The first and second Lemoine circles are well-known to geometers. According to this article the third Lemoine circle has been first discovered by Jean-Pierre Ehrmann in 2002. It is worth noting that the centres of these circles belong to a line that is going through the Lemoine point and the circumcenter of the triangle.

I would like to present the construction of another circle, which I believe might be called the 4th Lemoine circle. The idea behind its construction is slightly similar to that proposed by Ehrmann:

Symmedians AA', BB', CC' intersect each other at the point M (the Lemoine point of the triangle ABC). The circumcircle of the triangle A'B'C' was drawn (also known as the 'symmedial circle'). A'', B'', C'' are the first intersection points of the symmedians with the symmedial circle. Finally, the circumcircles of the triangles A''B''M, B''C''M, A''C''M always intersect the sides of the original triangle ABC at six points, that are conclycic.

Theorem illustration

As far as I can tell the point O (the center of our six point circle) is not included into the C.Kimberling's Encyclopedia, so it must be unknown or uncatalogued. O belongs to the line that contains the centres of the first three Lemoine circles, so by my reckoning, the dotted red circle that is shown in the picture above perfectly qualifies for being called the 4th Lemoine circle.

Could you please give a synthetic proof of this theorem ?

Geogebra dynamic sketch.

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    $\begingroup$ Interesting! Is this 4th Lemoine circle a Tucker circle? $\endgroup$ Aug 4, 2021 at 20:37
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    $\begingroup$ So, I've just checked on Euklid Dynageo that the circle is not a Tucker circle. This makes it even more surprising that its center seemingly lies on the Tucker line (i.e., on the line joining the circumcenter with the symmedian point)! $\endgroup$ Aug 5, 2021 at 7:30
  • $\begingroup$ @darijgrinberg, I'm not sure what specifically causes your surprise, so this may or may not affect it, but my analysis of Kimberling's ETC finds that 493 of the first 10000 triangle centres are on the Brocard axis, which makes it second only to the Euler line (960 of the first 10000). $\endgroup$ Aug 22, 2021 at 6:56
  • $\begingroup$ @darij grinberg mathoverflow.net/questions/402310/lemoine-lozada-circles $\endgroup$
    – A.Zakharov
    Aug 22, 2021 at 19:27

2 Answers 2


A notational preamble: Writing $K$ for the circumcenter, and $L_1$, $L_2$ (OP's $M$), $L_3$ for the centers of the First, Second, and (Ehrmann's) Third Lemoine Circles, all of which are collinear, we have this happy terminological coincidence: $$\frac{KL_i}{KL_1}=i \tag{1}$$ (This invites dubbing the circumcenter $L_0$, and the circumcircle itself the "Zero-th Lemoine Circle"; but I digress.)

To the problem at hand ...

Some ugly coordinate bashing confirms that OP's center $O$ lies on the line-of-$L$s, a rather remarkable property that provides the corresponding circle some "Numbered Lemoine Circle" credibility. Further bashing shows that the counterpart of $(1)$ is $$\frac{KO}{KL_1}=\frac{3 (a^2 + b^2 + c^2)^3 - (-a^2 + b^2 + c^2) (a^2 - b^2 + c^2) (a^2 + b^2 - c^2)}{8\,(a^2 + b^2) (a^2 + c^2) (b^2 + c^2)} \tag2$$ The dependence of this value upon the shape of the triangle distinguishes it from the $L_i$, perhaps so much so that OP's circle establishes a new category, "Numbered Non-Tucker Lemoine Circles".

I'll take this opportunity describe another point on the line-of-$L$s with a shape-independent ratio $(1)$:

enter image description here

For each vertex $V=A, B, C$, construct the circle through $V$ and $L_2$ whose center lies on $\overleftrightarrow{VK}$. The six "other" points where the three circles meet the triangle's side-lines are concyclic, and their circumcenter —known to Kimberling as $X(585)$wants to be denoted $L_{3/2}$, because $$\frac{KL_{3/2}}{KL_1} = \frac32 \tag3$$ This may-or-may-not earn the circle (which, incidentally, is a Tucker circle) the title of "$\frac32$-th Lemoine Circle".


By the Power of a Point theorem, $$B C_B \cdot B C_A = B B'' \cdot BM = B A_B \cdot B A_C$$ and thus again by the Power of a Point theorem, the points $C_B, C_A, A_B, A_C$ all lie on one circle, and symmetrically the points $A_B, A_C, B_A, B_C$ lie on one circle, and the points $B_A, B_C, C_B, C_A$ lie on one circle. If these are not all the same circle, their radical axes intersect at one point, that is the lines $AB, AC, BC$ intersect in one point which is a contradiction.

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    $\begingroup$ This is very nice, but does not show anything about the center of the circle. $\endgroup$ Aug 4, 2021 at 20:39
  • $\begingroup$ Thanks. I should have rather asked, whether it is a Tucker circle and what are its parameters or about the collinearity of the points O, R and M... $\endgroup$
    – A.Zakharov
    Aug 4, 2021 at 20:42
  • $\begingroup$ Oh sorry, I misread the question. I'll think about that $\endgroup$
    – Random
    Aug 4, 2021 at 21:17

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