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Can you provide a proof for the following proposition:

Proposition. Let $\triangle ABC$ be an arbitrary triangle with nine-point center $N$ and circumcenter $O$. Let $A',B',C'$ be a reflection points of the points $A,B,C$ with respect to the point $N$. Consider the three circles $k_1,k_2,k_3$ defined by the points $AOA'$ , $BOB'$ and $COC'$ , respectively. I claim that $k_1$,$k_2$ and $k_3$ meet at a common point $𝑃$.

enter image description here

GeoGebra applet that demonstrates this proposition can be found here.

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  • $\begingroup$ I have no time to check anything right now, but a quick back-of-the-envelope argument seems to show the following: Let $A'', B'', C''$ be the inverses of the points $A', B', C'$ in the circumcircle of triangle $ABC$. Let $XYZ$ be the triangle formed by the tangents to this circumcircle at $A, B, C$. Then, $A'', B'', C''$ are the midpoints of the segments $OX, OY, OZ$. Equivalently, $A'', B'', C''$ are the circumcenters of triangles $OBC, OCA, OAB$. By inversion in the circumcircle, it suffices to show that the lines $AA'', BB'', CC''$ concur. This is probably easy ... $\endgroup$ Nov 29, 2021 at 20:42
  • $\begingroup$ ... but is definitely known: It follows from locus property 2 of the Neuberg cubic, since X(3) lies on that cubic. $\endgroup$ Nov 29, 2021 at 20:43
  • $\begingroup$ Circumcentre of K1,K2,K3 are Collinear $\endgroup$
    – user423633
    Jun 19, 2022 at 7:36

4 Answers 4

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Such things are quick in complex numbers. Let $O=0$ be the origin, $ABC$ be the unit circle. The centroid of $ABC$ is $G=(A+B+C)/3$, the Euler circle is the image of the circle $ABC$ under homothety $f\colon X\to (3G-X)/2$ centered in $G$ with coefficient $-1/2$, thus $N=f(O)=(A+B+C)/2$. Next, $A'=2N-A=B+C$. The circle through $O,A,B+C$ has equation $g_A(z):=(A^2-BC)z\bar{z}+(B+C-A)z+A(BC-AB-AC)\bar{z}=0$. Such three circles have a common point different from $O$ if their equations are linearly dependent. They are: $(B-C)g_A+(C-A)g_B+(A-B)g_C=0$.

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Another way would be as following, though its incomplete at the end: For a triangle $\triangle{ABC}$ (assume $C\geq A \geq B$). Then the center of $k1$ is the intersection of the lines $l_{11}, l_{12}$ which passes bisects perpendicularly the lines $AO$ and $OA'$ respectively. As $A'$ is the reflection of $A$ about $N$, and $N$ is the middle of $HO$ ($H$ being the orthocenter), $AOA'H$ is a parallelogram, hence $OA'||AH$. So, $OA'$ is perpendicular to $BC$ and intersects $BC$ at the middle at $D$. Also we know, $AH=2OD$, which implies that $OD=DA'$. Hence, the center of $k1$ is the point where the perpendicular bisector of $AO$ cuts the line $BC$, let it be $\vec{C_1}=(x_1,y_1)$.

Let, $l_{12}$ upon extension intersects $AB$ at $T_1$. the, $\overline{AT_1}=\frac{R}{2}\csc{C}$ ($R$ is the circumradius of $\triangle{ABC}$). Also, $\angle{AT_{1}C_1}=C$. Then, similar calculation gives us the position vectors of $\vec{C_i}, i=1,2,3$, taking $B$ as the origin and $\vec{AB}$ line as $+x$ axis.

$\vec{C_1}=\left(\frac{(\frac{R}{2}\csc{C}-c)\cot{B}}{\cot{B}-\cot{C}},\frac{(\frac{R}{2}\csc{C}-c)}{\cot{B}-\cot{C}}\right)$

$\vec{C_2}=\left(-\frac{a\sin{A}-\frac{R}{2}}{\sin{(A-B)}} , 0\right)$

$\vec{C_3}=\left(-c-\frac{(\frac{R}{2}\csc{C}-c)\cot{A}}{\cot{A}-\cot{C}},\frac{(\frac{R}{2}\csc{C}-c)}{\cot{A}-\cot{C}}\right)$

As these circles have one common intersection point $O$, the other common would simply require that $C_1,C_2,C_3$ to be colinear. Hence, showing $\frac{y_3}{y_1}=\frac{x_3-x_2}{x_1-x_2}$ would prove it. $\frac{y_3}{y_1}=\frac{\cot{B}-\cot{C}}{\cot{A}-\cot{C}}$.

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Bringing together complex numbers and laziness. Following the answer by @FedorPetrov, we see that $N=(A+B+C)/2$ and so $A'=B+C$. Take inversion with respect to the unit circle (which is just a map $z\mapsto 1/\bar{z}$), then the circles are mapped to the lines connecting $A$ and $BC/(B+C)$ (and two analogous) and we want to prove they intersect at one point. If it was $A$ and $2BC/(B+C)$ then we would recover Lemoine point since $2BC/(B+C)$ is exactly the intersection of the tangent lines at $B$ and $C$ to the unit circle. So naturally we want to prove that for any $k\in\mathbb{R}$ the line $(A, k\cdot BC/(B+C))$ and two analogous intersect at one point. This condition is cubic in $k$ so we need to find four values of $k$ for which this is true. The value $k=2$ gives Lemoine point as discussed, value $k=0$ trivially gives the circumcenter $O$, the value $k=\infty$ gives the orthocenter $H$. Finally, for $k=-2$, after symmetry w.r.t. $O$ one obtains lines connecting $2BC/(B+C)$ and $-A$ which trivially intersect at the Nagel point of the triangle formed by the tangent lines to the unit circle at $A,B,C$.

All this just to avoid writing explicit formulas...

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It is straightforward to see that $A',B',C'$ are reflections of the circumcenter $O$ with respect to $BC, CA,AB$. Therefore, the center of $(AOA')$ is just the intersection of the mediatrix of $OA$ with $BC$.

We are left to prove that the mediatrix of $OA,OB,OC$ meet opposite sides at three collinear points (if the centers of three centers which meet are on a line and then the circles meet again). I'm pretty sure there must be a simple proof of this fact (make a parallel with the case where the tangent lines at $A,B,C$ meet the opposite sides at colinear points). I will post it if I find it.

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