, 2006

Let $T_{a} T_{b}$ intersect circle $\omega_{b}$ at $T_{b}$ and $U$, and let $T_{a} T_{c}$ intersect circle $\omega_{c}$ at $T_{c}$ and $V$. Further, let $U X$ be the tangent to $\omega_{b}$ at $U$, with $X$ on $A C$, and let $V Y$ be the tangent to $\omega_{c}$ at $V$, with $Y$ on $A B$. The homothety with centre $T_{b}$ and ratio $T_{b} T_{a} / T_{b} U$ maps the circle $\omega_{b}$ onto the circumcircle of $A B C$ and the line $U X$ onto the line tangent to the circumcircle at $T_{a}$, which is parallel to $B C$; thus $U X \| B C$. The same is true of $V Y$, so that $U X\|B C\| V Y$.

Let $T_{a} T_{b}$ cut $A C$ at $P$ and let $T_{a} T_{c}$ cut $A B$ at $Q$. The point $X$ lies on the hypotenuse $P M_{b}$ of the right triangle $P U M_{b}$ and is equidistant from $U$ and $M_{b}$. So $X$ is the midpoint of $M_{b} P$. Similarly $Y$ is the midpoint of $M_{c} Q$.

Denote the incentre of triangle $A B C$ as usual by $I$. It is a known fact that $T_{a} I=T_{a} B$ and $T_{c} I=T_{c} B$. Therefore the points $B$ and $I$ are symmetric across $T_{a} T_{c}$, and consequently $\angle Q I B=\angle Q B I=\angle I B C$. This implies that $B C$ is parallel to the line $I Q$, and likewise, to $I P$. In other words, $P Q$ is the line parallel to $B C$ passing through $I$.

Clearly $M_{b} M_{c} \| B C$. So $P M_{b} M_{c} Q$ is a trapezoid and the segment $X Y$ connects the midpoints of its nonparallel sides; hence $X Y \| B C$. This combined with the previously established relations $U X\|B C\| V Y$ shows that all the four points $U, X, Y, V$ lie on a line which is the common tangent to circles $\omega_{b}, \omega_{c}$. Since it leaves these two circles on one side and the circle $\omega_{a}$ on the other, this line is just the line $p_{a}$ from the problem statement.

Line $p_{a}$ runs midway between $I$ and $M_{b} M_{c}$. Analogous conclusions hold for the lines $p_{b}$ and $p_{c}$. So these three lines form a triangle homothetic from centre $I$ to triangle $M_{a} M_{b} M_{c}$ in ratio $1 / 2$, hence similar to $A B C$ in ratio $1 / 4$.