# Three circles intersecting at one point

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 $$𝑃$$. GeoGebra applet that demonstrates this proposition can be found here.

• 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 ... Nov 29, 2021 at 20:42
• ... but is definitely known: It follows from locus property 2 of the Neuberg cubic, since X(3) lies on that cubic. Nov 29, 2021 at 20:43
• Circumcentre of K1,K2,K3 are Collinear
– user423633
Jun 19, 2022 at 7:36

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: $$(B-C)g_A+(C-A)g_B+(A-B)g_C=0$$.

• Definitely one of those proofs from The Book Apr 24, 2021 at 6:45
• @მამუკაჯიბლაძე well, definitely not, but I let geometry lovers give a proof from the Book Apr 24, 2021 at 9:36
• What is "The Book" ? Nov 10, 2021 at 20:16
• Nov 10, 2021 at 22:37

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. $$\frac{y_3}{y_1}=\frac{\cot{B}-\cot{C}}{\cot{A}-\cot{C}}$$.

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...

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.