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Is this characterization of Neuberg cubic a new one?
Step-by-step mathematics solution: Is this characterization of Neuberg cubic a new one?
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1. What the problem is asking
For a given reference triangle (ABC) we do the following construction for an arbitrary point (M) in the plane.
- Let (I_{A},I_{B},I_{C}) be the mid‑points of the segments (MA,MB,MC).
- Draw the perpendicular bisector of (MA); it meets the side (BC) at a point we call (D).
(Analogously we obtain (E\in CA) from the perpendicular bisector of (MB) and (F\in AB) from the perpendicular bisector of (MC).)
The claim to be proved is
The triangle (DEF) is degenerate (the three points are collinear)
if and only if the point (M) lies on the Neuberg cubic of (\triangle ABC).
In other words, the locus of all points (M) for which the three perpendicular‑bisector intersections are collinear is precisely the Neuberg cubic.
We have to give a complete proof of this equivalence and indicate that the result is a known description of the Neuberg cubic.
2. Proof
2.1 A useful algebraic identity
Denote by (\vec{XY}) the vector (\overrightarrow{XY}).
If (I_{A}) is the midpoint of (MA) then
[ \boxed{\;2\;\vec{MA}!\cdot!\vec{BI_{A}} = AB^{2}-BM^{2}\;}\qquad(1a) ] [ \boxed{\;2\;\vec{MA}!\cdot!\vec{CI_{A}} = AC^{2}-CM^{2}\;}\qquad(1b) ]
Proof.
Write (\vec{BI_{A}}=\dfrac{\vec{BA}+\vec{BM}}{2}) and (\vec{CI_{A}}=
\dfrac{\vec{CA}+\vec{CM}}{2}). Then
[
\begin{aligned}
2\vec{MA}!\cdot!\vec{BI_{A}}
&=\vec{MA}!\cdot!(\vec{BA}+\vec{BM})
&=(\vec{MA}!\cdot!\vec{BA})+(\vec{MA}!\cdot!\vec{BM})
&=(\vec{CA}-\vec{CM})!\cdot!(\vec{CA}+\vec{CM})
\quad(\text{since }\vec{MA}=\vec{CA}-\vec{CM})
&=CA^{2}-CM^{2}=AB^{2}-BM^{2},
\end{aligned}
]
which is (1a). The second identity is obtained in the same way. ∎
These two relations will be used repeatedly.
2.2 The point (D) in terms of (M)
Because (D) lies on the perpendicular bisector of (MA) we have
[ \vec{DA}!\cdot!\vec{MA}=0\qquad\text{and}\qquad AD=MD . \tag{2} ]
The first condition can be written as
[ (\vec{DB}+\vec{BA})!\cdot!\vec{MA}=0, ]
i.e.
[ \vec{DB}!\cdot!\vec{MA}= -\vec{BA}!\cdot!\vec{MA}. \tag{3} ]
Write (\vec{DB}=t\,( \vec{BC})) where (t=\dfrac{BD}{BC}) (so (0<t<1) when (D) is between (B) and (C)). Using (3) we obtain
[ t\,\vec{BC}!\cdot!\vec{MA}= -\vec{BA}!\cdot!\vec{MA}. ]
Now (\vec{BC}= \vec{BI_{A}}-\vec{CI_{A}}); substituting this and employing the identities (1a)–(1b) gives
[ t\bigl[(AB^{2}-BM^{2})-(AC^{2}-CM^{2})\bigr]=AB^{2}-BM^{2}. ]
Consequently
[ \boxed{\displaystyle\frac{BD}{DC}= \frac{AB^{2}-BM^{2}}{AC^{2}-CM^{2}}}\qquad(4) ]
(and the analogous formulas for the other two sides).
2.3 Menelaus’ theorem for (\triangle ABC) and the transversal (D!-!E!-!F)
The three points (D\in BC,\;E\in CA,\;F\in AB) are collinear iff the Menelaus product equals (1):
[ \frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1 . \tag{5} ]
Insert the three ratios obtained in (4) (and the two cyclic permutations) into (5). After a short cancellation we obtain
[ \frac{AB^{2}-BM^{2}}{AB^{2}-AM^{2}}\; \frac{BC^{2}-CM^{2}}{BC^{2}-BM^{2}}\; \frac{CA^{2}-AM^{2}}{CA^{2}-CM^{2}}=1 . \tag{6} ]
Multiplying the numerators and denominators gives the equivalent condition
[ \boxed{(AB^{2}-BM^{2})(BC^{2}-CM^{2})(CA^{2}-AM^{2}) =(AB^{2}-AM^{2})(BC^{2}-BM^{2})(CA^{2}-CM^{2}) } . \tag{7} ]
Equation (7) is a cubic equation in the three numbers (MA^{2},\,MB^{2},\,MC^{2}); it is the analytic equation of the Neuberg cubic.
2.4 Recognising the Neuberg cubic
Let the barycentric coordinates of the variable point (M) with respect to (\triangle ABC) be ((x:y:z)) ((x+y+z=1)). The well‑known formulae for the squared distances from (M) to the vertices are
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