Angle between triangle orthocenter, center of inscribed circle and centroid is always obtuse

Question

If $S$ is the center of inscribed circle, $T$ centroid and $H$ orthocenter of a triangle, prove that $\angle TSH$ is always obtuse (i.e. $\gt 90^{\circ}$, with equilateral triangle being the only exception).

Or you can put it this way: Prove that $S$ always lies within the circle with diameter $TH$.

I was able to prove the theorem by using some vector geometry. It is sufficient to prove that:

$$ \begin{equation} \label{x1} \overrightarrow{ST} \cdot \overrightarrow{SH} \lt 0 \end{equation} $$

or:

$$(\overrightarrow{OT} - \overrightarrow{OS}) \cdot (\overrightarrow{OH} - \overrightarrow{OS}) \lt 0$$

...where $O$ denotes the center of circumscribed circle. It is also well know that:

$$ \overrightarrow{OH} = 3 \overrightarrow{OT} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} $$

$$\overrightarrow{OS} = \frac {1}{a+b+c} (a \overrightarrow{OA} + b \overrightarrow{OB} + c \overrightarrow{OC})$$

It is also possible to replace all dot products like $\overrightarrow{OA} \cdot \overrightarrow{OA}$ or $\overrightarrow{OA} \cdot \overrightarrow{OB}$ with scalar expressions involving $R$ (radius of circumscribed circle) and $a,b,c$ (triangle sides).

All this will eventually lead to a scalar inequality that is valid for any given $a$, $b$ and $c$. That inequality is not trivial and constitues a pretty interesting problem by itself.

Any ideas how to prove the theorem in a more elegant way?

Answer 1

We may exploit the following

Lemma. Given a triangle $ABC$ and some real number $d$, the locus of points $P$ such that the sum $\sigma(P)$ of the signed distances of $P$ from the sides of $ABC$ equals $d$ is a line orthogonal to the $OI$-line.

We trivially have $\sigma(I)=3r$ and $\sigma(O)=R+r$ holds by Carnot's theorem. By angle chasing $$\begin{eqnarray*}\sigma(H)&=&2R\left[\cos(A)\cos(B)+\cos(A)\cos(C)+\cos(B)\cos(C)\right]\end{eqnarray*}$$ hence $I$ lies between $O$ and the projection of $H$ on the $OI$-line. Additionally $O$ and $H$ are isogonal conjugates, hence $I$ lies on the angle bisector of $\widehat{HAO}$. The midpoint $N$ of $OH$ is the center of the nine-point-circle and by Feuerbach theorem $IN=\frac{R}{2}-r$. If we show that $OH>R-2r$ we are done, and it is well-known that $OH^2=9R^2-(a^2+b^2+c^2)$. We just need to prove

$$ (2R+2r)(4R-2r) > a^2+b^2+c^2 $$ which is not difficult.

We may even prove something stronger. By exploiting Poncelet's porism it is not diffucult to draw the set of triangles with a fixed circumcircle and incircle. For such triangles $H,N,G$ travel on fixed circles $\Gamma_{H},\Gamma_N,\Gamma_G$ (whose radii are not difficult to compute), hence $I$ is confined into the circle centered at $N$ with radius $\frac{R}{2}-r$ and $$\widehat{HIO}\geq 2 \arcsin\frac{OH}{2R-r}.$$

Answer 2

An idea: Say $O$ is an origin of position vectors, the we have $H=3T$ and now we have to prove $$(H-S)(T-S) = S^2-4T\cdot S +3T^2 <0$$ Now let $t=||T||$ and $x=||S||$ and let $\phi$ be the angle between $S$ and $T$. So we have to prove

$$ \cos \phi > {x^2+3t^2 \over 4tx} \geq {\sqrt{3}\over 2}$$

So it is enough to prove $\angle SOT< 30^{\circ}$.