Given a triangle $ABC$ how to find a point P such that the triangles $ABP, ACP$ and $BCP$ have equal perimeters

Question

WLOG, let $O = (0,0), A = (a,0)$ and $B = (b,c)$ be the vertices of a triangle and let $P = (x,y)$ be a point on the same plane such perimeters of the triangles $OAP, OBP$ and $ABP$ are equal. Then,

$\text{Perimeter}(\triangle OAP) = a + \sqrt{(x-a)^2 + y^2} + \sqrt{x^2 + y^2}\tag 1$

$\text{Perimeter}(\triangle OBP) = \sqrt{b^2 + c^2} + \sqrt{(x-b)^2 + (y-c)^2} + \sqrt{x^2 + y^2} \tag 2$

$\text{Perimeter}(\triangle ABP) = \sqrt{(a-b)^2 + c^2} + \sqrt{(x-a)^2 + y^2} + \sqrt{(x-b)^2 + (y-c)^2} \tag 3$

Theoretically, we can proceed by equating any two of the above equations. For equating $(1)$ and $(2)$ solving for $x$ gives incredibly complicated expression for $x$. Then by equating $(2)$ and $(3)$ and solving for $x$ we get another equally complicated expression for $x$. Equating these two expressions for $x$ we should get $y$ in terms or $a,b$ and $c$. In practice, But due to the complicated nature of the expressions involved, this approach seems hopeless.

My alternative approach was a brute force computational estimate of the coordinates $(x,y)$ which at the moment is scanning the entire solution space. This can be made more efficient if we can put bounds on $(x,y)$.

Question 1: For what triangles does $P$ exist? Are there any known estimates for $P$

Question 2: Is there a computationally efficient way to calculate $P$

Answer 1

Fig. 1 : $P=X(175)$, meeting point of the $3$ hyperbolas $H_A,H_B,H_C$ is the unique solution. Point $Q=X(176)$ is symmetrical of $P$ with respect to the circumcenter. Geogebra animation of this figure here. Explanations below.

Let us write the constraints under the form :

$$\begin{cases}a+v+w&=&p \\u+b+w&=&p\\u+v+c&=&p\end{cases}\tag{1}$$

using the following notations : $$\begin{cases}a=BC, \ b=CA, \ c=AB,\\ u=PA, \ v=PB, \ w=PC\\ p \ \text{the common perimeter}\end{cases} \tag{2}$$

Subtracting them in all possible ways, we get :

$$\begin{cases}v-w&=&b-c&(i)\\w-u&=&c-a&(ii)\\u-v&=&a-b&(iii)\end{cases}\tag{3}$$

Let us now consider the following constraints :

$$\begin{cases}|v-w|&=&|b-c|&(i')\\|w-u|&=&|c-a|&(ii')\\|u-v|&=&|a-b|&(iii')\end{cases}\tag{4}$$

which evidently imply constraints (3).

Constraints (4) describe hyperbolas $H_A,H_B,H_C$ with their 2 branches, due to the focal property of hyperbolas : the set of points $P$ defined by the absolute value of the difference of distances to two fixed points (its foci), whereas constraints (3) describe single branches.

Remark : As can be seen on figure 1, each vertex belongs to one of these hyperbolas (called for this reason : "Vertex-hyperbolas"). In order to understand it, let us focus our attention on hyperbola $(H_C)$ with equation (iii') : $C$ belongs to $(H)$ because $u-v=\color{red}{-}(a-b)$ $\ \iff \ PA-PB=CA-CB$. Please note that we have used a minus sign : $-(a-b)$ instead of $(a-b)$ : therefore the branch of $(H_C)$ to which $C$ belongs is not the branch we are interested in (defined by (iii)).

Little recap' : each equation (i), (ii) or (iii) describes a single branch of one of the hyperbolas $H_A,H_B,H_C$. The branches to be considered are those which do not pass through one of the vertices.

As a consequence, there is a unique solution $P$ which must be situated at the intersection of the $3$ valid hyperbola branches. One can wonder why the third hyperbola intersects the two first ones in one of their common points : it is plainly because, in relationship (3), by addition :

$$(i)+(ii)=(iii)$$

In fact, not all triangles have such a solution. There is a limitation on the angles (no angle must be larger than $\approx 106.26°$ : see the comments by YNK).

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I must pay a large tribute to @YNK who has indicated me that the point $P$ we were looking for is $X(175)$, i.e., a known and well studied point called the "isoperimetric point" in the "Encyclopedia of triangle centers" where you will learn its various properties.

Moreover, its twin point, $Q=X(176)$ is the intersection of the 3 other hyperbola branches !

Remarks :

1. Please note the particular case where, say, $C$ belongs to the perpendicular bissector of $AB$, otherwise said when $ACB$ is an isosceles triangle ; hyperbola $H_C$ degenerates into a (double) line which is the main altitude of the isosceles triangle, the two other hyperbolas $H_A$ and $H_B$ being symmetrical with respect to this altitude. As a consequence, $P$ belongs to the altitude issued from $C$.

2. I do not resist the pleasure of giving a figure that I have found in this article showing that the points of intersection of hyperbolas $H_A,H_B,H_C$ with the triangle sides are the points of contact of its incercle and excircles with these sides. Moreover, these 6 points belong to a same ellipse called Privalov's ellipse.

Fig. 2. (caution : point $O$ is not the circumcenter of triangle $ABC$).