BA}=\frac{BP}{BC}$. So $f(J)$ is fixed and therefore $K$ is fixed. Hence $K\equiv X$.

Let $\Gamma$ be the circumcircle of $\triangle ABC.$ Then, let the $A$-symmedian of $\triangle ABC$ intersect $\Gamma$ in $X_1$. We claim that if $X$ is the midpoint of $AX_1$ then $AQXR$ is cyclic for any $P.$ Let $Q_1,R_1$ be the reflections of $A$ across $Q$ and $R$ respectively. It suffices to show $AQ_1X_1R_1$ cyclic. WLOG let $P$ be closer to $B$ than $C$ which will mean that $Q_1$ is on the interior of $AC$ and $R_1$ is on the exterior of ray $AB$. We claim that $X_1$ is the Miquel Point of quadrilateral $ABPQ_1.$ Since $ACX_1B$ is already cyclic, to prove the claim we need to show that $CQ_1PX_1$ is cyclic. We will do this by proving that $\angle CQ_1X_1=\angle CPX_1.$ Let $M$ be the midpoint of $BC$. Note that since $AX_1$ is the $A$-symmedian of $\triangle ABC$, $\angle CAM=\angle X_1AB.$ Furthermore, $\angle ACM=\angle AX_1B$ so $\triangle ACM\sim \triangle AX_1B.$ Thus, \[\frac{AX_1}{BX_1}=\frac{AC}{CM}=2\cdot\frac{AC}{BC}=2\cdot \frac{RP}{BP}=2\cdot\frac{AQ}{BP}=\frac{AQ_1}{BP}.\]Therefore we have $\triangle AQ_1X_1\sim \triangle BPX_1$ by $SAS$ similarity. We can therefore conclude that $\angle CQ_1X_1=\angle CPX_1$ finishing our proof that $X_1$ is the Miquel Point of quadrilateral $ABPQ_1.$ Since $X_1$ is the Miquel Point, $AQ_1X_1R_1$ is cyclic, as desired. Since we already established that $X_1$ is on the symmedian, $\angle CAX=\angle CAX_1=\angle BAG.$

We will use barycentric coordinates. Let $P=(0,p,1-p)$. Then, $$R=(1-p,p,0),Q=(p,0,1-p).$$Let's find the equation for $(ARQ)$. Substituting $(1,0,0)$ gives us $w=0$, then using $(1-p,p,0)$ gives $v=c^2(1-p)$, and using $(p,0,1-p)$ gives $w=b^2p$. Therefore, the equation for $(ARQ)$ is $$(x+y+z)(c^2(1-p)y+b^2pz)=a^2yz+b^2xz+c^2xy.$$Note that the "fixed point" is just a solution $(x,y,z)$ that works regardless of $p$. To find such a triple, we want $p$ to cancel, so we want $c^2y=b^2z$. This leads to the solution $$(x,y,z)=(-a^2+b^2+c^2,b^2,c^2),$$which works regardless of $p$. Since the last two coordinates are $b^2$ and $c^2$, this lies on the A-symmedian, so we are done.

Here’s a pure angle chasing solution which only use the knowledge of dumpy point. We claim that the desired fixed point is the dumpy point $X=D$ which lies on the symmedian as required. Let $T$ be the midpoint of $AP$. Let $AD$ intersects $(ABC)$ again at $S$. By Miquel’s Theorem, it’s suffice to prove that $(BRD),(CQD)$ intersect on $BC$. We claim that the intersection point is $TD\cap BC=J$. We want $B,R,D,J$ to be concyclic. Let’s delete point $Q$ and introduce point $I=AS\cap AP$. We angle chase $$\angle BDI=\angle DBA+\angle DAB=\angle DAC+\angle DAB=\angle BAC=\angle BRP=\angle BRI$$so $B,R,D,I$ are cyclic. We now only need $B,D,I,J$ to be cyclic. But first, note that $B,I,P,S$ are concyclic by Reim’s Theorem on $\overline{AIS},\overline{BPC}$. Now let’s finish this problem by proving that $B,D,I,J$ are concyclic. $$\angle IBJ=\angle IBP=\angle ISP=\angle ADT=\angle IDJ.$$($DT\parallel SP$ because $DT$ is the $A$-midline of triangle $ASP$.)

