The problem and its normalization
Let \(ABC\) be an equilateral triangle inscribed in a circle \((O,R)\), and let \(P\) move on this circle. Denote by \(A’,B’,C’\) the perpendicular projections of \(P\) onto the lines \(BC,CA,AB\), respectively. These three feet lie on the Simson line of \(P\). For positions of \(P\) for which the triangles \(PAA’\), \(PBB’\), and \(PCC’\) are nondegenerate, their Euler lines are concurrent. Which is the exact real locus of their common point \(M\)? (Problem 231, proposed by Antreas Hatzipolakis, 2026)
The algebraic approach serves two purposes: to obtain an equation of the locus by Groebner basis elimination, and to determine which real points of that equation actually arise from the construction. The formulas below record the main algebraic certificates; their derivations are given in the full article.
A translation, a rotation, and a homothety allow us to take
$$
A=(1,0),\qquad
B=\left(-\frac12,\frac{\sqrt3}{2}\right),\qquad
C=\left(-\frac12,-\frac{\sqrt3}{2}\right).
$$
The circumcircle is now the unit circle centered at the origin. We write
$$
P=(u,v),\qquad u^2+v^2=1,\qquad M=(x,y).
$$
In particular, \(A’=(-1/2,v)\). The nondegeneracy conditions will be checked separately below.
Algebraic reduction and concurrence
Write the equations of the three Euler lines as \(L_A=0\), \(L_B=0\), and \(L_C=0\). They are obtained from the centroids and circumcenters of the corresponding triangles. With a consistent normalization of these equations, introduce
\begin{equation}
\label{eq:euler-reduced}
\begin{aligned}
p_2&=(5-u)x+vy-u,\\
p_3&=vx+(5+u)y-v.
\end{aligned}
\end{equation}
These simpler expressions arise from linear combinations of the three line equations. On \(u^2+v^2=1\), the following identities hold:
\begin{equation}
\label{eq:euler-certificates}
\begin{aligned}
L_A&=(1-u)(2u+5)p_2+(2u-3)vp_3,\\
L_B&=L_A-6p_2+2\sqrt3\,p_3,\\
L_C&=L_A-6p_2-2\sqrt3\,p_3.\\
\end{aligned}\qquad (1)
\end{equation}
They give the equivalence
\[
L_A=L_B=L_C=0\quad\Longleftrightarrow\quad p_2=p_3=0.
\]
The system \(p_2=p_3=0\) is linear in \(x,y\), with determinant
\[
(5-u)(5+u)-v^2=25-u^2-v^2=24.
\]
It therefore has a unique solution for every real \(P\) on the unit circle. For each admissible position of \(P\), the identities (1) show that this solution lies on all three Euler lines. This proves their concurrence and the uniqueness of \(M\).
Groebner basis elimination
Set \(p_1=u^2+v^2-1\) and consider
$$
J=\langle p_1,p_2,p_3\rangle\subseteq\mathbb Q[u,v,x,y].
$$
A Groebner basis computation with lexicographic order \(u>v>x>y\) gives
\begin{equation}
\label{eq:euler-elimination-ideal}
J\cap\mathbb Q[x,y]=\langle F(x,y)\rangle,
\end{equation}
where
\begin{equation}
\label{eq:euler-quartic}
\boxed{F(x,y)=24(x^2+y^2)^2-50(x^3-3xy^2)+27(x^2+y^2)-1.}
\end{equation}
In Maple, the same elimination ideal is obtained using \(\texttt{EliminationIdeal(J,{x,y})}\) from the \(\texttt{PolynomialIdeals}\) package. The set \({x,y}\) specifies the variables retained, while \(u,v\) are eliminated.
Every point of the geometric locus satisfies \(F=0\). The converse is not automatic: a real point of the quartic may correspond only to nonreal auxiliary coordinates, or to a degenerate configuration. These two possibilities must be distinguished by returning to the original system.
Real points and inverse reconstruction
Put \(s=x^2+y^2\). On \(F=0\), the polynomial identity
\begin{equation}
\label{eq:euler-radial}
(s-1)^2(16s-1)(36s-1)
=-2500y^2(3x^2-y^2)^2\le0
\end{equation}
shows that every real point of the quartic satisfies
\begin{equation}
\label{eq:euler-annulus}
\frac1{36}\le s\le\frac1{16}
\qquad\text{or}\qquad s=1.
\end{equation}
This restriction follows directly from the polynomial obtained by elimination.
For \(s\ne1\), solving \(p_2=p_3=0\) as a linear system in the auxiliary coordinates \(u,v\) gives
\begin{equation}
\label{eq:euler-inverse-formulas}
\boxed{
u=\frac{5(x-x^2+y^2)}{1-x^2-y^2},\qquad
v=\frac{5y(2x+1)}{1-x^2-y^2}
}\qquad (2)
\end{equation}
The common denominator is the determinant of that system. The recovered coordinates satisfy \(p_2=p_3=0\) and the identity
\begin{equation}
\label{eq:euler-inverse-certificate}
u^2+v^2-1=\frac{F(x,y)}{(1-x^2-y^2)^2}
\end{equation}
Consequently, every real point of \(F=0\) with \(s\ne1\) lifts to a unique real solution \((u,v)\) on the unit circle. It remains to check whether the corresponding geometric configuration is nondegenerate.
