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## The point of concurrency lies on the line ON

Problem: Let $ABC$ be a triangle, $O$ is the circumcenter, $\left(N\right)$ is the nine point circle and $F_{a}F_{b}F_{c}$ is the Feuerbach triangle. The perpendicular to the line $BC$ through the point $N$ intersects $\left(N\right)$ at the point $Q_{a}$ (the points $A$ and $Q_{a}$ lie on the same side of the line $BC$) and similarly construct the points $Q_{b}$, $Q_{c}$. Prove that the lines $F_{a}Q_{a}$, $F_{b}Q_{b}$, $F_{c}Q_{c}$ are concurrent and the point of concurrency lies on
the line $ON$.

My proof:
Construct the circumcircle of triangle $ABC$. Let $U$ be the midpoint of the arc $BC$ which does not contain $A$. Let $M_a$ be the midpoint of $BC$. $I_aM_a$ intersects the A-altitude of triangle $ABC$ at $X_a$, similar with $X_b, X_c$. Let $D, E, F$ be the tangencies of $(I_a)$ and $AC, BC, AB$ respectively.
Note that $U$ is the circumcenter of triangle $BCI_a$ then by Thales’s theorem: $\dfrac{UM_a}{AX_a}=\dfrac{I_aU}{I_aA}=\dfrac{UB}{I_aA}.$
But $\Delta UBM_a\sim \Delta I_aAD \Rightarrow \dfrac{UB}{I_aA}=\dfrac{UM_a}{I_aD}$. Therefore $\dfrac{UM_a}{AX_a}=\dfrac{UM_a}{I_aD}$ or $AX_a=r_a$.
Let $K$ be the reflection of $E$ wrt $I_a$ then $AX_aI_aK$ is a parallelogram. We get $AI_a$ intersects $KX_a$ at $T$ such that $T$ is the midpoint of $AI_a$. From my proof at the last corollary of Fontene’s theorem we can apply for the case $P\equiv I_a$ and get $F_a$ lies on the circle with diameter $ET$. If $F'_a$ is the projection of $E$ on $KX_a$ then $F'_a\in (ET)$ and $F'_a\in (I_a)$ so $F'_a\equiv F_a$, or $F_a\in X_aK$. But it is easy to see that $K, F_a, Q_a$ are collinear then $X_a, F_a, Q_a$ are collinear.
Let $Q'_a$ be the reflection of $Q_a$ wrt $N$. Construct the tangents of $(N)$ through $Q'_a, Q_b, Q_c$, they intersect each other and make triangle $A'B'C'$. Note that $A'B'//AB, B'C'//BC, C'A'//AC$ and $D,E,F; Q_b,F_a,Q_c$ are the tangencies of A and A’-excircle with their sides so $Q_bQ_c//DF$ or $Q_bQ_c//X_bX_c$. Similarly, we claim that $\Delta Q_aQ_bQ_c$ and $X_aX_bX_c$ have $Q_cQ_a//X_cX_a, Q_aQ_b//X_aX_b$. This means $X_aQ_a, X_bQ_b, X_cQ_c$ concur at their homothetic center. $(1)$
Let $L, P$ be the tangencies of $(I_b)$ and $BC, AB; J, Q$ be the tangencies of $(I_c)$ and $BC, AC$.
From my alticle at here we obtain $LP, QJ$ and A-altitude of triangle $ABC$ are concurrent and denote $X'_a$ as their concurrency. Because $BJ=CL=p-a$ then $M_a$ is the midpoint of $JL$.
Note that $\Delta BI_aC\sim \Delta LX'_aJ$ so by Thales’s theorem, $X'_aI_a$ must passes through the midpoint of $BC$, or $X'_a\equiv X_a$.
Let $G$ be the projection of $X_a$ on $I_bI_c, H_a$ be the projection of $A$ on $BC$. We have:
$\angle JX_aH_a=90^o-\angle X_aJC=\angle C/2=90^o-\angle ACI_b$
$=90^o-\angle X_aB_2C_2=\angle B_2X_aG$.
Note that $X_bX_c//I_bI_c$ then $X_aG$ is the altitude of triangle $X_aX_bX_c$, which follows that $X_aA$ passes through the circumcenter of triangle $X_aX_bX_c$. This means the orthocenter $H$ of triangle $ABC$ is the circumcenter of triangle $X_aX_bX_c. (2)$
From $(1)$ and $(2)$ we get the homothetic center of two triangles $X_aX_bX_c$ and $Q_aQ_bQ_c$ lies on $HN$. Our problem is solved.

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## Ten concyclic points

Problem (nsato): Let $A_1, A_2,A_3, A_4$ and $A_5$ be five concyclic points.  For $1 \le i < j \le 5$, let $X_{i,j}$ be the intersection of the Simson lines of $A_i$ and $A_j$ with respect to the triangle formed by the other three points.  Show that all ten such points $X_{i,j}$ are concyclic.

