Problem 1882

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 $I_aI_bI_c$, $DEF$, respectively.

Since $V$ is the circumcenter of triangle $I_aI_bI_c$, $\angle SVI_a = \angle LID$.
By Lemma 2, $\angle AKX = \angle LID$. Again according to Lemma 1, $H$, $S$, $V$ are collinear so $\angle AKX = \angle SVI_a = \angle AHV$.■

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 $\triangle XTL \sim \triangle AIO$.

*Proof.* Draw the diameter $AA'$ of $(O)$. $A'I$ 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 $DM \parallel XI_a$, we obtain $\angle JLX = \angle JI_aX = \angle AMK = \angle AA'I$.
We also have $\angle JXL = \angle JID = \angle XAI = \angle IAO$ so $\triangle XJL \sim \triangle AIA'$.
Since $O$, $J$ are the midpoints of $AA'$ 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 $VU \perp BC$. $X'$ is symmetrical to $X$ with respect to $AI$, $T$ is symmetrical to $A$ with respect to $EF$. $N$ is the midpoint of $XX'$. Draw $HW \perp UV$.
According to Lemma 3, we have $\angle A'AI = \angle XTA = \angle AHV$, therefore $\angle IAO = \angle HAI = \angle (A'A, HV) = \angle (A'A, JO)$.
Since $OA^2 = OI \cdot OS$, we obtain $\angle IAO = \angle ASO$.
So $\angle (SA, SO) = \angle (A'A, JO)$, from which $\angle A'AS = \angle JOS$. (1)
On the other hand, by Lemma 4, $\triangle AIO \sim \triangle XX'U$ so $\frac{SA}{SO} = \frac{AI}{AO} = \frac{XX'}{XU} = \frac{2XN}{HW} = \frac{2XT}{HV}$ (due to $\triangle XNT \sim \triangle HWV$) $= \frac{A'A}{JO}$ (2).

From (1) and (2), we deduce $\triangle SA'A \sim \triangle SJO$.
Similarly, we deduce $\triangle SA'A \sim \triangle SB'B \sim \triangle SC'C \sim \triangle SJO$.
So the two triangles $ABC$ and $A'B'C'$ are similar and have two circumcenters $O$ and $J$, respectively.