The congruent incircles point

Question

I have three vertices of a triangle $ A = (x_a, y_a) $, $ B = (x_b, y_b) $, and $ C = (x_c, y_c) $, and I need to find the coordinates of a point $ P $ such that the incircles of the triangles $ ACP $, $ BCP $, and $ ABP $ are congruent.

In barycentric coordinates, any point $ P $ can be expressed as:

$ P = \alpha A + \beta B + \gamma C $

where $ \alpha $, $ \beta $, and $ \gamma $ are proportional to the areas of triangles $ PBC $, $ PAC $, and $ PAB $, respectively, and satisfy $ \alpha + \beta + \gamma = 1 $.

Now, we know that the areas of these triangles should be proportional to the semiperimeter times the inradius. If the incircles of the triangles are congruent, this gives us the relations:

$ \text{Area of } PBC \propto (a + PB + PC), \quad \text{Area of } PAC \propto (b + PA + PC), \quad \text{Area of } PAB \propto (c + PA + PB) $

Here, $ a =BC$, $ b =AC$, and $ c =AB$ represent the side lengths of the triangle $ ABC $.

Then I define the following function for any point $ X $:

$ f(X) = w_a A + w_b B + w_c C $

where:

$ w_a = \frac{a + XB + XC}{a + b + c + 2(XA + XB + XC)}, \quad w_b = \frac{b + XA + XC}{a + b + c + 2(XA + XB + XC)}, \quad w_c = \frac{c + XA + XB}{a + b + c + 2(XA + XB + XC)} $

$P$ is the solution to $f(P)=P$ (it is a fixed point of the function).

However, I am stuck at this point due to the nonlinearity of the problem. The direct use of lengths complicates the calculations, leading to square roots and multiple variables in the equations. I would prefer a solution that involves other geometric properties, such as relevant angles or the inradius, which might lead to more elegant equations—even if these equations are only solvable numerically, perhaps involving trigonometric or higher-degree equations with just one variable. Could I approach this problem using such parameters to solve for $ \alpha $, $ \beta $, and $ \gamma $ instead of relying on the lengths?

Answer 1

Doing a $13-6-9$ search reveals that you are looking for $X(5394)$ in the Encyclopedia of Triangle Centres. Over there is listed

Barycentrics              (unknown)

…at least until I came.

---

Let $a,b,c$ be a triangle's side lengths. Let two points with normalised barycentric coordinates $x_1:y_1:z_1$, $x_2:y_2:z_2$ with respect to the triangle be given. Then the distance $d$ between them satisfies $$-d^2=a^2\Delta y\Delta z+b^2\Delta z\Delta x+c^2\Delta x\Delta y$$ Now let $x:y:z$ be the normalised barycentric coordinates of the desired triangle centre $P$. Applying the formula and the condition $x+y+z=1$ gives $$PC=\sqrt{(ay)^2+(bx)^2+(a^2+b^2-c^2)xy}$$ with similar expressions for $PA$ and $PB$. Then $$\frac{a+PB+PC}x=\frac{b+PC+PA}y=\frac{c+PA+PB}z=\frac{2\Delta(ABC)}r=k$$ where $r$ is the common inradius of the subtriangles. Adding the three equations together gives $a+b+c+2(PA+PB+PC)=k$, so define $t=PA+PB+PC$ and back-substitute $k$ into the three equations to get $$(1-2x)t-(a+b+c)x+a=PA$$ $$(1-2y)t-(a+b+c)y+b=PB$$ $$(1-2z)t-(a+b+c)z+c=PC$$ Upon squaring these equations to remove the roots over $PA,PB,PC$ and eliminating $x,y,z$, we find that $t$ satisfies the biquartic $$T:64t^8-16(5s_1^2-8s_2)t^6+4(11s_1^4-36s_1^2s_2+16s_2^2+24s_1s_3)t^4-s_1(11s_1^5-60s_1^3s_2+80s_1s_2^2+48s_1^2s_3-96s_2s_3)t^2+(s_1(s_1^3-4s_1s_2+12s_3))^2=0$$ where $s_1=a+b+c,s_2=ab+ca+bc,s_3=abc$. We thereby recover Peter Moses's 13 August 2019 formula for $r$; $T$'s largest root should be taken.

