Let $M$ and $N$ denote the midpoints of $\overline{AC}$ and $\overline{AB}$, respectively. [asy][asy] size(12cm); pair A = dir(97); pair B = dir(190); pair C = dir(350); pair M = midpoint(A--C); pair N = midpoint(A--B); pair G = extension(B, M, C, N); draw(A--G, blue); pair Y = A*dir((C-G)/(B-G))**2; pair X = A*dir((B-G)/(C-G))**2; pair S = extension(B, Y, C, G); pair R = extension(C, X, B, G); filldraw(A--B--C--cycle, invisible, blue); draw(B--M, blue); draw(C--N, blue); draw(R--A--S, lightred); draw(C--R, deepgreen); draw(B--S, deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$G$", G, dir(280)); dot("$S$", S, dir(270)); dot("$R$", R, dir(150)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ !size(12cm); A = dir 97 B = dir 190 C = dir 350 M = midpoint A--C N = midpoint A--B G 280 = extension B M C N A--G blue !pair Y = A*dir((C-G)/(B-G))**2; !pair X = A*dir((B-G)/(C-G))**2; S 270 = extension B Y C G R 150 = extension C X B G A--B--C--cycle / 0.1 lightblue / blue B--M blue C--N blue R--A--S lightred C--R deepgreen B--S deepgreen */ [/asy][/asy] From the given condition that $\measuredangle ACR = \measuredangle CGM$, we get that \[ MA^2 = MC^2 = MG \cdot MR \implies \measuredangle RAC = \measuredangle MGA. \]Analogously, \[ \measuredangle BAS = \measuredangle AGN. \]Hence, \[ \measuredangle RAS + \measuredangle BAC = \measuredangle RAC + \measuredangle BAS = \measuredangle MGA + \measuredangle AGN = \measuredangle MGN = \measuredangle BGC. \]

Not that hard for a P1. Haven't checked for config issues yet tho

nice and simple hopefully this is correct, pretty similar to v_enhances solution so it should be Let $M$, $N$ be the midpoints of $AC$, $AB$ respectively. Claim: $\measuredangle RAC = \measuredangle MGA$ and $\measuredangle BAS = \measuredangle AGN$. Proof. We show that $\measuredangle RAC = \measuredangle MGA$; the other equality can be proved in a similar manner. By the given angle conditions, we have that $\measuredangle MGC = \measuredangle RCA$; hence \[\triangle MGC \sim RMC \implies \frac{MC}{MG} = \frac{MR}{MC} \implies MC^2 = MG \cdot MR; \]but by definition $MA = MC$ hence \[MA^2 = MG \cdot MR \implies \triangle MAG \sim \triangle MRA\]and the angle equality follows. $\blacksquare$ Now it is clear that \[ \measuredangle RAS+\measuredangle BAC= \measuredangle RAC + \measuredangle BAS = \measuredangle MGA + \measuredangle AGN = \measuredangle MGN = \measuredangle BGC\]as desired. $\square$

We make no synthetic observations. Assume WLOG that $A, B, C$ are counter-clockwise. Translate such that $g=0$. Scale such that $b\bar b = 1$. Let $k = c \bar c$. Taking $S'$ on the other side of $B$ as $S$, $\angle ABS' = \angle BGC = \arg(c/b)$. Thus the line $BS$ is, for real $x$, \[\frac{z-b}{a-b}=x\frac cb \implies z = b+x(a-b)\frac cb = b+x(-2b-c)\frac cb = b-2cx-\frac{c^2}{b}x.\]$G, S, C$ are collinear so $\frac sc = \overline{\left(\frac sc\right)}$. Thus \[\frac bc - 2x - \frac cbx = \frac{1/b}{k/c}-2x-\frac{k/c}{1/b}x,\]giving $x=-\frac 1k$. Then $s = b+\frac{2c}{k}+\frac{c^2}{bk}$. Similarly, $CR$ the line \[\frac{z-c}{a-c}=x\frac bc \implies z=c+x\frac bc (-b-2c) = c-2bx-\frac{b^2}{c}x.\]Now $\frac{r}{b} = \overline{\left(\frac rb\right)}$, so \[\frac cb - 2x - \frac bc x = \frac{k/c}{1/b}-2x-\frac{1/b}{k/c}x,\]giving $x=-k$. Then $r = c+2bk+\frac{b^2k}{c}$. The desired condition is equivalent to $\frac{s-a}{r-a} \frac{c-a}{b-a} \frac{b}{c} \in \mathbb R$. But \[\frac{s-a}{r-a} \frac{c-a}{b-a} \frac{b}{c} = \frac{\frac{1}{bk}(2b^2k+2bc+c^2+cbk)(-2c-b)(b)}{\frac{1}{c}(2c^2+2bck+b^2k+bc)(-2b-c)(c)} = \frac{1}{k} \cdot \frac{(c+bk)(c+2b)(b+2c)}{(c+bk)(b+2c)(c+2b)}=\frac{1}{k} \in \mathbb R.\]