My proof:

Lemma 1: Given 4 points $A_1, A_2, A_3, A_4$ on the circle $(O)$. Then the Simson lines $d_1, d_2, d_3, d_4$ of $A_1, A_2, A_3, A_4$ wrt $\Delta A_2A_3A_4, A_1A_3A_4, A_1A_2A_4, A_1A_2A_4$, respectively, concur at the orthocenter $X$ of $A_1A_2A_3A_4$. Let $G$ be the centroid of $A_1A_2A_3A_4$ then $\vec{OG}=\frac{1}{2}\vec{OX}$.

Proof: Denote $H_2, H_3$ the orthocenters of the triangles $A_1A_3A_4$ and $A_1A_2A_4$. Then $A_2H_3//=A_3H_2$. We get $A_2H_3H_2A_3$ is a parallelogram. Therefore $A_2H_2, A_3H_3$ concur at the midpoint $X$ of $A_2H_2$ and $A_3H_3$. It’s easy to see that $X$ also lies on $d_2$ and $d_3$ Similarly we get $d_1,d_2,d_3,d_4$ concur at $X$.
Let $M, N$ be the midpoints of $A_1A_3$ and $A_2A_4$ then $G$ is the midpoint of $MN.$
We have $XN//=\frac{1}{2}A_4H_2//=OM$ so $XNOM$ is a parallelogram. We get $G$ is the midpoint of $OX$.

Lemma 2: Given 5 points $A_1, A_2, A_3, A_4, A_5$ on the circle $(O)$. Denote $H_1, H_2, H_3, H_4 , H_5$ the orthocenters of $A_2A_3A_4A_5, A_1A_3A_4A_5, A_1A_2A_4A_5, A_1A_2A_3A_5, A_1A_2A_3A_4,$ respectively. Then 5 points $H_1, H_2, H_3, H_4, H_5$ are concyclic.
Proof: Let $G_1, G_3,G_4$ be the centroids of $A_2A_3A_4A_5, A_1A_2A_4A_5, A_1A_2A_3A_5$, respectively. According to lemma 1 we obtain $G_1G_4//H_1H_4$. Let $M, N, P$ be the midpoints of $A_3A_5, A_2A_4, A_1A_2$ then $G_1G_4$ is the midline of the triangle $MNP$. This means $G_1G_4//NP//A_1A_4$. Similarly, $G_3G_4//A_3A_4$. Therefore $\angle H_1H_4H_3=\angle A_1A_4A_3$.
Similarly $\angle H_1H_5H_3=\angle A_1A_5A_3=\angle A_1A_4A_3=\angle H_1H_4H_3$. So $H_1,H_3,H_4,H_5$ are concyclic. Similarly we are done.

Lemma 3: Given a triangle $ABC$ with its circumcircle $(O)$. Let $E, F$ be arbitrary points on $(O)$ then the angle between the Simson lines of two points $E,F$ is half the measure of the arc $EF$.

This lemma is trivival, so I will leave it to you.

Back to our problem: Let $H_1,H_2, H_3, H_4, H_5$ be the orthocenters of 5 triangles $A_iA_jA_k$ ($1\leq i). Applying the lemma 2 we obtain $H_1H_2H_3H_4H_5$ are cyclic. Applying the lemma 1 we get the Simson lines $d, l$ of $A_3, A_1$ wrt $\Delta A_2A_4A_5, A_2A_4A_5$ passes through $H_1, H_3$. Let $F$ be the intersection of $d$ and $l$ then applying the lemma 3 : $\angle H_1FH_3=\angle A_1A_4A_3=\angle H_1H_4H_3$, which follows that $F$ lies on $(H_1H_2H_3H_4H_5)$. Similarly we are done.

Another corollary of lemma 2:

Problem 2 (Nguyen Van Linh): Given a cyclic pentagon $ABCDE$. Let $A',B',C',D',E'$ be the second intersections of the 9-point circles of triangles $ABC,BCD,CDE,DEA,EAB$. Prove that $A',B',C',D',E'$ are concyclic.

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## From coaxal circles to the generalization of Steinbart’s theorem

Problem 1(Tran Quang Hung): Let $ABC$ be a triangle with its circumcircle $(O)$. Let $P$ and $Q$ be arbitrary points such that $P, O, Q$ are collinear. $A_1B_1C_1$ is the pedal triangle of $P$ wrt $\Delta ABC, A_2B_2C_2$ is the circumcevian triangle of $Q$ wrt $\Delta ABC$. Show that $(PA_1A_2), (PB_1B_2), (PC_1C_2)$ are coaxal.
Proof:

