Equal detour point

The equal detour point is a triangle center with the Kimberling number X(176). It is characterized by the equal detour property, that is if you travel from any vertex of a triangle ${\displaystyle ABC}$ to another by taking a detour through some inner point ${\displaystyle P}$ then the additional distance travelled is constant. This means the following equation has to hold:[1]

${\displaystyle {\begin{aligned}&|AP|+|PC|-|AC|\\=&|AP|+|PB|-|AB|\\=&|BP|+|PC|-|BC|\end{aligned}}}$

isoperimetric point ${\displaystyle Q}$, incircle center ${\displaystyle I}$,
Gergonne point ${\displaystyle G}$, equal detour point ${\displaystyle P}$
equal detours:
${\displaystyle h_{A}+h_{C}-b=h_{A}+h_{B}-c=h_{B}+h_{C}-a}$harmonic range:${\displaystyle {\frac {|QI|}{|PI|}}={\frac {|QG|}{|PG|}}}$

The equal detour point is the only point with the equal detour property if and only if the following inequality holds for the angles ${\displaystyle \alpha ,\beta ,\gamma }$ of the triangle ${\displaystyle ABC}$:[2]

${\displaystyle \tan \left({\frac {\alpha }{2}}\right)+\tan \left({\frac {\beta }{2}}\right)+\tan \left({\frac {\gamma }{2}}\right)\leq 2}$

If the inequality does not hold, then the isoperimetric point possesses the equal detour property as well.

The equal detour point, isoperimetric point, the incenter and the Gergonne point of a triangle are collinear, that is all four points lie on a common line. Furthermore, they form an harmonic range as well (see graphic on the right).[3]

The equal detour point is the center of the inner Soddy circle of a triangle and the additional distance travelled by the detour is equal to the diameter of the inner Soddy Circle.[3]

The barycentric coordinates of the equal detour point are[3]

${\displaystyle \left(a+{\frac {\Delta }{s-a}}:b+{\frac {\Delta }{s-b}}:c+{\frac {\Delta }{s-c}}\right).}$

and the trilinear coordinates[1]

${\displaystyle \left(\sec \left({\frac {\alpha }{2}}\right)\cos \left({\frac {\beta }{2}}\right)\cos \left({\frac {\gamma }{2}}\right)+1:\sec \left({\frac {\beta }{2}}\right)\cos \left({\frac {\gamma }{2}}\right)cos\left({\frac {\alpha }{2}}\right)+1:\sec \left({\frac {\gamma }{2}}\right)\cos \left({\frac {\alpha }{2}}\right)\cos \left({\frac {\beta }{2}}\right)+1\right)}$