# Triangle formed by circumcenter, orthocenter and incenter

I'm trying to solve the following question from my analytical geometry course:

In a right-angled triangle $$T$$ the circumcenter $$O$$, the orthocenter $$H$$ and the incenter $$I$$ are the vertex of the triangle $$T'$$. The sides sizes of $$T$$ are $$a$$, $$b$$ and $$c$$ and its centroid is $$G$$.

• What relation must a, b and c satisfy so that the segment $$I$$ $$G$$ is height of the triangle $$T'$$?

• What relation must a, b and c satisfy so that the segment $$I$$ $$G$$ is angle bissector of the triangle $$T'$$?

I've found the Euler Line (the line that contains $$O$$, $$G$$ and $$C$$) equation of $$T$$,

$$y = x\cdot-\frac bc + \frac{b(b+c)}{a+b+c}$$

and the equation of the segment $$I$$ $$G$$ line,

$$y = x\cdot \frac{c(a-2b+c)}{b(a+b-2c)} + \frac{c(b-c)}{a+b-2c}$$

but then I can't finish the exercise. Can someone help me?

If I understand your question correctly, I should get the following picture (with $\triangle T = \triangle XHY$ and $\triangle T’ = \triangle IHO$) .

If I slide X along the x-axis (or Y along the y-axis or both at the same time), $\angle IGH$ can never be $90^0$.

The second part is easier to answer. If GI bisects $\angle HIO$, then by angle bisector theorem, HI : IO = 2 : 1 since HG : GO = 2 : 1.

• Thanks Mick, your figure illustrates the question very well. How can I prove that $\angle IGH$ can never be $90^0$? Commented May 5, 2017 at 17:35
• If $GI$ bisects $\angle HIO$, then $HI$ : $IO$ = 2 : 1 or if and only if it? Because if the reciprocal is true I think that we have only two equations, $d(HI)$ = $2d(IO)$ and $a^2$ = $b^2$ + $c^2$. Makes sense? Commented May 5, 2017 at 19:26
• At this time, I can only answer your first question. I play with it (not proving it) using Geogebra . As I have mentioned, I slice X along the x-axis, (i. e. stretching or shortening HX), I cannot get $GI \bot HO$.
– Mick
Commented May 6, 2017 at 0:20

Spoiler: the answer to both questions is: there is no such triangle.

Using usual notation for the right-angled $$\triangle ABC$$ with side lengths $$a,b,c:\ c^2=a^2+b^2$$, semiperimeter $$\rho=\tfrac12(a+b+c)$$, inradius $$r=\tfrac12(a+b-c)$$, circumradius $$R=2c$$, incenter $$I$$, centroid $$G$$, circumcenter $$O$$ and orthocenter $$H=C$$.

We can use known relations:

\begin{align} |GI|^2&=\tfrac19\,(\rho^2+5r^2-16rR) \tag{1}\label{1} ,\\ |GO|^2&=R^2-\tfrac29\,(\rho^2-r^2-4rR) \tag{2}\label{2} ,\\ |OI|^2&=R\,(R-2r) \tag{3}\label{3} ,\\ |IH|^2&=(r+2R)^2+2r^2-\rho^2 \tag{4}\label{4} ,\\ |CG|&=2|GO| \tag{5}\label{5} . \end{align}

For the first question we must have

\begin{align} |GI|^2+|GO|^2-|OI|^2&=0 \tag{6}\label{6} ,\\ \text{or }\quad \rho&=\sqrt{7r^2+10rR} \tag{7}\label{7} . \end{align}

For the second question we must have

\begin{align} |IH|-2\,|OI|&=0 \tag{8}\label{8} ,\\ \text{or }\quad |IH|^2-4\,|OI|^2&=0 \tag{9}\label{9} ,\\ \text{or }\quad \rho&=\sqrt{3r^2+12rR} \tag{10}\label{10} . \end{align}

To check if the condition \eqref{7} or \eqref{10} is achievable for a valid right-angled triangle, let's consider a map of all possible shapes of triangles in terms of two dimensionless parameters: $$u=\rho/R$$ and $$v=r/R$$.

It is known, that for a valid triangle $$v\in(0,\tfrac12)$$, and $$u(v)\in\Big(u_{\min}(v),u_{\max}(v)\Big)$$, where

\begin{align} u_{\min}(v)&= \sqrt{27-(5-v)^2-2\sqrt{(1-2v)^3}} \tag{11}\label{11} ,\\ u_{\max}(v)&= \sqrt{27-(5-v)^2+2\sqrt{(1-2v)^3}} \tag{12}\label{12} . \end{align}

Moreover, any point $$(v,u)$$ of the area $$\mathcal{T}$$, bounded by \eqref{11}, \eqref{12} and the vertical line $$v=0$$, corresponds to a unique valid triangle with $$R=1,\ \rho=u,\ r=v$$. Boundary curves \eqref{11}, \eqref{12} correspond to isosceles shapes, and the point $$(\tfrac12,\tfrac{3\sqrt3}2)$$ corresponds to the equilateral shape.

The condition for the right-angled triangles is given by \begin{align} u(v)&=v+2 \tag{13}\label{13} , \end{align}

this is a straight line, crossing the area of all valid triangular shapes $$\mathcal{T}$$.

Conditions \eqref{7} and \eqref{10} combined with \eqref{13} result in two values of $$v$$,

\begin{align} v_1&=\tfrac{\sqrt{33}}6-\tfrac12 \tag{14}\label{14} \\ \text{and }\quad v_2&=\sqrt6-2 ,\quad\text{ respectively} \tag{15}\label{15} , \end{align}

with corresponding $$u_1=v_1+2$$ and $$u_2=v_2+2$$.

Both points $$(v_1,u_1),\ (v_2,u_2)$$ are located outside the area of valid triangular shapes, below the lower boundary \eqref{11}.

Thus the answer to both questions is: there is no any valid triangle with given properties.

The sides $$a,b,c$$ of the triangle with $$R=1$$ and given pair $$(v,u)$$ can be found as the roots of cubic equation

\begin{align} x^3-2u\,x^2+(u^2+v^2+4v)\,x-4\,uv &=0 \tag{16}\label{16} , \end{align}

and for all points $$(v,u)$$ from the validity region $$\mathcal{T}$$, all the roots are positive and correspond to a valid triangle.

It's easy to check that both solutions of \eqref{16} with given pairs $$(v_1,u_1),\ (v_2,u_2)$$ result one side length equal $$2$$ (the hypotenuse), and the other two roots are complex.

This is the illustration of the map, where the blue and red are the boundary curves, the black line correspond to the right triangles, and green and orange lines correspond to the conditions of the first and the second question, respectively.