THE Tenth STARS OF MATHEMATICS COMPETITION

The three lines in question are concurrent at the center of the nine-point circle $\omega$ of the triangle $ABC$. In what follows, polarity always refers to $\omega$.

To prove that the center of $\omega$ lies on the perpendicular through $A$ to the line $S_A{T}_A$, it is sufficient to show that the latter is the polar of $A$; a similar argument applies to $B$ and $C$.



Showing that $T_A$ also lies on the polar of $A$ involves only elements relative to vertex $A$ and the subscript $A$ is henceforth dropped out to write $M,P,T$ instead of $M_A,P_A,T_A$, respectively. Let the line $AM$ meet $\omega$ again at $K$, and let $L$ be the midpoint of the line segment joining $A$ to the orthocenter of the triangle $ABC$. Completion of the cyclic quadrangle $KLMP$ shows that the polar of $A$ passes through the point where the lines $KL$ and $MP$ meet, so it is sufficient to show $K,L,T$ collinear. To this end, we show that the lines $KL$ and $LT$ are both perpendicular to the line $AM$. Recall that $L$ is the antipode of $M$ in $\omega$, and $LM$ is parallel to the circumradius through $A$. The former implies that the lines $AKM$ and $KL$ are perpendicular, and the latter implies that so are the lines $AT$ and $LM$. Since the lines $ALP$ and $MPT$ are also perpendicular, $L$ is the orthocenter of the triangle $AMT$, so the lines $AM$ and $LT$ are perpendicular.

Remark. Since the polar of $A$ passes through $T_A$, the polar of $T_A$ passes through $A$. Now let the tangents of $\omega$ at $M_A$ and $P_A$ meet at $X_A$; the points $X_B$ and $X_C$ are defined similarly. The line $M_A{P}_A$ through $T_A$ is the polar of $X_A$, so the polar of $T_A$ passes through $X_A$. Hence the line $A{X}_A$ is the polar of $T_A$; similarly, the lines $B{X}_B$ and $C{X}_C$ are the polars of $T_B$ and $T_C$, respectively. Since $T_{A}B/T_{A}C=(AB/AC{)}^2$ and the like, the points $T_A,T_B,T_C$ are collinear, so their polars, $A{X}_A,B{X}_B,C{X}_C$, are concurrent; that is, the triangles $ABC$ and $X_A{X}_B{X}_C$ are in perspective.

Notice that $B$ and $C$ also lie on the polar of $X_A$ to infer that the polars of $B$ and $C$ meet at $X_A$; similarly, the polars of $C$ and $A$ meet at $X_B$, and the polars of $A$ and $B$ meet at $X_C$. Consequently, $S_A,T_A,X_B,X_C$ all lie on the polar of $A$ and the like.