Does $\triangle ABC$ exist such that $\triangle ABC \sim \triangle DEF$, with $D, E, F$ being the incentre, centroid, orthocentre of $\triangle ABC$?

Question:

Does $\triangle ABC$ exist such that $\triangle ABC \sim \triangle DEF$, with $D, E, F$ being the incentre, centroid, orthocentre of $\triangle ABC$, resp.?

For such a triangle to exist, it must be obtuse. Besides that, I have no idea how to prove or disprove it. For the case of $D, E, F$ being the orthocentre, centroid, circumcentre, it's impossible as they lie on the same line (Euler's line). That's the motivation of the problem. I have a feeling that brute force methods are needed (coordinate geometry). But I hate such an ugly approach. Any idea?

Solution 1:

Yes, this triangle exists. (Found using brute force approximation.)

Coordinates: $$A\approx(0.182,0.260)\quad B=(0,0)\quad C=(1,0)\\ D\approx(0.229,0.120)\quad E\approx(0.394,0.087)\quad F\approx(0.182,0.571)$$

Angles: $$a=d\approx107.2957\quad b=e\approx55.0744\quad c=f\approx 17.6299$$

Solution 2:

Here's what I mean by a brute force approach：

Without loss of generality, let's choose an appropriate scale such that $AB=1$.

Set

$$A=(0,0)$$ $$B=(1,0)$$ $$C=(x,y)$$

Then the coordinate of the centroid $E$ is

$$E=\left(\frac {x+1}{3},\frac {y}{3}\right)$$

To find out the coordinate of the incentre $D$, let's calculate the side lengths and perimeter $P$ first

$$|AB|=1$$ $$|AC|=\sqrt{x^2+y^2}$$ $$|BC|=\sqrt{(x-1)^2+y^2}$$ $$P=AB+AC+BC=1+\sqrt{x^2+y^2}+\sqrt{(x-1)^2+y^2}$$

Then the coordinate of the incentre D is

\begin{align} D &=\frac{|BC| A + |AC| B + |AB| C} { P } \\ &= \left (\frac { \sqrt { x ^ { 2 } + y ^ { 2 } } + x} {1 + \sqrt { x ^ { 2 } + y ^ { 2 } } + \sqrt { ( x - 1 ) ^ { 2 } + y ^ { 2 } }} , \frac{y}{1 + \sqrt { x ^ { 2 } + y ^ { 2 } } + \sqrt { ( x - 1 ) ^ { 2 } + y ^ { 2 } }} \right) \end{align}

The coordinate of the orthocentre $F(a,b)$ satisfies

$$\left\{ \begin{aligned} \frac{b}{a-1} &=-\frac{x}{y} \\ a&=x \end{aligned} \right.$$

Hence $$F=\left(x, \frac{x-x^2}{y}\right)$$

By similarity

$\frac{DE}{BC}= \sqrt { \frac { \left( \frac { x + 1 } { 3 } - \frac { \sqrt { x ^ { 2 } + y ^ { 2 } + x } } { 1 + \sqrt { x ^ { 2 } + y ^ { 2 } } + \sqrt { ( x - 1 ) ^ { 2 } + y ^ { 2 } } } \right) ^ { 2 } + \left( \frac { y } { 3 } - \frac{y}{1 + \sqrt { x ^ { 2 } + y ^ { 2 } } + \sqrt { ( x - 1 ) ^ { 2 } + y ^ { 2 } }} \right) ^ { 2 } } { ( x - 1 ) ^ { 2 } + y ^ { 2 } } }$

$=\frac{EF}{AC}=\sqrt{\frac{\left(\frac { x + 1 } { 3 }-x \right)^2 + \left(\frac { y } { 3 }-\frac { x - x ^ { 2 } } { y } \right)^2}{ x ^ { 2 } + y ^ { 2 } }}$

$=\frac{DF}{AB}=\sqrt{\left(\frac { \sqrt { x ^ { 2 } + y ^ { 2 } } + x } { 1 + \sqrt { x ^ { 2 } + y ^ { 2 } } + \sqrt { ( x - 1 ) ^ { 2 } + y ^ { 2 } } }-x\right)^2+\left(\frac { y } { 1 + \sqrt { x ^ { 2 } + y ^ { 2 } } + \sqrt { ( x - 1 ) ^ { 2 } + y ^ { 2 } } }-\frac { x - x ^ { 2 } } { y }\right)^2}$

And I've no idea how to solve it with bare hands. Feel free to complete my answer. I'm curious whether there is a purely geometric approach (maybe non-constructive).