By alternate segment theorem since $\measuredangle MCR=\measuredangle RGC$ we have $MC$ is tangent to circle $CGR$. By power of a point, $$MA^2 = MC^2 = MG\times MR$$so $MA$ is tangent to circle $AGR$ so $$\measuredangle GBA=\measuredangle MBA=\measuredangle RAG$$and by symmetry $\measuredangle ACG=\measuredangle GAS$. Thus, $$\measuredangle RAS = \measuredangle RAG +\measuredangle GAS =\measuredangle GBA+\measuredangle ACG =\measuredangle BGC-\measuredangle BAC$$

Let $M,N$ be the midpoints of $AB,AC$. Then, $\angle MBG=\angle ABS-\angle GBS=180^\circ-\angle BGC-\angle GBS=\angle MSB,$ and so $MA^2=MB^2=MG \cdot MS,$ hence $\angle BAG=\angle ASG$. Similarly, $\angle CAG=\angle ARG,$ and so $\angle RAS+\angle BAC=\angle RGS-\angle ARG-\angle ASG+\angle BAC=\angle BGC-\angle CAG-\angle BAG+\angle BAC=\angle BGC,$ as desired.

Alternatively, you can do what I did in contest and use the similar triangles to deduce the coordinates of $r,s$ and then complex bash :skull:

Denote $M,N$ to be the midpoints of $AC$ and $AB$ respectively. Note that, $$\angle ACR = \angle CGM = \angle ABS = \angle BGN$$ Now notice that this implies that $MC$ is tangent to $(CGR)$ and $NB$ is tangent to $(BGS)$. Now notice that $M$ lies on the radical axis of $(ARG)$ and $(CGR)$, which implies $MA$ is tangent to $(ARG)$. Using a similar argument, we get $NA$ is tangent to $(ASG)$. To finish, note that \begin{align*} \angle RAS+\angle BAC \\ = \angle RAG + \angle SAG + \angle BAG + \angle CAG \\ = 360 - \angle CAG - \angle BGA - \angle BAG - \angle AGC + \angle BAG + \angle CAG \\ = \angle BGC \end{align*}

Let $X\in CN,Y\in BM$,satisfy $AX//BG,AY//CG$ respectively Consider that $A,X,G,B$and $A,G,C,Y$ are parallelogram So $\angle AXG=\angle XGB=\angle ABS$ So $A,X,B,S$ are cyclic Hence $\angle BAS=\angle BXS$ Similarly $\angle RAC=\angle BYC$ Note that $\angle BAC+\angle RAS=\angle BAS+\angle RAC=\angle BXC+\angle BYC$ Notice that $BX//AG//CY$ So $\angle BYC=\angle XBG$ Hence$\angle \angle BXC+\angle BYC=\angle BGC$

