System of Equations: Distances from the Torricelli Point to the Vertices of a Triangle

Question

I am calculating the length between the Torricelli point and the triangle apex. With the law of cosines I get this system:

$$\left\{\begin{array}{cc} x^2+y^2+xy&=a^2\,,\\x^2+z^2+xz&=b^2\,,\\z^2+y^2+zy&=c^2\,.\end{array}\right.$$

Please help me solve it.

Answer 1

Here is some progress. This might be useful if your real task is not about finding the distances from the Torricelli point to the vertices, but the sum of the distances.

Let $ABC$ be a triangle whose internal angles are smaller than $\dfrac{2\pi}{3}$ (which means the Torricelli point is not one of its vertices, otherwise the problem is trivial). Suppose that $a=BC$, $b=CA$, and $c=AB$. The point $O$ is the Torricelli point of the triangle $ABC$. Write $p=OA$, $q=OB$, and $r=OC$. (Note that $x=r$, $y=q$, and $z=p$ in your notations.) Then, $$a^2=q^2+qr+r^2\,,$$ $$b^2=r^2+rp+p^2\,,$$ and $$c^2=p^2+pq+q^2\,.$$

Draw an equilateral triangle $BCB'$ such that $B'$ and $A$ are on the opposite sides of $BC$. Suppose that $O'$ is on the line segment $OB'$ such that $O'O=r$. It follows that $O'B'=q$, making $$p+q+r=AB'=\sqrt{AC^2+B'C^2-2\cdot AC\cdot B'C\cos(\angle ACB')}\,.$$ We have $AC=b$, $B'C=a$, and $\angle ACB'=\angle ACB+\dfrac{\pi}{3}$, whence $$\cos(\angle ACB')=\cos\left(\angle ACB+\frac{\pi}{3}\right)=\frac{\cos(\angle ACB)-\sqrt{3}\,\sin(\angle ACB)}{2}\,.$$ Since $$\cos(\angle ACB)=\frac{a^2+b^2-c^2}{2ab}\text{ and }\sin(\angle ACB)=\frac{2\,\Delta}{ab}\,,$$ where $\Delta$ is the area of the triangle $ABC$, we get $$p+q+r=\sqrt{\frac{a^2+b^2+c^2}{2}+2\sqrt{3}\,\Delta}\,.\tag{*}$$ Because $$a^2+b^2+c^2=2\,(p+q+r)^2-3\,(rp+pq+qr)\,,$$ we get $$rp+pq+qr=\dfrac{4}{\sqrt{3}}\,\Delta\,.\tag{#}$$ (This equality can be found using an argument relating the areas of the triangles $BOC$, $COA$, $AOB$, and $ABC$.) We also have $$a^2(q-r)+b^2(r-p)+c^2(p-q)=(q^3-r^3)+(r^3-p^3)+(p^3-q^3)=0\,.$$ According to WolframAlpha, the expressions for $p$, $q$, and $r$ are terrible. Here is an expression for $p$: $$p = \small\sqrt{\frac{-a^6+2b^6+2c^6 + 4 a^4( b^2 +c^2) - 5 a^2( b^4+c^4) +b^2c^2(b^2+c^2)- 3 a^2 b^2 c^2-\sqrt{3\,\Xi_a}}{6\,(a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2)}}\,,$$ where $$\begin{align}\Xi_a&:=\small-a^{12} + 4 a^10 (b^2+c^2) - 6 a^8( b^4+c^4) - 10 a^8 b^2 c^2 + 4 a^6 (b^6+c^6)\\&\small\ \ \ \ \ + 10 a^6 b^2 c^2(b^2+c^2) - a^4 (b^8+c^8) - 6 a^4 b^2 c^2(b^4+c^4) - 3 a^4 b^4 c^4 \\&\small\ \ \ \ \ \ \ \ \ \ + 2 a^2 b^2 c^2 (b^8+c^6)- b^4 c^4(b^4+c^4) + 2 b^6 c^6\,.\end{align}$$ For $q$ and $r$, the expressions are similar. Apparently, $$\Xi_a=16\Delta^2(a^2-b^2)^2(a^2-c^2)^2\,.$$