Now append $T$ to the three equations with $t$ and eliminate $t,y,z$. The result is an octic $G$ in $x$ whose coefficients are degree-$9$ homogeneous polynomials in $a,b,c$. We have not yet solved the problem of determining barycentric coordinates because $x$'s relationship to the determined $t$ has not been elucidated.

---

To do that we run integer relation on $\{t^ix^j|0\le i\le4,0\le j\le1\}$ (we need at least $9$ dimensions because $t$ is algebraic of degree $8$) for many choices of $a,b,c$ and interpolate the coefficients found. We find that $$f_0 + f_2 t + f_4 t^2 + f_6 t^3 + f_8 t^4 + (f_1 + f_3 t + f_5 t^2 + f_7 t^3 + f_9 t^4) x=0$$ where $$f_0 = (a - b - c) (a + b + c) (c^3 - (a+b) c^2 + (-a^2+6 a b-b^2) c + (b^3-a b^2-a^2 b+a^3))\\ f_1 = -3 (a + b + c)^2 (c^3 - (a+b) c^2 + (-a^2+6 a b-b^2) c + (b^3-a b^2-a^2 b+a^3))\\ f_2 = 2 (a + b + c) (2 a + b + c) (c^2 + (-4 a+2 b) c + (7 a^2-4 a b+b^2))\\ f_3 = -12 (a + b + c) (c^3 - b c^2 + (a^2+2 a b-b^2) c + (2 a^3+a^2 b+b^3))\\ f_4 = 4 (2 c^3 + (2 b+a) c^2 + (2 b^2-6 a b+2 a^2) c + (2 b^3+b^2 a+2 b a^2+7 a^3))\\ f_5 = 48 (a + b + c) (b c - a^2)\\ f_6 = -8 (a + b + c) (2 a + b + c)\\ f_7 = 48 (b + c) (a + b + c)\\ f_8 = -16 (2 a + b + c)\\ f_9 = 48 (a + b + c).$$ This can be verified quickly by taking the resultant with $T$ eliminating $t$ and checking that $G$ exactly divides into the resultant. So we have $$x = -\frac{f_0 + f_2 t + f_4 t^2 + f_6 t^3 + f_8 t^4}{f_1 + f_3 t + f_5 t^2 + f_7 t^3 + f_9 t^4}$$ but it turns out that the formula becomes even simpler upon substituting $t=(k-s_1)/2$. The final formula for the first normalised barycentric coordinate $x$ of $X(5394)$ is $\operatorname{num}(a,b,c)/\operatorname{den}(a,b,c)$ where $$\operatorname{num}(a,b,c)=(2a+m)k^4 - 3(a+m)(2a+m)k^3 + (-a^3+13a^2m+11am^2+m^3+8ap+4mp)k^2 - 2(a+m)(a^3-m^3+8ap+4mp)k + 6a(a+m)((a-m)(a+m)^2+4mp)$$ $$\operatorname{den}(a,b,c)=3(a+m)k (k^3 - 2(2a+m)k^2 + 2(a^2+3am+2p)k - 12ap)$$ where $m=b+c$ and $p=bc$.

---

All these calculations are performed in the various files of a gist I have created to accompany this answer.

I was made aware of a $50 reward for computing this. I sent it to Clark Kimberling to claim it, and I accepted a donation of the reward to the OEIS in my name in lieu of a cheque to me, as I am a Singaporean.

Answer 2

It looks like any two in-circles are tangent on the side common to them at the same point. Assuming this is true, then $ P $ would be the Fermat point of the triangle; i.e. $\angle APB = \angle BPC = \angle APC = 120$. This can be shown by drawing the angle bisectors and perpendicular from incentre to side in each of the triangles $ABP, BCP, ACP$ and proving that the six resulting triangles around $P$ subtend the same angle at $P$.

However, I was unable to prove the assumption is actually true.