[asy][asy] unitsize(3cm); pair A = dir(95); pair B = dir(190); pair C = dir(350); pair M=(A+C)/2; pair N=(A+B)/2; pair G=(A+B+C)/3; pair P=IP(Line(G, M, 10), circumcircle(A, C, G), 1); pair R = 2M-P; pair Q=IP(Line(G, N, 10), circumcircle(A, B, G), 1); pair S = 2N-Q; draw(A--B--C--cycle,heavygreen); draw(circumcircle(A,C,G),red); draw(B--P, heavygreen); draw(C--N,heavygreen); draw(A--P,red); draw(C--P,red); draw(A--G,heavygreen); draw(A--R--S--cycle,red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$G$",G,NE); dot("$N$",N,dir(N)); dot("$M$",M,dir(M)); dot("$R'$",P,dir(P)); dot("$R$",R,dir(R)); dot("$S$", S, dir(S)); [/asy][/asy] Let $M$ be the midpoint of $\overline{AC}$ and $N$ be the midpoint of $\overline{AB}$. Let $R'$ denote the reflection of $R$ over $M$, so that $\triangle AR'M \cong \triangle CRM$. Then $AR'CR$ is a parallelogram, and we can angle chase to get $$\measuredangle CAR' = \measuredangle ACR = \measuredangle CGR',$$so quadrilateral $AR'CG$ is cyclic. Thus, we have $$\measuredangle GAC = \measuredangle GR'C = \measuredangle GRA.$$Analogously, we have $\measuredangle BAG = \measuredangle ASG$. Thus, to finish, note that $$\measuredangle BAC + \measuredangle RAS = \measuredangle BAG + \measuredangle GAC + \measuredangle RAS = \measuredangle ASG + \measuredangle GRA + \measuredangle RAS = - \measuredangle SGR = \measuredangle BGC,$$as desired.

Let $M$ and $N$ denote the midpoints of $AC$ and $AB$ respectively. We have $\angle BSN= 180^\circ -\angle BNS- \angle NBS= \angle BGC- \angle BNS= \angle NBG $, and since point $S$ lies on ray $GC$, we infer that the line $AB$ is tangent to circle $(B, G, S)$. Also, $\angle CRM= 180^\circ -\angle CMR- \angle MCR= \angle BGC- \angle CMR= \angle MCG$ and $R$ lies on ray $GB$, therefore the line $AC$ is tangent to circle $(C, G, R)$. From the above, we infer that $\bigtriangleup CRM \sim GCM\Rightarrow \frac{CR}{CM}= \frac{GC}{GM}\Rightarrow \frac{CR}{CA}= \frac{GC}{GB}$. Also, $\bigtriangleup BSN \sim GBN\Rightarrow \frac{BS}{BN}= \frac{GB}{GN}\Rightarrow \frac{BS}{BA}= \frac{GB}{GC}$. Therefore, $\frac{CR}{CA}= \frac{BA}{BS}$ but we also know that $\angle ACR= \angle ABS$. Thus, $\bigtriangleup ABS\sim RCA \Rightarrow \angle BAS= \angle ARC$. Finally, we examine 2 cases: 1) If point $S$ lies on segment $GC$ and point $R$ lies outside of the segment $GB$, we have : $\angle BGC= \angle RAC+ \angle ARC= \angle RAS+ \angle SAC+ \angle BAS= \angle RAS+ \angle BAC$ which is what we wanted to prove. 2) If point $S$ lies outside of the segment $GC$ and point $R$ lies on segment $GB$, we have: $\angle BGC= \angle RAC+ \angle ARC= \angle RAS-\angle SAC+ \angle BAS= \angle RAS+ \angle BAC$ which again is the result.

Let $N$ and $K$ be the midpoints of $AC$ and $AB$ respectively. $\angle CNR=\angle CNG$ $\angle RCN=180 - \angle RGC=\angle NGC$ $\Rightarrow$ $\bigtriangleup CNR \sim \bigtriangleup GNC$ So $\frac{CN}{GN}=\frac{NR}{NC}$ $\Leftrightarrow$ $NG.NR=NC^2=NA^2$ $\Leftrightarrow$ $NA$ is tangent to $(AGR)$ $\Leftrightarrow$ $\angle ARN=\angle GAN$ Similarly $\bigtriangleup BSK \sim \bigtriangleup GBK$ and $KG.KS=KB^2=KA^2$ ,so $KA$ is tangent to $(AGS)$ and $\angle KAG=\angle KSA$. Now: $\angle RAS + \angle BAC= \angle RAG + \angle GAS + \angle BAG + \angle GAC= \angle RAG + \angle ARG + \angle GAS + \angle GSA=\\ =180 - \angle RGA + 180 - \angle AXS= 360- \angle RGA + \angle AXS= \angle BGC$

