Team selection tests

a) We will prove that $SM$ and $SN$ are isogonal in $\angle ASB$, since $(SMN)$ touches $(SAB)$ and $J$ belongs to the line connecting $S$ and the center of $(SAB)$. Indeed, according to Steiner's theorem for pairs of isogonals, we need to show that $$\frac{MA}{MB}\cdot \frac{NA}{NB}=\frac{S{A}^2}{S{B}^2}.$$ On the other hand, since $HM$ and $KN$ are symmedian of triangles $HAB$ and $KAB$, let $Z$ be the intersection of $HK$ with $AB$. The left hand side of the above equation can be calculated by $$\frac{MA}{MB}\cdot \frac{NA}{NB}=\frac{H{A}^2}{H{B}^2}\cdot \frac{K{A}^2}{K{B}^2}=\frac{Z{A}^2}{Z{B}^2}=\frac{A{F}^2}{B{E}^2}.$$ Next, suppose that $SR,AR,BR$ meet $AB,RB,RA$ at $R^{\prime },F^{\prime },E^{\prime }$ respectively. According to Thales's theorem and Ceva's theorem, we have $$\frac{AF}{BE}=\frac{AF}{SR}\cdot \frac{SR}{BE}=\frac{A{F}^{\prime }}{S{F}^{\prime }}\cdot \frac{B{E}^{\prime }}{S{E}^{\prime }}=\frac{A{R}^{\prime }}{B{R}^{\prime }}=\frac{SA}{SB}.$$ In short, we get $$\frac{MA}{MB}\cdot \frac{NA}{NB}=\frac{A{F}^2}{B{E}^2}=\frac{S{A}^2}{S{B}^2}.$$ so $SM,SN$ is isogonal in $\angle ASB$ and the center of $(SMN)$ lies on the fixed line.

b) By the lemma in Problem 3, $SE$ and $SF$ are isogonal with respect to $\angle ASB$. We will prove that the median at $G$ of the triangle $GPQ$ passing through the fixed point $L$ is the midpoint of the arc $CD$ that does not contain $S$ of $(SCD)$. Rewriting the problem in a more compact form as follows: Let $SAB$ be a triangle with $I$ is the midpoint $AB$, any two points $C,D$ on $SB,SA$. Two points $U,V$ belong to the circumcircle of triangle $SAB$ such that $SU,SV$ is isogonal in $\angle ASB$ and $G$ is the intersection of $AU$ with $BV$. Let $d$ be the angle bisector $\angle ASB$, on the line through $B$ and $A$ parallel to $d$, take the points $P$ and $Q$ satisfying $CP\parallel SU$, and $DQ\parallel SV$. Let $T$ be the midpoint of $PQ$. Prove that $GT$ passes through the midpoint $L$ of arc $CD$ that does not contain $S$ of the circumcircle of triangle $SCD$. Let $K$ be the second intersection of $SL$ and the circumcircle of triangle $SAB$, let $J$ be the second intersection of the circumcircle of triangle $SCD$ with the circumcircle of triangle $SAB$. It is easy to see that $TI$ is the midline of the trapezoid $AQPB$, so $TI\parallel AQ\parallel SL$. Therefore, by Thales's theorem, we only need to prove that $$\frac{TI}{LK}=\frac{GI}{GK}.$$



First of all, we have $$\begin{array}{rl}\frac{GI}{GK}& =\frac{GI}{GA}\cdot \frac{GA}{GK}=\frac{\sin GKA}{\sin GAK}\cdot \sin GAI\\ & =\frac{\sin UAB}{\sin UAK}\cdot \sin GKA=\frac{UB}{UK}\cdot \sin GKA.\end{array}$$ On the other hand, since $$\angle QAD=\angle KSA=\angle AVK\text{ and }\angle QDA=\angle VSA=\angle VKA$$ then $△BUK=△AVK\sim △QAD$ and similarly they are similar to $△PBC$. On the other hand, by the rotation predicate we have $$△SAD\sim △SKL\sim △SBC.$$ From the above pairs of similar triangles, we have the ratio transformation $$\frac{UB}{UK}=\frac{AQ}{AD}=\frac{BP}{BC}=\frac{AQ+BP}{AD+BC}.$$ Finally, since $AQ+BP=2LK$, according to the property of the midline of the trapezoid and the Ptolemy's theorem, $$KA(JA+JB)=JK\cdot AB$$ so we get the following $$\begin{array}{rl}\frac{TI}{LK}& =\frac{AQ+BP}{2LK}\\ & =\frac{AD+BC}{2LK}\cdot \frac{UB}{UK}\\ & =\frac{JA+JB}{2JK}\cdot \frac{UB}{UK}\\ & =\frac{AB}{2KA}\cdot \frac{UB}{UK}\\ & =\frac{UB}{UK}\cdot \sin GKA\\ & =\frac{GI}{GK}.\end{array}$$ So the equality is proved and $GT$ passes through $L$. Therefore, the median of vertex $G$ in triangle $GPQ$ passes through the midpoint of arc $CD$ which does not contain $S$ of the circumcircle of triangle $SCD$, which is a fixed point. $\square$