Answer 2

In order for me to remember which variable corresponds to which geometric object. I will relabel them as follows:

Let $A,B,C$ be the 3 vertices and $a,b,c$ be the 3 sides of a triangle in usual notation. Let $u, v, w$ be the distance between the Torricelli point and $A,B,C$ respectively. Let $\Delta$ be the area of the triangle. The problem at hand becomes:

Given $$\begin{align} \alpha &\stackrel{def}{=} a^2 = v^2 + vw + w^2\\ \beta &\stackrel{def}{=} b^2 = w^2 + wu + u^2\\ \gamma &\stackrel{def}{=} c^2 = u^2 + uv + v^2 \end{align}$$ How to solve for $u,v,w$?

Let $\delta = uv + vw + wu$, it is not hard to verify

$$\begin{array}{rcrcrcrl} \alpha &+& \beta &+& \gamma &+& 3\delta &= 2(u+v+w)^2\\ -\alpha &+& \beta &+& \gamma &+& \delta &= 2u(u+v+w)\\ \alpha &-& \beta &+& \gamma &+& \delta &= 2v(u+v+w)\\ \alpha &+& \beta &-& \gamma &+& \delta &= 2w(u+v+w) \end{array}$$ This implies

$$(u,v,w) = \left(\frac{P-2a^2}{Q}, \frac{P-2b^2}{Q}, \frac{P-2c^2}{Q}\right)$$

where $\;P = a^2 + b^2 + c^2 + \delta\;$ and $\;Q = \sqrt{2(a^2 + b^2 + c^2 + 3\delta)}\;$.

Everything comes down to the computation of $\delta$. Notice the area of triangle can be expressed in terms of $\delta$,

$$\Delta = \frac12(uv + vw + wu)\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{4}\delta$$

With help of Heron's formula, $\delta$ can be computed as follows:

$$\delta = \frac{4}{\sqrt{3}}\Delta = \frac{1}{\sqrt{3}}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}$$

Answer 3

In @Batominovski's symmetric notation, we have $$\begin{align} a^2&=q^2+r^2+q r\\ b^2&=r^2+p^2+r p\\ c^2&=p^2+q^2+p q \end{align} \tag{1}$$ Using Mathematica's Resultant function (or, equivalently, Groebner bases), we can eliminate, say, $q$ and $r$ to get $$\begin{align} 0 &= 9 p^4 (a^4 - a^2 b^2 + b^4 - a^2 c^2 - b^2 c^2 + c^4) \\[4pt] &+3 p^2 (a^6 - 4 a^4 b^2 + 5 a^2 b^4 - 2 b^6 - 4 a^4 c^2 + 3 a^2 b^2 c^2 - b^4 c^2 + 5 a^2 c^4 - b^2 c^4 - 2 c^6) \\[4pt] &+ \phantom{9p^4}(a^2 - b^2 - b c - c^2)^2 (a^2 - b^2 + b c - c^2)^2 \\ &= 9H (p^2)^2 + 3J (p^2) + K \end{align} \tag{2}$$ This is ugly, but at least it's a quadratic in $p^2$ that can be readily solved. Conveniently, the discriminant factors nicely. $$\begin{align} (3J)^2-4(9H)K &= 27 (a^2 - b^2)^2 (a^2 - c^2)^2 (-a + b + c) (a + b - c) (a - b + c) (a + b + c) \\[4pt] &= 27 (a^2-b^2)^2(a^2-c^2)^2\cdot 16|\triangle ABC|^2 \tag{3} \end{align}$$ so that we have $$p^2 = \frac{-J \pm 4\,|\triangle ABC|\,|a^2-b^2|\,|a^2-c^2|\,\sqrt{3}}{6H} \tag{4}$$

I suspect there's an clean way to write $(4)$, and perhaps even a clever way to derive it, but that'll take a bit more thinking.