Let $M,N$ be the midpoints of $AB,AC$. Notice by linear motion, we have $\frac{MR}{AB} = -\frac{NQ}{AC}$ while lengths are directed. Now take a homothety with scale factor $2$. Let $S$ be the intersection of the cirumcircle of $ABC$ and the symmedian point. I claim that this point, the images of $R,Q$ under the homothety, and $A$ are cyclic, which implies the problem statement. Let $R',Q'$ be the images of $R,Q$ under this homothety. This is equivalent to showing that $AS$ is the radical axis of the circumcircles of $(ABC), (AR'Q')$. Now consider the function taking points $X$ and returning the power of $X$ relative to circle $(AR'Q')$ minus the power of $X$ relative to circle $(ABC)$. It is well known that this function is linear. Then evaluated at $B$, the function gives $BR' \cdot BA - 0$, evaluated at $C$ this gives $CQ' \cdot CA - 0$. Let $AS$ meet $BC$ at $K$. By symmedian properties, $\frac{BK}{CK} = \frac{AB^2}{AC^2}$. Then the value of the function at $k$ is $\frac{AC^2 (BR')(AB) + AB^2 (AC)(CQ')}{AC^2 + AB^2} = 0$, thus $AK$ is the radical axis and since $AKS$ collinear we are done.

We invoke barycentric coordinates with $\triangle ABC$ as the reference triangle. Let $P = (0,p,1-p)$. We can see that $Q = (p,0,1-p)$ and $R = (1-p,p,0)$. Pluggin $A$ into the general circle we find that $u = 0$. Plugging in $Q$ we find that $w = b^2p$, and plugging in $R$ we find that $v = c^2(1-p)$. Thus we can see that the circle is given by $$-a^2yz -b^2xz - c^2xy + (c^2(1-p)y + b^2pz)(x+y+z) = 0$$. We find that the point $(-a^2+b^2+c^2:b^2:c^2)$ is a point which always satisfies this regardless of $p$. We can also see that it is parametrized by $(a^2:b^2:c^2)$ which is the isogonal conjugate of $G$. $\blacksquare$

Shameless me... T__T [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. */ pair A = (-72.37299,66.07399); pair B = (-98.21072,-33.37530); pair C = (27.39746,-33.13605); pair T = (1.55974,-132.58535); pair X = (37.23381,4.72399); pair Y = (-71.90471,-59.53355); pair P = (-23.11266,-68.25658); pair Q = (-145.83160,-216.66787); pair R = (81.53006,-86.96458); import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); draw(B--C, linewidth(0.5)); draw(circle((-35.47646,3.40690), 72.72220), linewidth(0.5)); draw(B--T, linewidth(0.5)); draw((-44.30101,-86.98220)--(-48.88588,-87.56842), linewidth(0.5)); draw((-44.30101,-86.98220)--(-43.74061,-82.39409), linewidth(0.5)); draw((-48.32549,-82.98032)--(-52.91036,-83.56655), linewidth(0.5)); draw((-48.32549,-82.98032)--(-47.76509,-78.39222), linewidth(0.5)); draw(T--X, linewidth(0.5)); draw((20.11035,-61.18410)--(22.92808,-64.84814), linewidth(0.5)); draw((20.11035,-61.18410)--(15.86547,-63.01321), linewidth(0.5)); draw(X--Q, linewidth(0.5) + red); draw(R--Y, linewidth(0.5) + red); draw(Q--R, linewidth(0.5) + linetype("4 4") + blue); draw(Q--B, linewidth(0.5)); draw(B--A, linewidth(0.5)); draw((-84.57827,19.09591)--(-81.76055,15.43188), linewidth(0.5)); draw((-84.57827,19.09591)--(-88.82316,17.26680), linewidth(0.5)); draw(A--C, linewidth(0.5)); draw((-18.46328,12.46709)--(-23.04815,11.88086), linewidth(0.5)); draw((-18.46328,12.46709)--(-17.90289,17.05519), linewidth(0.5)); draw((-22.48776,16.46897)--(-27.07263,15.88274), linewidth(0.5)); draw((-22.48776,16.46897)--(-21.92737,21.05707), linewidth(0.5)); draw(C--R, linewidth(0.5)); dot("$A$", A, NW); dot("$B$", B, dir(180)); dot("$C$", C, dir(0)); dot("$T$", T, SE); dot("$X$", X, dir(0)); dot("$Y$", Y, SW); dot("$P$", P, SE); dot("$Q$", Q, SW); dot("$R$", R, SE); [/asy][/asy] I claim that the fixed point is the $A$-dumpty point. Perform a $\sqrt{bc}$-inversion. Let $X=P^*Q^*\cap\odot(ABC)$ and $Y=P^*R^*\cap\odot(ABC)$. Then from the condition $\measuredangle BAC=\measuredangle AQP=\measuredangle AP^*Q^*=\measuredangle AP^*X=\measuredangle ABX$ which gives us that $X$ is the intersection of the line parallel through $C$ to $AB$ with $\odot(ABC)$ (due to isosceles trapezium). Similarly $Y$ is the intersection $B$ to $AC$ with $\odot(ABC)$. So we can redefine $Q^*$ and $R^*$ as $XP^*\cap AB$ and $YP^*\cap AC$ respectively. Also note that the image of the $A$-dumpty point under the inversion is the reflection of $A$ over the midpoint of $BC$. Call this point $T$. We now finally need to show that $Q^*R^*$ passes through $T$. Firstly, it is clear that $T=CX\cap BY$. So by Pascal on $ACXP^*YB$ we are done. And thus finally inverting back, we have our claim proved which finishes.