Denote $O_a, O_b, O_c$ the centers of $(PA_1A_2), (PB_1B_2), (PC_1C_2)$; $X$ the intersection of the line through $A_2$ and perpendicular to $PA_2$. Similarly we define $Y, Z$.
Note that $O_a$ is the midpoint of $PX$ therefore $(O_a), (O_b), (O_c)$ are coaxal iff $O_a, O_b, O_c$ are collinear iff $X, Y, Z$ are collinear. $(1)$
We have $\dfrac{XB}{XC}=\dfrac{S_{XBA_2}}{S_{XCA_2}}=\dfrac{\sin \angle XA_2B. A_2B}{\sin \angle XA_2C. A_2C}=\dfrac{\sin \angle XA_2B}{\sin \angle XA_2C}.\dfrac{\sin \angle BAQ}{\sin \angle CAQ}$
But $\dfrac{\sin \angle XA_2B}{\sin \angle XA_2C}=\dfrac{\cos \angle BA_2P}{\cos \angle CA_2P}$.
Let $A_3$ be the intersection of $A_2P$ and $(O), A_4$ be the intersection of $A_3O$ and $(O)$. Similarly we define $B_3, B_4, C_3, C_4$.
We have $\dfrac{\cos \angle BA_2P}{\cos \angle CA_2P}=\dfrac{\cos \angle BA_4A_3}{\cos \angle CA_4A_3}=\dfrac{\sin \angle BA_3A_4}{\sin \angle CA_3A_4}=\dfrac{\sin \angle BAA_4}{\sin \angle CAA_4}$
Hence $\dfrac{XB}{XC}=\dfrac{\sin \angle BAQ}{\sin \angle CAQ}.\dfrac{\sin \angle BAA_4}{\sin \angle CAA_4}.$
Do the same with $\dfrac{YC}{YA}, \dfrac{ZA}{ZB}$ then applying Ceva-sine theorem we obtain $(1)$ $\Leftrightarrow AA_4, BB_4, CC_4$ are concurrent.
On the other side, according to Pascal’s theorem for 6 points $A, B, A_3, B_3, A_2, B_2$ we claim $M, P, Q$ are collinear ($M$ is the intersection of $AB_3$ and $BA_3$)
Applying Pascal’s theorem again for 6 points $A, B, A_3, B_3, A_4, B_4$ we get the intersection of $AA_4$ and $BB_4$ lies on $PQ$. Similarly the intersection of $CC_4$ and $BB_4$ lies on $PQ$. This means $AA_4, BB_4, CC_4$ are concurrent. We are done.

Corollaries:

(1) (Nguyen Van Linh): Given triangle $ABC$ and its circumcircle $(O)$. Denote $M_a, M_b, M_c$ the midpoints of $BC, CA, AB$, respectively, $P$ an arbitrary point inside $\Delta ABC, A',B',C'$ the second intersections of $AP, BP, CP$ and $(O)$. Prove that $(A'OM_a), (B'OM_b), (C'OM_c)$ concur at two points.

(2) ( 77ant) : For $\triangle ABC$ with its orthocenter $H$, $\triangle DEF$ is its orthic triangle.
$AK$, $BL$, $CM$ are perpendicular to $EF$, $FD$, $DE$ respectively.
Prove that three circumcircles of $\triangle KHD$, $\triangle LHE$, $\triangle FHM$ are coaxal.

But it is not the end of the line, I found that the generalization of Steinbart’s theorem is the application of problem 1:
Problem 2 (Nguyen Van Linh): Given a triangle $ABC$. Let $P$ be an arbitrary point in the plane, $A_1B_1C_1$ be the pedal triangle of $P$ wrt $\Delta ABC$. Let $O$ be the center of $(A_1B_1C_1)$,$L$ be an arbitrary point on $PO$. $A_1L, B_1L, C_1L$ intersects $(A_1B_1C_1)$ at $A_2,B_2,C_2$, respectively. Prove that $AA_2, BB_2, CC_2$ are concurrent.

Proof:

Let $Q$ be the isogonal conjugate of $P$ wrt $\Delta ABC. X_3, Y_3, Z_3$ be the projections of $P$ onto $B_2C_2, C_2A_2, A_2B_2$.
Consider the homothetic with center $P$ and ratio 2: $O\mapsto Q, A_2\mapsto X, B_2\mapsto Y, C_2\mapsto Z, A_1\mapsto X_2$, $B_1\mapsto Y_2, C_1\mapsto Z_2, X_3\mapsto X_1, Y_3\mapsto Y_1, Z_3\mapsto Z_1; L\mapsto T.$
Then $Q$ is the center of $(XYZ)$ and $X_1Y_1Z_1$ is the pedal triangle of $P$ wrt $\Delta XYZ$.
We have $T, L, P$ are collinear therefore $T, P, Q$ are collinear. Moreover, $T$ is the intersection of $XX_2, YY_2, ZZ_2$.
By problem 1, we get $(PX_1X_2), (PY_1Y_2), (PZ_1Z_2)$ are coaxal, which follows that the intersections of $B_2C_2$ and $BC, A_2C_2$ and $AC, A_2B_2$ and $AB$ are collinear. According to Desargues’s theorem we claim $AA_2, BB_2, CC_2$ are concurrent.

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