Let the midpoint of $\overline{AB}$ be $M$ and let the midpoint of $\overline{AC}$ be $N$. Notice by the condition we have \[\angle MBS = \angle BGM \implies \triangle BMG \sim \triangle SMB.\] This implies \[\frac{MG}{MB} = \frac{MB}{MS} \implies MA^2=MB^2=MG \cdot MS.\] Hence, we have \[\frac{MA}{MS} = \frac{MG}{MA} \implies \triangle MAS \sim \triangle GAM\]\[\implies \angle MAG = \angle MSA.\] Similarly, $\angle NRA = \angle GAN$. Thus, we have \begin{align*} &\phantom{=} \angle BAC + \angle RAS = \angle MAG + \angle NRA + \angle RAS \\ &= \angle RAS + \angle GRA + \angle GSA = \angle BGC. \ \square \end{align*} [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.288651162790703, xmax = 21.47172093023256, ymin = -2.7780465116279114, ymax = 15.636465116279071; /* image dimensions */ /* draw figures */ draw((5,12)--(-4.169023255813958,-0.12167441860465383), linewidth(1)); draw((-4.169023255813958,-0.12167441860465383)--(14.853302325581398,-0.25674418604651433), linewidth(1)); draw((14.853302325581398,-0.25674418604651433)--(5,12), linewidth(1)); draw((-4.169023255813958,-0.12167441860465383)--(9.926651162790698,5.871627906976743), linewidth(1)); draw((-1.2177286743345868,1.1331786769882735)--(14.853302325581398,-0.25674418604651433), linewidth(1)); draw((10.549253914691349,1.5903140779807183)--(-4.169023255813958,-0.12167441860465383), linewidth(1)); draw((0.41548837209302114,5.939162790697673)--(14.853302325581398,-0.25674418604651433), linewidth(1)); draw((5,12)--(-1.2177286743345868,1.1331786769882735), linewidth(1)); draw((5,12)--(10.549253914691349,1.5903140779807183), linewidth(1)); draw((5,12)--(5.2280930232558145,3.8738604651162776), linewidth(1)); /* dots and labels */ dot((5,12),dotstyle); label("$A$", (5.083255813953487,12.214697674418606), NE * labelscalefactor); dot((-4.169023255813958,-0.12167441860465383),dotstyle); label("$B$", (-4.4789767441860505,0.10344186046511367), NE * labelscalefactor); dot((14.853302325581398,-0.25674418604651433),dotstyle); label("$C$", (14.893348837209305,-0.03162790697674684), NE * labelscalefactor); dot((9.926651162790698,5.871627906976743),linewidth(4pt) + dotstyle); label("$N$", (10.013302325581394,6.046511627906976), NE * labelscalefactor); dot((5.2280930232558145,3.8738604651162776),linewidth(4pt) + dotstyle); label("$G$", (5.228372093023254,4.042976744186045), NE * labelscalefactor); dot((10.549253914691349,1.5903140779807183),linewidth(4pt) + dotstyle); label("$S$", (10.643627906976745,1.769302325581393), NE * labelscalefactor); dot((-1.2177286743345868,1.1331786769882735),linewidth(4pt) + dotstyle); label("$R$", (-1.6299534883720963,1.219069767441858), NE * labelscalefactor); dot((0.41548837209302114,5.939162790697673),linewidth(4pt) + dotstyle); label("$M$", (0.0133953488372065,6.1140465116279055), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $M_{B}$, and $M_{C}$ denote the midpoints of $AC$, and $AB$ respectively. The angle conditions imply $$\triangle M_{B}GC \sim \triangle M_{B}CR,$$$$\triangle M_{C}BS \sim \triangle M_{C}GB$$Which means $$M_{B}C^{2}=M_{B}G \cdot M_{B}R$$$$M_{C}B^{2}=M_{C}G \cdot M_{C}S$$Since $M_{B}C=M_{B}A$, and $M_{C}B=M_{C}A$. $$M_{B}A^{2}=M_{B}G \cdot M_{B}R$$$$M_{C}A^{2}=M_{C}G \cdot M_{C}S$$So $$\triangle M_{B}GA \sim \triangle M_{B}AR,$$$$\triangle M_{C}AS \sim \triangle M_{C}GA$$So $$\measuredangle SAR+\measuredangle CAB= \measuredangle CAR + \measuredangle SAB = \measuredangle AGM_{C} + \measuredangle M_{B}GA = \measuredangle M_{B}GM_{C} = \measuredangle CGB$$$\square$

Easy for a P1 Let X and Y be midpoints of AB and AC Observe XBG is similar to XSG YGC similar to YCR Then observe 2 more similar triangles using lengths and then finish by angle chasing

