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Hello it's related to my answer for Prove the inequality $\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq 3\frac{a+b+c}{ax+by+cz}$

My answer fails but I don't know why ... So I was thinking a generalization of the following formula:

$$\frac{IA^2}{CA\cdot AB}+\frac{IB^2}{BC\cdot AB}+\frac{IC^2}{CA\cdot BC}=1$$

I know that it's related to the Steiner inellipse and we have for a triangle ABC and the ellipse of foci $P$ and $Q$:

$$\frac{PA\cdot QA}{BA\cdot CA}+ \frac{PB\cdot QB}{CB\cdot AB}+ \frac{PC\cdot QC}{BC\cdot AC}=1$$

But in my proof I have also use the following formula:

\begin{align} \frac{1}{IA^2}+\frac{1}{IB^2}+\frac{1}{IC^2} &= \frac{1}{r^2}-\frac{1}{2rR} \\ IA^2+IB^2+IC^2 &= s^2+r^2+8rR \\ CA\cdot AB+BC\cdot AB+CA\cdot BC &= s^2+(4R+r)r \\ \frac{1}{CA\cdot AB}+\frac{1}{BC\cdot AB}+\frac{1}{CA\cdot BC} &= \frac{1}{2rR} \end{align}

So what's the new expression of:

\begin{align} \frac{1}{BA\cdot CA}+\frac{1}{CB\cdot AB}+\frac{1}{BC\cdot AC} &= ? \\ \frac{1}{PA\cdot QA}+\frac{1}{PB\cdot QB}+\frac{1}{PC\cdot QC} &= ? \\ PA\cdot QA+PB\cdot QB+PC\cdot QC &= ? \\ BA\cdot CA+CB\cdot AB+BC\cdot AC &= ? \end{align}

In function of the parameters of the inellipse and the triangle $ABC$ like the area and the side of the triangle or the semi major semi minor axes of the ellipse?

Edit: I have a good news

The centroid $M$ of the triangle $ABC$ correspond to the centre of the inellipse and we have the following relation for $P$ any interior point related to the triangle $ABC$:

$$PA^2+PB^2+PC^2=MA^2+MB^2+MC^2+3MP^2$$

enter image description here Thanks a lot.

Ng Chung Tak
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2 Answers2

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A triangle $\triangle ABC$ with its centroid at the origin is the image of an origin-centered equilateral under a linear transformation. The transformation carries the equilateral's incircle to $\triangle ABC$'s Steiner inellipse. With an appropriate rotation, and by considering similar triangles (and/or ellipses) equivalent, we may assume that the transformation is a simple vertical scaling. This being so, ...

Consider the origin-centered equilateral $\triangle A_\star B_\star C_\star$ with vertex coordinates $$A_\star := 2 r \operatorname{cis}\theta \qquad B_\star := 2 r\operatorname{cis}(\theta+120^\circ) \qquad C_\star := 2 r \operatorname{cis}(\theta-120^\circ)$$ where $\operatorname{cis}(\cdot) := (\cos(\cdot), \sin(\cdot))$. The (origin-centered) incircle of this triangle has radius $r$. For some non-negative $s<r$, apply the vertical-scaling transformation $$(x,y) \to \left(\;x, \frac{ys}{r}\;\right)$$ chosen so as to turn the incircle of $\triangle A_\star B_\star C_\star$ into an origin-centered, axis-aligned ellipse with (horizontal) major radius $r$ and (vertical) minor radius $s$.

Let $A$, $B$, $C$ be the images of $A_\star$, $B_\star$, $C_\star$ under the transformation. One readily derives these relations: $$\begin{align} a^2 &= 12 \left( r^2 \sin^2\theta\phantom{\left(-60^\circ\right)}\; + s^2 \cos^2\theta \right) \\ b^2 &= 12 \left( r^2 \sin^2\left(\theta-60^\circ\right)+s^2 \cos^2\left(\theta-60^\circ\right) \right) \\ c^2 &= 12 \left( r^2 \sin^2\left(\theta+60^\circ\right)+s^2 \cos^2\left(\theta+60^\circ\right) \right) \\ \end{align}$$

Without too much trouble, one can derive $$r^2 = \frac{1}{36}\left(a^2 + b^2 + c^2 + 2 t^2 \right) \qquad s^2 = \frac{1}{36}\left(a^2+b^2+c^2-2t^2\right)$$ where $$t^4 = a^4 + b^4 + c^4 - b^2 c^2 - c^2 a^2 - a^2 b^2$$ and also $$ \tan 2\theta = \frac{\sqrt{3}(c^2-b^2)}{-2a^2+b^2+c^2} \quad\to\quad \cos 2\theta = \frac{-2a^2+b^2+c^2}{2t^2} $$

Now, the inellipse's foci are the points $P_{\pm} := (\pm\sqrt{r^2-s^2},0)$, and we have (after some manipulation) $$\begin{align} |\overline{AP_{\pm}}|^2 &= \frac{1}{9}\left(-a^2+2b^2+2c^2+t^2\mp 12 r t \cos\theta \right) \end{align}$$ I haven't found a particularly-insightful way to write the right-hand side. However, the product of the "$+$" and "$-$" forms gives a convenient difference of squares that reduces quite nicely to this:

$$|\overline{AP_{+}}||\overline{AP_{-}}| = \frac{1}{3}bc \quad\text{thus, also}\quad |\overline{BP_{+}}||\overline{BP_{-}}| = \frac{1}{3}ca \quad |\overline{CP_{+}}||\overline{CP_{-}}| = \frac{1}{3}ab \tag{$\star$}$$

Consequently, it's no wonder that $$\frac{|\overline{AP_{+}}||\overline{AP_{-}}|}{|\overline{AB}||\overline{AC}|} + \frac{|\overline{BP_{+}}||\overline{BP_{-}}|}{|\overline{BC}||\overline{BA}|}+ \frac{|\overline{CP_{+}}||\overline{CP_{-}}|}{|\overline{CA}||\overline{CB}|} = 1$$ which is a somewhat weaker statement. In any case, $(\star)$ would seem to be the key to the relations you seek.

Blue
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Sums $$CA\cdot AB+BC\cdot AB+CA\cdot BC = s^2+(4R+r)r$$ and $$\frac{1}{CA\cdot AB}+\frac{1}{BC\cdot AB}+\frac{1}{CA\cdot BC}=\frac{1}{2rR}$$ are not related with Steiner inellipse.

Two remaining sums are based on the products of distances from the foci $P$ and $Q$ of the Steiner inellipse of the triangle $\Delta ABC$ to its vertices, which can be found as follows. Let the vertices $A$, $B$, and $C$ and the foci $P$ and $Q$ have the complex coordinates $z_A$, $z_B$, $z_C$, $z_P$, and $z_Q$, respectively. Accoriding to Steiner’s Theorem [MP, Th. 2.1], $z_P$ and $z_Q$ are given by the equality,

$$g\pm \sqrt{g^2-\frac f3},$$

where $g=\frac 13\left(z_A+z_B+z_C\right)$ is the centroid and $f=z_Az_B+ z_Bz_C+ z_Az_C$.

Then, for instance,

$$|PA\cdot QA|=|z_P-z_A||z_Q-z_A|=\left|(g –z_A)^2- g^2+\frac f3\right|= \left|2gz_A+z_A^2+\frac f3\right|.$$

I don’t know whether we can further simplify the expressions for two remaining sums.

References

[MP] D. Minda, S. Phelps Triangles, ellipses, and cubic polynomials, American Mathematical Monthly, 115 (8) (2008), 679-689.

Alex Ravsky
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