Problem 1950

First, we present two following lemmas. Let $ABC$ be a triangle inscribed in $(O)$ and $N$ be the nine-point center of $\triangle ABC$.

Lemma 1. Let $K$ be the center of $(BOC)$, then $AN$, $AK$ are isogonal with respect to $\angle BAC$.

*Proof.* Let $X$ be the reflection of $O$ through $BC$. Let $H$ be the orthocenter of $\triangle ABC$. We have $AH = OX$, and $AH$ is parallel to $OX$. Hence, $AX$ cuts $OH$ at the midpoint of $OH$, which is $N$, and thus $N$ is the midpoint of $AX$. We have $\angle OXB = \angle KOB = \angle KBO$, so $\triangle OKB \sim \triangle OBX$, implying that $\frac{OK}{OB} = \frac{OB}{OX}$, or
$$OX \cdot OK = OB^2 = OA^2.$$
Hence, we have $\frac{OK}{OA} = \frac{OA}{OX}$, implying that $\triangle OKA \sim \triangle OAX$. Therefore, $\angle OKA = \angle XAO = \angle DAK$.
Since $AH$, $AO$ are isogonal with respect to $\angle BAC$, we have $\angle BAK = \angle NAC$, implying that $AN$, $AK$ are isogonal with respect to $\angle BAC$.

Lemma 2. Let $B'$, $C'$ be the reflections of $B$, $C$ through $AC$, $AB$, respectively, then $AK$ is perpendicular to $B'C'$.
*Proof.* Let $Y$, $Z$ be the reflection of $O$ through $CA$, $AB$. Similar to the argument above, we have $N$ is the midpoint of $BY$ and $CZ$. Let $N_b$, $N_c$ be the reflection of $N$ through $CA$, $AB$. Since $OC'$ is the reflection of $ZC$ through $AB$, and $N$ is the midpoint of $ZC$, then $N_c$ is the midpoint of $OC'$. Similarly, $N_b$ is the midpoint of $OB'$. Hence, $N_bN_c$ is parallel to $B'C'$. We have
$$\angle KAN_b = \angle N_bAC + \angle KAC = \angle NAC + \angle NAB = \angle BAC,$$
and similarly, $\angle KAN_c = \angle BAC$. On the other hand, we have $AN_b = AN = AN_c$, and since $\angle KAN_b = \angle KAN_c$, we get $AK$ is perpendicular to $N_bN_c$. So $AK \perp B'C'$, as desired. $\Box$

Back to the main problem, let $O_a$, $O_b$, $O_c$ be the circumcenters of $\triangle NBC$, $\triangle NCA$, $\triangle NAB$ and $K_a$, $K_b$, $K_c$ be the circumcenters of $\triangle O_aBC$, $\triangle O_bCA$, $\triangle O_cAB$. By Lemma 2, we have
$$m_a \equiv NK_a,\quad m_b \equiv NK_b,\quad m_c \equiv NK_c$$

a) If $N$ is the orthocenter $H$, let $m'_a$, $m'_b$ and $m'_c$ be the reflections of $m_a$, $m_b$ and $m_c$ through the bisectors of angles $BHC$, $CHA$ and $AHB$. Let $J$ be the nine-point center of $ABC$. Note that $(J)$ is also the nine-point circle of $BHC$. By Lemma 1, we have $HK_a$, $KJ$ are isogonal with respect to $\angle BHC$, so $m'_a$ goes through $J$. Similarly, $m'_b$, $m'_c$ also go through $J$, and thus $m'_a = m'_b = m'_c = HJ$.

b) Assume that $N$ is the nine-point center of $\triangle ABC$. Let $N_a$, $N_b$ and $N_c$ be the reflections of $N$ through $BC$, $CA$ and $AB$. Let $(X)$, $(Y)$ and $(Z)$ be the nine-point circles of $\triangle NBC$, $\triangle NCA$ and $\triangle NAB$. We will show that the reflections of $m_a$, $m_b$ and $m_c$ through $BC$, $CA$ and $AB$ concur on $N_aN_bN_c$.
Denote $(N)$ the nine-point circle of $\triangle ABC$. Let $P$ be the Poncelet point of the four points $A$, $B$, $C$ and $N$; $E$, $F$ be the midpoints of $AC$, $AB$. We have $P$, $E$ are the intersections of $(N)$ and $(Y)$, and $P$, $F$ are the intersections of $(N)$ and $(Z)$. Hence, $PE \perp NY$, $PF \perp NZ$, and we have
$$\angle YNZ = 180^\circ - \angle FPE = 180^\circ - \angle BAC.$$
Let $S$, $T$ be the intersections of $NK_b$, $NK_c$ with $CA$, $AB$, respectively. Let $K$ be the center of $N_aN_bN_c$. Denote the intersection of the reflection line of $m_b$, $m_c$ through $AC$, $AB$ as $L$. We have $BK \perp N_aN_c$, $CK \perp N_bN_c$, and $K$ is the isogonal conjugate of $N$ in $\triangle ABC$.

We have then
$$\begin{align*} \angle LN_bN_a &= 180^\circ - \angle RSN_b - \angle N_bRA \\ &= 180^\circ - \angle NSC - (90^\circ - \angle ACK) \\ &= 90^\circ - \angle NSC + \angle ACK. \end{align*}$$
And similarly, $\angle LN_cN_a = 90^\circ - \angle NTB + \angle ABK$. It suffices to show that $L \in (N_aN_bN_c)$, and to show that, we need to show $\angle ACK + \angle ABK = \angle NSC + \angle NTB$. Since $NK_b$, $NY$ are isogonal with respect to $\angle ANC$, we have
$$\angle NSC = 180^\circ - \angle K_bNC - \angle NCA = \angle YNA - \angle NCA,$$
and similarly, $\angle NTB = \angle ZNA - \angle NBA$. Thus,
$$\angle NSC + \angle NTB = \angle YNZ - (\angle NBA + \angle NCA).$$
So we need to show that
$$\angle B + \angle C = \angle NBA + \angle NCA + \angle ABK + \angle ACK,$$
but this is true since, $\angle B = \angle ABN + \angle CBN = \angle ABN + \angle ABK$ and $\angle C = \angle ACN + \angle BCN = \angle ACN + \angle ACK$. The proof is completed. $\square$