IMO 2006 Shortlisted Problems

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 $UX$ be the tangent to ${\omega }_b$ at $U$, with $X$ on $AC$, and let $VY$ be the tangent to ${\omega }_c$ at $V$, with $Y$ on $AB$. The homothety with centre $T_b$ and ratio $T_b{T}_a/T_{b}U$ maps the circle ${\omega }_b$ onto the circumcircle of $ABC$ and the line $UX$ onto the line tangent to the circumcircle at $T_a$, which is parallel to $BC$; thus $UX\Vert BC$. The same is true of $VY$, so that $UX\Vert BC\Vert VY$.

Let $T_a{T}_b$ cut $AC$ at $P$ and let $T_a{T}_c$ cut $AB$ at $Q$. The point $X$ lies on the hypotenuse $P{M}_b$ of the right triangle $PU{M}_b$ 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 $ABC$ 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 $\angle QIB=\angle QBI=\angle IBC$. This implies that $BC$ is parallel to the line $IQ$, and likewise, to $IP$. In other words, $PQ$ is the line parallel to $BC$ passing through $I$.



Clearly $M_b{M}_c\Vert BC$. So $P{M}_b{M}_{c}Q$ is a trapezoid and the segment $XY$ connects the midpoints of its nonparallel sides; hence $XY\Vert BC$. This combined with the previously established relations $UX\Vert BC\Vert VY$ 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 $ABC$ in ratio $1/4$.