Reflect $G$ about the midpoints of $AB, AC$ to $X,Y$ respectively, denote the midpoint of $BC$ as $Z$. By properties of parallelograms, we have $AGBX, AGCY$ are parallelograms, $\angle{AXG}=\angle{XGB}=180-\angle{BGC}=\angle{ABS}\implies AXBS$ is a cyclic quadrilateral. Similarly, we have $ARCY$ is a cyclic quadrilateral. Thus, we have $\angle{BAS}=\angle{BXS}=\angle{AGX}=\angle{CGZ}; \angle{RAC}=\angle{RYC}=\angle{AGY}=\angle{BGZ}$. As $\angle{BAS}+\angle{RAC}=\angle{BAC}+\angle{RAS}=\angle{BGC}$, the problem is done.

I like this one! Let $E$ and $F$ be the midpoints of $AC$ and $AB$, respectively. Note that by the given angle conditions, $\triangle FBS \sim \triangle FGB$ and $\triangle ECR \sim \triangle EGC$. It follows that $EA^2 = EC^2 = EG \cdot ER$, so $\angle ARG = \angle CAG$. Thus $\angle RAG = 180^{\circ} - \angle ARG - \angle AGR = 180^{\circ} - \angle CAG - \angle AGR$. In a similar fashion we derive $\angle SAG = 180^{\circ} - \angle BAG - \angle AGS$, so $\angle RAS = \angle RAG + \angle SAG = 360^{\circ} - (\angle CAG + \angle BAG) - (\angle AGR + \angle AGS )$. But $\angle CAG + \angle BAG = \angle BAC$ and $360^{\circ} - (\angle AGR + \angle AGS) = \angle BGC$. Thus $\angle RAS + \angle BAC = \angle BGC$ and we are done.

Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively. The given condition implies $\angle MGB \cong \angle MBS$. So, $\triangle MGB \sim \triangle MBS$ and \begin{align*} \dfrac{MG}{MB} = \dfrac{BM}{SM} \quad \longrightarrow \quad MG = MS = AM^2 \quad \longrightarrow \quad \triangle GMA \sim \triangle AMS. \end{align*}Suppose $P$ is the midpoint of $BC$, then $\angle PGC \cong \angle MGA \cong \angle MAS$. Similarily, $\angle PGB \cong \angle NAR$. Summing yields \begin{align*} \angle BGC &= \angle BGP + \angle PGC = \angle RAN + \angle MAS = \angle RAS + \angle BAS, \end{align*}as desired.

The angle conditions imply that $EB$ is tangent to $(BGS)$ and $DC$ is tangent to $(CGR)$. Hence $EB^2 = EG \cdot ES \implies EA^2 = EG \cdot ES \implies EA$ is tangent to $(AGS) \implies \measuredangle{BAS} = \measuredangle{EAS} = \measuredangle{AGE}$. Similarly, $\measuredangle{RAC} = \measuredangle{DGA}$. Therefore, $\measuredangle{RAS} + \measuredangle{BAC} = \measuredangle{RAC} + \measuredangle{BAS} = \measuredangle{AGE} + \measuredangle{DGA} = \measuredangle{BGC}$.

Unfortunately I didn't realize the tangency Let $E$ and $F$ be the midpoints of $AC$ and $AB$ respectively. Then angle condition directly implies $\triangle EGC \sim \triangle ECR$ and $\triangle FGB \sim \triangle FBS$. Then $\frac{CR}{AC/2} = \frac{CR}{CE} = \frac {GC}{GE} = \frac{2FG}{BG/2}$ and $\frac{BS}{AB/2} = \frac{BS}{FB} = \frac{BG}{FG}$. Multiplying both equations we get $\frac{BS}{AB} = \frac{AC}{CR}$ implying $\triangle ABS \sim \triangle RCA$. Hence $\angle BGC = 180 - \angle ABS = \angle BAS + \angle BSA = \angle BAS + \angle RAC = \angle BAC + \angle RAS$.

Let $D, E$ be the midpoints of $AC, AB$ respectively. Notice that $\triangle DCR \sim \triangle DGC,$ so we have that $$\frac{GD}{AD} = \frac{GD}{CD}=\frac{CD}{RD} = \frac{AD}{RD} \implies \triangle ADR \sim \triangle GDA.$$Therefore, $\angle RAC = \angle AGD,$ and similarly $\angle BAS = \angle AGE.$ Adding these up yields the desired result. QED