Team selection tests

We state the following familiar lemma: LEMMA. Given triangle $ABC$ and two points $X,Y$. Suppose $BX$ cuts $CY$ at $Z$ and $BY$ cuts $CX$ at $T$. Then if $AX$ and $AY$ are congruent in $\angle BAC$ then so are $AZ$ and $AT$.

We first reduce the problem to a simpler form through pairs of similar triangles. Let $D$ be the projection of $H$ on $BC$. It is easy to see that $$△MF{E}^{UK}\sim △OB{C}^{UD}$$ so we think of constructing triangle $TBC$ similar to triangle $RFE$ where $RE$ and $RF$ are tangent to $(PBF)$ and $(QCE)$.

Let $X,Y$ be the intersection of $AP$ and $AQ$ with $EF$. Firstly $$\angle XPF=90^{\circ }-\angle OAX=\angle PBF$$ infer $X$ is on $(BPF)$ or $$\angle RFE=\angle BXA\text{ and }\angle REF=\angle YCA.$$



Therefore, if $△TBC\sim △RFE$ then $BT$, $BX$ is isogonal in $△ABC$ and $CY$, $CT$ is isogonal in $△ACB$. In short, if $S$ is the intersection of $BX$ and $CY$, then $S$, $T$ is isogonal conjugate in triangle $ABC$.

Let $L$ be the intersection of $BY$ with $CX$ and $J$ the isogonal conjugate of $L$ in triangle $ABC$. We will prove that $O$ and $D$ lie on $JT$. Indeed, $$\begin{array}{rl}B(JT,OC)& =B(LS,HA)=(YX,EF)=C(YX,EF)\\ & =C(XY,FE)=C(LS,HA)=C(JT,OB)\end{array}$$ so $J$, $O$, $T$ are collinear.



Next, since $AX$ and $AY$ are isogonal in $\angle BAC$, according to the lemma, $AS$, $AL$ isogonal in $\angle BAC$ leads to $A$, $L$, $T$ and $A$, $J$, $S$ are collinear. Furthermore, in the complete quadrilateral $BLSC.XY$, if $XY$ intersects $BC$ at $D^{\prime }$ then $$L(CB,S{D}^{\prime })=-1=L(CB,D{D}^{\prime }),$$ in other words, $SL$ goes through $D$. Finally, to prove that $JT$, $SL$ and $BC$ are concurrent, we use Desargues' theorem for the pair of triangles $BLT$ and $CSJ$. Since $BT$, $BX$ and $CJ$, $CX$ are isogonal pairs in triangle $ABC$, the intersection $Z$ of $BT$ and $CJ$ is the isogonal conjugate of $X$ in the triangle $ABC$, thus $Z$ is on $AY$. Therefore, the intersection of $(BT,CJ)$, $(BL,CS)$ and $(TL,JS)$ are collinear or $JT$ passes through the point $D$.

Therefore, $T$ belongs to $OD$ so $R$ belongs to $MK$. This finishes the proof. $\square$