Team selection tests

We state some lemmas as follows.

Lemma 1. Given triangle $ABC$ inscribed in $(O)$, altitudes $AD$, $BE$, $CF$. $L$ is the Lemoine point of triangle $ABC$. $K$ is the orthocenter of triangle $DEF$. Then $O$, $L$, $K$ are collinear.



Proof. Let $H$ be the orthocenter of triangle $ABC$. $X$, $Y$, $Z$ are the midpoints of $EF$, $DF$, $DE$ respectively. $N$ is the Euler center of triangle $ABC$, $J$ is the incenter of triangle $XYZ$, $G$ is the centroid of triangle $DEF$. Since $N$ is the circumcenter of triangle $DEF$ so $K$, $G$, $N$ are collinear and $\frac{\overline{GK}}{\overline{GN}}=-2$. The homothety with center $G$ and ratio $-2$ maps triangle $XYZ$ into triangle $DEF$, so it maps $J$ into $H$. So $J$ is the midpoint of $KO$. Note that $XJ\parallel DA$ implies $XJ\perp BC$. Similarly, we deduce that the straight lines passing through $X$, $Y$, $Z$ and perpendicular to $BC$, $CA$, $AB$, respectively, concur at $J$. The straight lines through $A$, $B$, $C$ and perpendicular to $YZ$, $XZ$, $XY$, respectively, concur at $O$ and $AX$, $BY$, $CZ$ concur at $L$ so according to Sondat's theorem, $L$, $J$, $O$ are collinear. So $K$, $L$, $O$ are collinear. ■

Lemma 2. Given triangle $ABC$ inscribed in a circle $(O)$, the tangent at $B$ and $C$ of $(O)$ intersect at $T$. $H$ is the projection of $T$ on the tangent at $A$ of $(O)$. $S$ is symmetrical to $T$ through $BC$. $L$ is the Lemoine point of triangle $ABC$. Then $\angle HST=\angle AOL$.

Proof. (Truong Tuân Nghĩa, student K53 high school specializing in Natural Sciences) $AH$ meets $BC$ at $X$. The line through $X$ perpendicular to $OL$ intersects $OT$ at $K$. $KO$ meets $AH$ at $P$. $AT$ meets $BC$ at $R$. We have $O(AR,LT)=-1$ and $XP\perp OA$, $XT\perp OR$, $XK\perp OL$, $XM\perp OT$ so $(PT,KM)=-1$. Again, $M$ is the midpoint of $ST$, so applying the same formula as Maclaurin and Newton, we obtain $\overline{PS}\cdot \overline{PK}=\overline{PM}\cdot \overline{PT}=\overline{PH}\cdot \overline{PX}$. We deduce that $XHSK$ is cyclic. It follows that $\angle HST=180^{\circ }-\angle HXK=\angle AOL$. ■



Lemma 3. Let the triangle $ABC$ be inscribed in a circle $(O)$ and circumscribed about a circle $(I)$. $(I)$ touches $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. $H$ is the orthocenter of triangle $ABC$. $AH$ meets $BC$ at $X$. $K$ is symmetrical to $A$ through $EF$. $V$ is a point symmetrical to $I$ through $O$. Then $\angle AHV=\angle AKX$.

Proof. Let $I_a,I_b,I_c$ be the excenters of triangle $ABC$ with respect to $A$, $B$, $C$ respectively. Let $S$, $L$ be the Lemoine points of triangle $I_a{I}_b{I}_c$, $DEF$, respectively.



Since $V$ is the circumcenter of triangle $I_a{I}_b{I}_c$, $\angle SV{I}_a=\angle LID$. By Lemma 2, $\angle AKX=\angle LID$. Again according to Lemma 1, $H$, $S$, $V$ are collinear so $\angle AKX=\angle SV{I}_a=\angle AHV$.■

Lemma 4. Let the triangle $ABC$ be inscribed in a circle $(O)$ and circumscribed about a circle $(I)$. $(I)$ is tangent to $BC$ at $D$. $AX\perp BC$. $T$ is symmetrical to $X$ through $AI$. $L$ is the point of contact of the excircle ($I_a$) with $BC$. Then $△XTL\sim △AIO$.



Proof. Draw the diameter $A{A}^{\prime }$ of $(O)$. $A^{\prime }I$ meets $(O)$ at $K$. $AI$ meets $(O)$ at $M$. $KM$ meets $AX$ at $N$. $J$ is the midpoint of $XT$. We have the well-known result that $K$, $D$, $M$ are collinear and $INXD$ is a rectangle. It follows that $DM\parallel X{I}_a$, we obtain $\angle JLX=\angle J{I}_{a}X=\angle AMK=\angle A{A}^{\prime }I$. We also have $\angle JXL=\angle JID=\angle XAI=\angle IAO$ so $△XJL\sim △AI{A}^{\prime }$. Since $O$, $J$ are the midpoints of $A{A}^{\prime }$ and $XT$ respectively, we get $\Delta XTL\sim \Delta AIO$. ■

Back to the problem.



Let $S$ be the inverse of $I$ with respect to $(O)$. $J$ is the midpoint of $HI$. $V$ is a point symmetrical to $I$ through $O$. Draw $VU\perp BC$. $X^{\prime }$ is symmetrical to $X$ with respect to $AI$, $T$ is symmetrical to $A$ with respect to $EF$. $N$ is the midpoint of $X{X}^{\prime }$. Draw $HW\perp UV$. According to Lemma 3, we have $\angle A^{\prime }AI=\angle XTA=\angle AHV$, therefore $\angle IAO=\angle HAI=\angle (A^{\prime }A,HV)=\angle (A^{\prime }A,JO)$. Since $O{A}^2=OI\cdot OS$, we obtain $\angle IAO=\angle ASO$. So $\angle (SA,SO)=\angle (A^{\prime }A,JO)$, from which $\angle A^{\prime }AS=\angle JOS$. (1) On the other hand, by Lemma 4, $△AIO\sim △X{X}^{\prime }U$ so $\frac{SA}{SO}=\frac{AI}{AO}=\frac{X{X}^{\prime }}{XU}=\frac{2XN}{HW}=\frac{2XT}{HV}$ (due to $△XNT\sim △HWV$) $=\frac{A^{\prime }A}{JO}$ (2).

From (1) and (2), we deduce $△S{A}^{\prime }A\sim △SJO$. Similarly, we deduce $△S{A}^{\prime }A\sim △S{B}^{\prime }B\sim △S{C}^{\prime }C\sim △SJO$. So the two triangles $ABC$ and $A^{\prime }{B}^{\prime }{C}^{\prime }$ are similar and have two circumcenters $O$ and $J$, respectively.