any solution?

It's an old problem but since no one has posted a solution, I will post a solution using complex coordinates. The problem statement is very nice, so I think it deserves a synthetic proof. However I cannot find the official solutions for 2010 Balkan MO shortlists, and I don't even know if they have been published.

) We have $a_1+bc\bar{a_1}=b+c$ and $\bar{a_1}=ka_1+m$ so we get $a_1=\frac{b+c-bcm}{bck+1}$. Also $\bar{a_1}=\frac{kb+kc+m}{bck+1}$. The projections of $A_1$ onto lines $AB$ and $AC$ are $\frac{1}{2}(a+b+a_1-ab\bar{a_1})$ and $\frac{1}{2}(a+c+a_1-ac\bar{a_1})$ respectively, therefore $$2a'=\frac{2a+b+c}{2}+a_1-\frac{1}{2}a(b+c)\bar{a_1}$$$$=\frac{2a+b+c}{2}+\frac{2b+2c-2bcm-(ab^2+2abc+ac^2)k-(ab+ac)m}{2(bck+1)}$$ For part (a), our objective is to prove that $\frac{\bar{a'}-\bar{b'}}{a'-b'}=\frac{\bar{b'}-\bar{c'}}{b'-c'}$. So we calculate: $$2(a'-b')=\frac{a-b}{2}+\frac{M}{2(bck+1)(ack+1)}$$where the numerator $M$ is $$M=(2b+2c-2bcm-(ab^2+2abc+ac^2)k-(ab+ac)m)(ack+1)-(2a+2c-2acm-(a^2b+2abc+bc^2)k-(ab+bc)m)(bck+1)$$$$=(b-a)(ac^3k^2+bc^3k^2+2abc^2k^2+abcmk+ac^2mk+bc^2mk-abk-c^2k-cm+2)$$Therefore we get $$2(a'-b')=\frac{a-b}{2}+\frac{M}{2(bck+1)(ack+1)}$$$$=\frac{(b-a)(c^2(ab+bc+ca)k^2+(cm(ab+bc+ca)-c^2-ab-bc-ca)k-cm+1)}{(bck+1)(ack+1)}$$$$=\frac{(b-a)(c^2k+cm-1)((ab+bc+ca)k-1)}{(bck+1)(ack+1)}$$Observe that $\overline{c^2k+cm-1}=\frac{1}{c^2k}-\frac{m}{ck}-1=-\frac{c^2k+cm-1}{c^2k}$, so $$\frac{\bar{a'}-\bar{b'}}{a'-b'}=(-\frac{1}{ab})(-\frac{1}{c^2k})(bck)(ack)\frac{\overline{(ab+bc+ca)k-1}}{(ab+bc+ca)k-1}$$$$=k\frac{\overline{(ab+bc+ca)k-1}}{(ab+bc+ca)k-1}$$which is symmetric in terms of $a,b,c$. Therefore $\frac{\bar{a'}-\bar{b'}}{a'-b'}=\frac{\bar{b'}-\bar{c'}}{b'-c'}$ as desired. For part (b) of the problem, this is the case when $m=0$. The center of the Euler circle is $\frac{a+b+c}{2}$, so we want to prove that $\frac{\bar{a'}-\overline{\frac{a+b+c}{2}}}{a'-\frac{a+b+c}{2}}=\frac{\bar{a'}-\bar{b'}}{a'-b'}$. So we calculate: $$2(a'-\frac{a+b+c}{2})=\frac{2a+b+c}{2}+\frac{2b+2c-(ab^2+2abc+ac^2)k}{2(bck+1)}-(a+b+c)$$$$=\frac{(b+c)(1-(ab+bc+ca)k)}{2(bck+1)}$$We can finish: $$\frac{\bar{a'}-\overline{\frac{a+b+c}{2}}}{a'-\frac{a+b+c}{2}}=\frac{1}{bc}bck\frac{\overline{(ab+bc+ca)k-1}}{(ab+bc+ca)k-1}=k\frac{\overline{(ab+bc+ca)k-1}}{(ab+bc+ca)k-1}$$which is equal to $\frac{\bar{a'}-\bar{b'}}{a'-b'}$ as we already calculated. Therefore the center of the Euler circle is on $\ell'$ as desired.

