Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

I remember posting this a long time ago . Consider the inversion wrt the incircle. It invariates the incircle and $\Omega_a,_b,_c$, so it turns $X$ into $X'$, for any $X\in\{A,B,C\}$. At the same time, if the incircle touches $BC,CA,AB$ in $D,E,F$, then the inversion turns $A$ into the midpoint of $EF$ and the like, so $A',B',C'$ are the midpoints of $EF,FD,DE$ respectively, meaning that the circumcircle of $A'B'C'$ is the nine-point circle of $D EF$, thus having half its circumradius, which is the inradius of $ABC$.

OFFICIAL SOLUTION Denote by $I$ the incenter, by $r$ the inradius, by $D,E,F$ the contacts of the incircle with $BC,CA,AB$ respectively and by $P,Q,R$ the midpoints of the segments $[EF],[FD],[DE]$. We will prove that $\Omega_{a}$ is the circle $(B,C,Q,R)$. We firstly notice that from the right-angled triangles $IBD$ and $IDQ$ we get $IQ \cdot IB = ID^{2} = r^{2}$ and in the same way $IR \cdot IC = r^{2}$, therefore the points $B,C,R,Q$ are on a circle $\Gamma_{a}$. The points $Q$ and $R$ belong to the segment $(IB)$ and $(IC)$, so $I$ is exterior to the circle $(B,Q,R,C)$ and $I's$ power with respect to the circle $(B,Q,R,C)$. In the same way $\Gamma_{b}$ is the circle $(C,R,P,A)$ and $\Gamma_{c}$ is the circle $(A,P,Q,B)$. It follows that $A',B',C'$ coincide with $P,Q,R$ and the required conclusion is not obvious. Another Solution: Let $O_{a}$ be the center of $\Omega_{a}$ and $M$ be the midpoint of $(BC)$. Denote, using oriented segments, $\overline{M,O_{a}} = x$ (the positive sense on the perpendicular bisector of $(BC)$ being $\overrightarrow{ID}$. The radius of $\Omega_{a}$ is $x^{2} + \frac{a^{2}}{4}$ and \[IO^{2}_{a} = DM^{2} + (\overline{ID}+ \overline{MO_{a}})^{2} = \frac{a}{2} - \left(\frac{p-b}{2}\right)^{2} + (r+x)^{2}.\] The condition of $\Omega$ and $\Omega_{a}$ being orthogonal is $O_{a}I^{2} = O_{b}B^{2}+r^{2}$, that is \[x^{2} + \frac{a^{2}}{4} + r^{2} = \left(\frac{b-c}{2}\right)^{2} + (r+x)^{2} \Leftrightarrow \frac{(p-b)(p-c)}{2r} \Leftrightarrow X = 2R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}.\] It follows that \[OO_{a} = \overline{OM} + \overline{MO_{a}} = 2R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} = R + \frac{r}{2}.\] Therefore $O_{a},O_{b},O_{c}$ are on the circle of the center $O$ and radius $R + \frac{r}{2}$. Let $\{N\} = O_{a}O_{c} \cap BC$. The angle $\widehat{O_{a}OO_{c}}$ has measure $\pi - b$ (regardless of the position of $B$), so $\widehat{O_{a}OO_{c}} = \frac{B}{2}$ and \[MN = x \tan \frac{B}{2} = 2R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} = \frac{p-c}{2},BN = \frac{a}{2} - \frac{p-c}{2} = \frac{p-b}{2}.\] Hence $BN = \frac{BD}{2}$ and, because $MN < BM, N \in (BM)$. The same holds for the common point of $AB$ and $O_{a}O_{c}$, therefore the reflection of $B$ in $O_{a}O_{c}$ is on $DF$. This proves that $B'$ - the second common point of $\Omega_{a}$ and $\Omega_{b}$ - is the midpoint of $(DF)$. The conclusion follows easily.

Just for the sake of completeness, the other thread where the problem was discussed is http://www.mathlinks.ro/Forum/viewtopic.php?t=4592 . darij

Let $r$ be the radius of $\Omega$, then we have that the power of $I$ with respect to $\Omega_a, \Omega_b, \Omega_c$ is $r^2$. Thus, $I$ is the radical center of $\Omega_a, \Omega_b, \Omega_c$, so $AA', BB', CC'$ concur at $I$. Also, \[r\cdot AB = 2[AIB] = IA \cdot IB \cdot \sin \angle AIB\] where $[AIB]$ is the area of triangle $AIB$. But \[\angle AIB = 180^{\circ} - \angle IAB - \angle IBA = 180^{\circ} - \angle IAC - \angle IBC. \] By the similarity of $ICA$ and $IA'C'$ and of $ICB$ and $IB'C'$, this equals \[180^{\circ} - \angle IC'A' - \angle IC'B' = 180^{\circ} - \angle A'C'B',\] so \[r\cdot AB = IA \cdot IB \cdot \sin \angle A'C'B'.\] By the similarity of $IAB$ and $IB'A'$, \[A'B' = \frac{IB' \cdot AB}{IA} = \frac{r^2 AB}{IA \cdot IB}\] (because the power of $I$ with respect to $\Omega_a$ is $r^2$). And now, we use the (extended) law of sines: Let $R$ be the circumradius of $A'C'B'$. Then \[R = \frac{A'B'}{2 \sin \angle A'C'B'}=\frac{r^2 AB}{2IA \cdot IB \sin\angle A'C'B'}.\] Since $r\cdot AB = IA \cdot IB \cdot \sin \angle A'C'B',$ this is equal to $r/2$ and we are done.

