Geometry problem with triangle Given a triangle $ABC$ with side lengths $a$, $b$, and $c$. There exists a unique point $P$ such that
$$d^2 = AB^2+AP^2+BP^2 = AC^2+AP^2+CP^2=BC^2+BP^2+CP^2.$$
Question: How to express $d^2$ in terms of $a$, $b$, and $c$, and the circumradius $R$ of the triangle $ABC$?

What I tried: $R$ can be obtained using Heron's formula: $R=\frac{abc}{4S}$, with $S=\sqrt{s(s-a)(s-b)(s-c)}$, $s=(a+b+c)/2$.
I tried to estimate $d^2$ for some specific values.

*

*For example, when $a=b=c$, then $d^2=5R^2=5a^2/3$.

*In the extreme cases of $a=b+c$, $b=a+c$, or $c=a+b$ (with $a>0$, $b>0$, $c>0$), we have $d^2=32R^2$.

*And if $ABC$ is a right triangle (e.g., if $a^2+b^2=c^2$), we have $d^2=8R^2=a^2+b^2+c^2$.

I have no idea how to come to a general expression for $d^2$ in terms of $a$, $b$, $c$, and $R^2$ and I would be curious to see solutions.

$\begin{array}{} AB=c & BC=a & AC=b \\ AP=f & BP=g & CP=h \end{array}$
$\begin{array}{} d^2=c^2+f^2+g^2 & (1)  \\ d^2=b^2+f^2 +h^2 & (2)  \\ d^2=a^2+g^2+h^2 & (3) \ \end{array}$
$\begin{array}{} A=(p,q) & B=(a,0) & C=(0,0) \end{array}$
where:
$\begin{array}{} p=\frac{a^2+b^2-c^2}{2a} & q=\frac{2S}{a} & S=\sqrt{s(s-a)(s-b)(s-c)} & s=\frac{a+b+c}{2}\end{array}$
$\begin{array}{} f^2=(x-p)^2+(y-q)^2 \\ g^2=(x-a)^2+y^2 \\ h^2=x^2+y^2 \\ \end{array}$
$\begin{array}{} (1)(3) & c^2+f^2-a^2-h^2=0 & ⇒ & f^2=a^2+h^2-c^2 \\ (2)(3) & b^2+f^2-a^2-g^2=0 & ⇒ & f^2=a^2+g^2-b^2 \\ & ⇒  a^2+h^2-c^2=a^2+g^2-b^2 & ⇒ & h^2-c^2-g^2+b^2=0  \\ \end{array}$
$\begin{array}{} x^2+y^2-c^2-[(x-a)^2+y^2]+b^2=0 \\ -c^2+2ax-a^2+b^2=0 \\ x_{P}=\frac{a^2-b^2+c^2}{2a} \end{array}$
$f^2=a^2+h^2-c^2$
$(x-p)^2+(y-q)^2=a^2+x^2+y^2-c^2$
$y=\frac{-1}{2q}(2px-p^2-q^2+a^2-c^2)$
replace $p$, $q$ and $S^2$
$y_{P}=\frac{1}{8·aS}(b^4-2b^2c^2+c^4-3a^4+2a^2c^2+2a^2b^2)$
P was identified as L (de Longchamps point).
The distance from the vertex of a triangle to the de Longchamps point is:
$\begin{array}{} AL^2=f^2=16R^2-2(b^2+c^2)-a^2 \\ BL^2=g^2=16R^2-2(c^2+a^2)-b^2 \\ CL^2=h^2=16R^2-2(a^2+b^2)-c^2 \end{array}$
$\begin{array}{}  d^2=c^2+f^2+g^2 \\ d^2=c^2+16R^2-2(b^2+c^2)-a^2+16R^2-2(c^2+a^2)-b^2 \\ d^2=32R^2-3(a^2+b^2+c^2) \end{array}$
$\begin{array}{}  d^2=c^2+f^2+g^2 \\ d^2=c^2+16R^2-2(b^2+c^2)-a^2+16R^2-2(c^2+a^2)-b^2 \\ d^2=32R^2-3(a^2+b^2+c^2) \end{array}$
