geometrytriangles
I am trying to prove the following:
Let $T$ be the Gergonne point (the intersection of the lines that connect the points of tangency of the incircle with the vertices of the triangle) of $\triangle ABC$. Show that if $T$ coincides with the incenter or the circumcenter or the orthocenter or the centroid of $\triangle ABC$, then the triangle must be equilateral.
I would appreciate any help. Thank you!
Circle with the center $O$ is inscribed in the equilateral triangle $ABC$ with the height $CF$ and inradius $OF$, circle with the center $R$ is inscribed in the equilateral triangle $DEC$ with the height $CG$ and inradius $RG$,
\begin{align}
|CF|&=3|CG|
,
\end{align}
hence, the radius
\begin{align} |OF|&=3\cdot|RG|=3\cdot4=12 , \end{align}
and the side length of $\triangle ABC$ \begin{align} |AC|&= \frac{|CF|}{\tfrac {\sqrt3}2} =\frac{3\cdot|OF|}{\tfrac {\sqrt3}2} =2\sqrt3\cdot|OF|=24\sqrt3 . \end{align}
Proof. Note that the incenter, $I$ is determined by the intersection of the internal angle bisectors of the triangle. We will consider the $A$-excircle, i.e. the excircle whose center, let's call it $X$, is determined by the intersection of the external angle bisectors at vertices $B$ and $C$. Note that $X$ lies on the line $AI$ as well.
Let $\angle BAC = \alpha$, $\angle ABC = \beta$ and $\angle ACB = \gamma$. Furthermore, let $AX \cap (ABC) = M$, so we need to show that $MI = MX$. We claim that $BICX$ is a cyclic quadrilateral centered at $M$. Firstly, we have $\angle BAM = \angle CAM = \frac{\alpha}{2}$, so $MB = MC$. Next, we show that $\triangle MCI$ is isosceles. Indeed, we have $\angle MIC = \angle IAC + \angle ACI = \frac{\alpha}{2} + \frac{\gamma}{2} = \angle BAM + \angle BCI = \angle BCM + \angle BCI = \angle MCI$.
Since $MB = MC = MI$, $M$ is the circumcenter of $\triangle BCI$, it remains to show that $X ∈ (BCI)$. We have $\angle BIC = 180° - \frac{\beta}{2} - \frac{\gamma}{2}$. On the other hand, $\angle BXC = 180° - \angle CBX - \angle BCX = 180° - \frac{180° - \beta}{2} - \frac{180° - \gamma}{2} = \frac{\beta}{2} + \frac{\gamma}{2}$.
Therefore, $\angle BIC + \angle BXC = 180°$, i.e. $X ∈ (BCI)$ and $(BCI)$ is centered at $M$, so $MI = MX$, as desired.
Best Answer
Case 1
$T=H$ coincides with orthocenter of $ABC.$ Let $T_a,$ $T_b,$ $T_c$ be the points where incircle touches $BC,$ $AC$ and $AB$ respectively. Since $TT_a$ is perpendicular to $BC,$ $TT_b$ is perpendicular to $AC,$ $TT_c$ is perpendicular to $AB$ it follows that $T$ is also an incenter of $ABC.$ Hence, all bisectors are simultaneously altitudes of the triangle which implies that $ABC$ is equilateral.
Case 2
$T=M$ coincides with a centroid of $ABC.$ Then, $BT_a=CT_a$ or using well-known formulas $BT_a=\frac{a+c-b}{2}=CT_a=\frac{a+b-c}{2}.$ Last equality implies that $b=c.$ In exactly the same way one can show that $a=c.$ Hence, $ABC$ is again equilateral.
Case 3
$T=O$ coincides with the circumcenter of $ABC.$ IMHO, the most natural way is to "compute" a bit. More precisely, $\angle T_cAB=90-\angle ACB=90-\gamma,$ $\angle T_aAC=90-\angle ABC=90-\beta.$ Note that the ratio of the areas of $\triangle AT_aB$ and $AT_aC$ is equal to $$\frac{BT_a}{CT_a}=\frac{p-b}{p-c}=\frac{AB\cos\gamma}{AC\cos\beta}=\frac{c\cos\gamma}{b\cos\beta}=\frac{c^2(b^2+a^2-c^2)}{b^2(a^2+c^2-b^2)}.$$
After that one can get the same expressions for the other sides and play with equalities a little bit to get that $a=b=c.$