Let $Q_A$ be the $A$-dumpty point of $\triangle ABC$. We will show that $AQQ_AR$ is always cyclic, which will finish since $Q_A$ is fixed and lies on the $A$-symmedian. It's well-known that $Q_A$ is the center of the spiral similarity taking $\overline{BA} \mapsto \overline{AC}$. But from the parallel lines we have $$\frac{BR}{RA} = \frac{BP}{PC} = \frac{AQ}{QC},$$so this spiral similarity also takes $R \mapsto Q$. If $M$, $N$ are the midpoints of $\overline{BA}$ and $\overline{AC}$ then in fact $Q_A$ is the center of the spiral similarity taking $\overline{MR} \mapsto \overline{NQ}$, so it must also be the center of the spiral similarity taking $\overline{MN} \mapsto \overline{RQ}$. Therefore $Q_A = (ARQ) \cap (AMN)$ and we are done.

We employ barycentric coordinates! Set $\triangle ABC$ be the reference triangle. Now, let $P=(0:r:s)$. Then, it is easy to see that $Q=(r:0:s)$ and $R=(s:r:0)$ due to the given parallel condition. Now, let $X$ be the second intersection of $(AQR)$ and the $A-$symmedian. Since $X$ lies on the $A-$symmedian, we immediately have $X=(t:b^2:c^2)$ for some real number $t$. Consider the equation of circle $(AQR)$, \[-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0\]which since $A$ lies on this circle must have $u=0$. Further, since $R$ lies on this circle we have \begin{align*} -a^2yz-b^2xz-c^2xy+(vy+wz)(x+y+z)&=0\\ -c^2rs + (vr)(r+s) &= 0\\ v&=\frac{c^2s}{r+s} \end{align*}and similarly, since $Q$ lies on this circle we also have $w=\frac{b^2r}{r+s}$. Thus, the equation of the circle is simply, \[-a^2yz-b^2xz-c^2xy+(\frac{c^2s}{r+s}y+\frac{b^2r}{r+s}z)(x+y+z)=0\]which since $X$ lies on this circle implies, \begin{align*} -a^2b^2c^2 - tb^2c^2 - tb^2c^2 + \left(\frac{b^2c^2s + b^2c^2r}{r+s}\right)(t+b^2+c^2) &=0\\ -a^2b^2c^2 -2tb^2c^2 + b^2c^2(t+b^2+c^2) &= 0\\ tb^2c^2 &= b^2c^2(b^2+c^2-a^2)\\ t &= b^2+c^2-a^2 \end{align*}Thus, $X=(b^2+c^2-a^2:b^2:c^2)$ which is clearly a fixed point which does not depend on the position of $P$ on $\overline{BC}$.