Thinking a bit more, I think there is a simple explanation using vectors. It is not purely synthetic, but there are no heavy calculations, and it kinda explains why the problem works, unlike the solution with complex coordinates. Define $D,E,F$ as the projections of $A,B,C$ onto lines $BC,CA,AB$ respectively. Define $O$ as the circumcircle, $H$ as the orthocenter, $N$ as the center of the Euler circle. Define $P,Q,R$ as the midpoints of $AD, BE, CF$ respectively. Define $S,T$ as the projections of $A_1$ onto lines $AB, AC$ respectively. There are real numbers $u,v$ such that $\vec{A_1}=u\vec{B}+v\vec{C}$ and $u+v=1$. Then $\vec{S}=u\vec{B}+v\vec{F}$ and $\vec{T}=u\vec{E}+v\vec{C}$. So $\vec{A'}=\frac{1}{2}(\vec{S}+\vec{T})=u(\frac{\vec{B}+\vec{E}}{2})+v(\frac{\vec{C}+\vec{F}}{2})=u\vec{Q}+v\vec{R}$. So $A'$ is the point on $QR$ such that $\vec{QA'}:\vec{A'R}=\vec{BA_1}:\vec{A_1C}$. Similar properties hold for $B'$ and $C'$. Now part (a) is clear. Because of Menelaus theorem, we have $\frac{\vec{A_1C}}{\vec{BA_1}}\frac{\vec{B_1A}}{\vec{CB_1}}\frac{\vec{C_1B}}{\vec{AC_1}}=-1$. So we get $\frac{\vec{A'R}}{\vec{QA'}}\frac{\vec{B'P}}{\vec{RB'}}\frac{\vec{C'Q}}{\vec{PC'}}=-1$. Because of Menelaus theorem on triangle $PQR$, this implies $A',B',C'$ are collinear. For part (b), if $O$ is on line $\ell$ we have $\vec{O}=k\vec{A_1}+l\vec{B_1}$ for some real numbers $k,l$ such that $k+l=1$. We previously defined $u,v$ such that $\vec{A_1}=u\vec{B}+v\vec{C}$ and $u+v=1$. Similarly define $x,y$ such that $\vec{B_1}=x\vec{A}+y\vec{C}$ and $x+y=1$. We have $\vec{O}=lx\vec{A}+ku\vec{B}+(kv+ly)\vec{C}$. Also $\vec{O}=\frac{(sin2A)\vec{A}+(sin2B)\vec{B}+(sin2C)\vec{C}}{sin2A+sin2B+sin2C}$. (This is a basic barycentric coordinate formula) Since $lx+ku+(kv+ly)=k(u+v)+l(x+y)=k+l=1$, We have $lx=\frac{sin2A}{sin2A+sin2B+sin2C}$, $ku=\frac{sin2B}{sin2A+sin2B+sin2C}$, $kv+ly=\frac{sin2C}{sin2A+sin2B+sin2C}$.

Therefore $k\vec{A'}+l\vec{B'}=\frac{1}{2}\vec{O}+\frac{1}{2}\vec{H}=\vec{N}$, so $N$ is on $\ell'$. Following the lines of this proof, much more can be said: For part (a), $\vec{A'B'}:\vec{B'C'}=\vec{A_1B_1}:\vec{B_1C_1}$. For part (b), $\vec{A'B'}:\vec{B'C'}:\vec{C'N}=\vec{A_1B_1}:\vec{B_1C_1}:\vec{C_1O}$. i.e. the ratio of distances between $A_1,B_1,C_1$(and $O$ for part (b)) is preserved for $A',B',C'$(and $N$ for part (b)). This can also be checked easily in the complex coordinate solution.