Exceptional points and the exact locus
Three isolated real points
When \(s=1\), the equation \(F=0\) reduces to
$$
x^3-3xy^2=1
$$
Using $y^2=1-x^2$, this becomes
$$
(x-1)(2x+1)^2=0
$$
Thus the points of the quartic on the unit circle are precisely \(A,B,C\). All other real points satisfy \(OM\le1/4\), so these three vertices are isolated real points of the algebraic curve.
They do not belong to the geometric locus. For example, at \(M=A=(1,0)\) the original system forces
$$
u=\frac52,\qquad v=\pm\frac{i\sqrt{21}}2
$$
The real point \(A\) therefore has only nonreal auxiliary solutions. The same holds for \(B\) and \(C\) by symmetry.
Six degenerate configurations
The triangle \(PAA’\) is degenerate exactly when \(v(u+1/2)=0\). Considering all three triangles, the excluded positions of \(P\) are the vertices and their antipodes:
$$
P\in{A,B,C,-A,-B,-C}
$$
On the unit circle, their exclusion is expressed by
$$
v(4u^2-1)\ne0
$$
The six corresponding limiting values of $M$ are
\begin{equation}
\label{eq:euler-six-points}
\left(\frac14,0\right),\quad
\left(-\frac18,\pm\frac{\sqrt3}{8}\right),\quad
\left(-\frac16,0\right),\quad
\left(\frac1{12},\pm\frac{\sqrt3}{12}\right)
\end{equation}
They are exactly the points of \(F=0\) with \(s=1/16\) or \(s=1/36\). Indeed, substitution of these two values reduces the quartic to
\begin{aligned}
s=\frac1{16}:&\qquad (4x-1)(8x+1)^2=0,\\
s=\frac1{36}:&\qquad (6x+1)(12x-1)^2=0
\end{aligned}
The uniqueness in (2) also shows that none of these six points can be generated by another, admissible position of \(P\).
Now suppose that \(F(x,y)=0\) and \(1/36<x^2+y^2<1/16\). The inverse formulas recover a unique real \(P\) on the unit circle. This \(P\) cannot be one of the six excluded positions, since their images lie on the two boundary circles. Hence all three triangles are nondegenerate, and Section Euler concurrence, shows that their Euler lines meet at \(M\).
This completes the converse and proves that the exact real locus is
If the construction is explicitly extended by continuity to its degenerate positions, the six limiting points are included and the strict inequalities become non strict. The isolated points \(A,B,C\) remain excluded.
Encoding nondegeneracy in the ideal
The condition \(v(4u^2-1)\ne0\) can be imposed algebraically by introducing an auxiliary variable \(z\) and considering
$$
\widetilde J=\langle p_1,p_2,p_3,zv(4u^2-1)-1\rangle
\subseteq\mathbb Q[z,u,v,x,y]
$$
Nevertheless, a further elimination computation gives
\begin{equation}
\label{eq:euler-augmented-elimination}
\widetilde J\cap\mathbb Q[x,y]=\langle F\rangle
\end{equation}
The six degenerate inputs have been removed from the enlarged system, but their limiting images still satisfy every polynomial equation of the locus. In addition, the nonreal auxiliary solutions above \(A,B,C\) satisfy the nondegeneracy condition and therefore remain present. Encoding nondegeneracy does not replace the separate verification that the auxiliary coordinates are real.
Geometric interpretation and methodological significance
The geometric solution of Francisco Javier Garcia Capitan identifies the small closed curve as the isogonal image of the circle centered at \(O\) with radius \(R/5\)1. The isogonal conjugate of an interior point is the common point of the lines obtained by reflecting its joins to the vertices across the corresponding internal angle bisectors. For an equilateral triangle, isogonal and isotomic conjugacy coincide.
More precisely, if \(\iota\) denotes isogonal conjugacy with respect to \(ABC\), then
$$
M=\iota(T),\qquad
\overrightarrow{OT}=-\frac15\overrightarrow{OP}
$$
In the normalization \(R=1\), the image of the entire small circle is the closed curve obtained by adjoining the six limiting points to topos \(\mathscr{P}\). The exact locus of the original, nondegenerate construction omits these six points. For general circumradius \(R\), all coordinates scale by \(R\), and the radial restrictions become
$$
\frac{R^2}{36}<OM^2<\frac{R^2}{16}
$$
The main advantage of the polynomial formulation is that it supports both exact elimination and a systematic investigation of the resulting equation. The Groebner basis computation produces the quartic inverse reconstruction and the geometric conditions determine its admissible real points. This example separates two distinct sources of exceptional points, nonreal auxiliary solutions and degenerate geometric configurations, and shows how the exact real locus can be recovered without a parametrization of the curve.
\(\square\)
- Problem 231, proposed by Antreas Hatzipolakis, with a solution by Francisco Javier Garcia Capitan, dated October 23, 2026. ↩︎