USA Tstst

Solution 1 using power of a point From the given condition that $\angle ACR = \angle CGM$, we get that
$$MA^2 = MC^2 = MG \cdot MR \Rightarrow \angle RAC = \angle MGA.$$
Analogously,
$$\angle BAS = \angle AGN.$$
Hence,
$$\angle RAS + \angle BAC = \angle RAC + \angle BAS = \angle MGA + \angle AGN = \angle MGN = \angle BGC.$$

Solution 2 using similar triangles As before, $\triangle MGC \sim \triangle MCR$ and $\triangle NGB \sim \triangle NBS$. We obtain
$$\frac{|AC|}{|CR|} = \frac{2|MC|}{|CR|} = \frac{2|MG|}{|GC|} = \frac{|GB|}{2|NG|} = \frac{|BS|}{2|BN|} = \frac{|BS|}{|AB|}$$
which together with $\angle ACR = \angle ABS$ yields
$$\triangle ACR \sim \triangle SBA \Rightarrow \angle BAS = \angle CRA.$$

Hence
$$\angle RAS + \angle BAC = \angle RAC + \angle BAS = \angle RAC + \angle CRA = -\angle ACR = \angle BGC,$$
which proves the statement.

Solution 3 using parallelograms Let $M$ and $N$ be defined as above. Let $P$ be the reflection of $G$ in $M$ and let $Q$ the reflection of $G$ in $N$. Then $AGCP$ and $AGBQ$ are parallelograms.

Claim — Quadrilaterals APCR and AQBS are concyclic.
Proof. Because $\angle APR = \angle APG = \angle CGP = -\angle BGC = \angle ACR$. $\square$

Thus from $\overline{PC} \parallel \overline{GA}$ we get
$$\angle RAC = \angle RPC = \angle GPC = \angle PGA$$
and similarly
$$\angle BAS = \angle BQS = \angle BQG = \angle AGQ.$$
We conclude that
$$\angle RAS + \angle BAC = \angle RAC + \angle BAS = \angle PGA + \angle AGQ = \angle PGQ = \angle BGC.$$

Solution 4 also using parallelograms, by Ankan Bhattacharya Construct parallelograms ARCK and ASBL. Since
$$\angle CAK = \angle ACR = \angle CGB = \angle CGK,$$
it follows that AGCK is cyclic. Similarly, AGBL is also cyclic.

Finally, observe that
$$\begin{align*} \angle RAS + \angle BAC &= \angle BAS + \angle RAC \\ &= \angle ABL + \angle KCA \\ &= \angle AGL + \angle KGA \\ &= \angle KGL \\ &= \angle BGC \end{align*}$$
as requested.

Solution 5 using complex numbers, by Milan Haiman Note that $\angle RAS + \angle BAC = \angle BAS + \angle RAC$. We compute $\angle BAS$ in complex numbers; then $\angle RAC$ will then be known by symmetry.
Let $a, b, c$ be points on the unit circle representing $A, B, C$ respectively. Let $g = \frac{1}{3}(a + b + c)$ represent the centroid $G$, and let $s$ represent $S$.

Claim — We have
$$\frac{s-a}{b-a} = \frac{ab-2bc+ca}{2ab-bc-ca}.$$
Proof. Since $S$ is on line $CG$, which passes through the midpoint of segment $AB$, we have that
$$s = \frac{a+b}{2} + t(c-g)$$
for some $t \in \mathbb{R}$.

By the given angle condition, we have that
$$\frac{(s-b)/(b-a)}{(c-g)/(g-b)} \in \mathbb{R}.$$
Note that
$$\frac{s-b}{b-a} = t \frac{c-g}{b-a} - \frac{1}{2}.$$
So,
$$t \frac{g-b}{b-a} - \frac{g-b}{2(c-g)} \in \mathbb{R}.$$

$$t = \frac{\operatorname{Im}\left(\frac{g-b}{2(c-g)}\right)}{\operatorname{Im}\left(\frac{g-b}{b-a}\right)} = \frac{1}{2} \cdot \frac{\left(\frac{g-b}{c-g}\right) - \overline{\left(\frac{g-b}{c-g}\right)}}{\left(\frac{g-b}{b-a}\right) - \overline{\left(\frac{g-b}{b-a}\right)}}$$
Let $N$ and $D$ be the numerator and denominator of the second factor above.
We want to compute
$$\frac{s-a}{b-a} = \frac{1}{2} + t \frac{c-g}{b-a} = \frac{(b-a) + 2t(c-g)}{2(b-a)} = \frac{(b-a)D + (c-g)N}{2(b-a)D}$$
We have
$$\begin{align*} (c-g)N &= g-b-(c-g)\overline{\left(\frac{g-b}{c-g}\right)} \\ &= \frac{a+b+c}{3} - b - \left(c - \frac{a+b+c}{3}\right) \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{3}{b}}{\frac{3}{c} - \frac{1}{a} - \frac{1}{b} - \frac{1}{c}} \\ &= \frac{(a+c-2b)(2ab-bc-ca) - (2c-a-b)(ab+bc-2ca)}{3(2ab-bc-ca)} \\ &= \frac{3(a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2)}{3(2ab-bc-ca)} \\ &= \frac{(a-b)(b-c)(a-c)}{2ab-bc-ca} \end{align*}$$

$$\begin{align*} (b-a)D &= g-b-(b-a)\overline{\left(\frac{g-b}{b-a}\right)} \\ &= \frac{a+b+c}{3} - b - (b-a) \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{3}{b}}{\frac{3}{b} - \frac{3}{a}} \\ &= \frac{(a+c-2b)c + (ab+bc-2ca)}{3c} \\ &= \frac{ab-bc-ca+c^2}{3c} \\ &= \frac{(a-c)(b-c)}{3c} \end{align*}$$

$$\frac{s-a}{b-a} = \frac{\frac{1}{3c} + \frac{a-b}{2ab-bc-ca}}{\frac{2}{3c}} = \frac{2ab-bc-ca+3c(a-b)}{2(2ab-bc-ca)} = \frac{ab-2bc+ca}{2ab-bc-ca}.$$

$$\frac{r-a}{c-a} = \frac{ab-2bc+ca}{2ca-ab-bc}.$$

$$\frac{s-a}{b-a} \div \frac{r-a}{c-a} = \frac{2ab-bc-ca}{2ca-ab-bc}$$

We also have that $\angle BGC$ is the argument of
$$\frac{b-g}{c-g} = \frac{2b-a-c}{2c-a-b}.$$

Note that these two complex numbers are inverse-conjugates, and thus have the same argument. So we're done.