This problem also has a solution like 2009 ISL G3's solution. Solution Let $O_a$, $O_b$ and $O_c$ denote the centres of circles $\Omega_a$, $\Omega_b$ and $\Omega_c$, respectively. Let $A_1$ and $A_2$ denote the intersections of $\Omega_a$ and $\Omega$. Define $B_1$, $B_2$, $C_1$ and $C_2$ analogously. Let $D, E$ and $F$ denote the intersections of $\Omega$ with $BC$, $CA$ and $AB$, respectively. Let $P$, $Q$ and $R$ denote the midpoints of $EF$, $DF$ and $DE$, respectively. By the definition of orthogonality, lines $O_aA_1$ and $O_aA_2$ are tangent to circle $\Omega$. Since $O_aB=O_aA_1=O_aA_2=O_aB$, it follows that $O_a$ lies on the radical axes of circles $B$ and $\Omega$ and circles $C$ and $\Omega$. Note that these radical axes are the midlines of the triangles $\triangle{BDF}$ and $\triangle{CDE}$. By the same argument, $O_b$ lies on the midlines of triangles $\triangle{CDE}$ and $\triangle{AEF}$ and $O_c$ lies on the midlines of triangles $\triangle{BDF}$ and $\triangle{AEF}$. Reflecting points $A$, $B$ and $C$ in the midlines of triangles $\triangle{AEF}$, $\triangle{BDF}$ and $\triangle{CDE}$ respectively yields that $A'=P$, $B'=Q$ and $C'=R$. Therefore, $\triangle{A'B'C'}$ is the medial triangle of $\triangle{DEF}$ and has a circumradius equal to half of that of circle $\Omega$.

Let $D,E,F$ be the points of tangency of $\Omega$ with $BC,CA,AB$. Do an inversion at $\Omega$. Notice that $\Omega_i$ maps to itself $\forall i \in \{ a,b,c \}$. Then, $ \{C',C\} = \Omega_a \cap \Omega_b \rightarrow \Omega_a \cap \Omega_b = \{ C', C\} $. However, $C$ is not its own inverse and thus $C'$ is the inverse of $C$, i.e. the midpoint of $DE$. An analogous thing occurs with $A$ and $B$. Notice $ R(A'B'C')=R(DEF)/2 = R(\Omega)/2$ since $A'B'C'$ is the medial triangle of $DEF$. Done.

Inversion wrt $\Omega$ maps $A,B,C$ to $A',B',C',$ so $A'B'C'$ is the medial triangle of $ABC-\text{tangential triangle},$ thus done.

Take an inversion with respect to $\Omega$. We have that $\Omega_A$, $\Omega_B$, and $\Omega_C$ stay constant from the orthogonal condition. This means that $(A'B'C')$ is the inverse of $(ABC)$ wrt to $\Omega$. Let $(ABC)=\omega$. Let $OI \cap \omega = P,Q$ with $P$ and $I$ on opposite sides of $O$. Let $R$ and $S$ be the intersection of $\omega$ with the line through $I$ perpendicular to $OI$. We have that $PI=\frac{r^2}{R-\sqrt{R^2-2Rr}}$, $QI=\frac{r^2}{R+\sqrt{R^2-2Rr}}$, and $RI=SI=\frac{r^2}{\sqrt{2Rr}}$. Let $X$ be the center of $(PQRS)$. We have that $X$ lies on $PQ$, so $XI=\frac{r\sqrt{R^2-2Rr}}{2R}$. This means the radius of the new circle is $$\sqrt{XI^2+RI^2}=\sqrt{\frac{r^2(R^2-2Rr)}{4R^2}+\frac{r^3}{2R}} = \sqrt{\frac{r^2R^2-2Rr^3}{4R^2}+\frac{2r^3R}{4R^2}} = \frac{r}{2},$$as desired so we are done.

Let $A_1B_1C_1$ be the contact triangle of $ABC$. Now, we redefine the points $A', B', C'$ so that $A'B'C'$ is the medial triangle of $A_1B_1C_1$. Because $A$ and $A'$ are clearly inverses wrt $\Omega$, we know $A'$ belongs to $\Omega_b$ and $\Omega_c$. Using analogous reasoning allows us to recover the original definitions of $B'$ and $C'$ as well. Thus, since $(A'B'C')$ is the Nine-Point Circle of $A_1B_1C_1$, the desired result follows readily. $\blacksquare$

Consider the inversion about $\Omega$. Since $\Omega_{a},\Omega_{b},\Omega_{c}$ are orthogonal, they are fixed. Now, intersections are preserved, so $A\to A'$ and $A'\to A$. Thus, $A'B'C'$ is the medial triangle of the contact triangle. We are done.

Let $D$, $E$, and $F$ be the tangency points of $\Omega$ with $BC$, $CA$, and $AB$. I claim that $A'$ is the midpoint of $EF$, and similarly for $B'$ and $C'$. This is simple to prove: inversion about $\Omega$ swaps $\Omega_b$, hence the image of $A$ (which is the midpoint of $EF$) lies on $\Omega_b$. The same holds for $\Omega_c$. Now the circumcircle of $\triangle A'B'C'$ is the nine-point circle of $\triangle DEF$ and therefore has half the radius of $\Omega$. $\blacksquare$

Note that $\Omega_a$ and $\Omega_b$ intersect at $C$ and $C'$. Since they are both orthogonal to $\Omega$, we get that $C$ and $C'$ are inverses with respect to $\Omega$. Now note that the inverses of the vertices of a triangle with respect to the incircle are the midpoints of the intouch triangle, and the radius of the nine-point circle is half the radius of the circumcircle. $\blacksquare$