## New answers tagged plane-geometry

3

The "short answer" is as follows.
The point $P$ is in general not constructible, for instance, for the triangle with sides $6$, $9$, $13$
constructing $P$ implies we can solve a polynomial equation $\Pi(K)=0$, where
$\Pi=a_7K^7+a_6K^6+a_5K^5+a_4K^4+a_3K^3+a_2K^2 + a_1K+a_0$
is an irreducible polynomial of degree seven with rational coefficients,
...

3

Not an answer. I'm just expanding a comment about @PeterTaylor's observation that the known pseudovertices $X(4)$, $X(74)$, $X(1138)$ lie on the Neuberg cubic ...
Bernard Gibert's "Pairs and Triads of points on the Neuberg Cubic
connected with Euler Lines and Brocard Axes Isometric Parallel Chords" Proposition 1 characterizes the Neuberg cubic of $\...

0

Here is a proof that is pretty standard for existence and unicity in such geometric setting by convergence, in the figure the initiale cevians intersect at $P$. Arrange the segments to the sides in increasing order, here $PF\le PE\le PG$. It is easy to see that the orange parts are smaller than their corresponding segments $(PE,PF)$ and the green ones are ...

0

This is a comment, but too long for that. One can try to apply the $p,q$ method, i.e., assume that the three vertices are $(0,0)$, $(1,0)$ and $(p,q)$ respectively. Then if $P$ has barycentric coordinates $(\lambda_1,\lambda_2,\lambda_3)$ with respect to the vertices, the equality of the squares of the lengths from $P$ to the vertices of the cevian ...

4

This is a report on an unsuccessful computational approach which is rather too long for a comment.
I work with complex numbers to represent the points in the obvious way.
It suffices to consider $\mu(z) = t(z,0,1)$ because this can be extended under the invariants to the full $t(z,z',z'')$. Since multiplication by a complex number is just rotation and ...

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