Maximize Euler

If a triangle's incentre lies on its Euler line, it must be isosceles (and vice-versa); so in a sense this problem is asking for the "most scalene" triangle with perimeter $16$ and area $11$ .

I didn't get very far analytically (but would be interested if someone did); the basis of my numerical approach was as follows:

Call the triangle's area $T$ and its semiperimeter $s$ . Let its sidelengths be $a,b,c$ (as usual opposite vertices $A,B,C$ respectively).

We know $T$ and $s$ ; let's make $a$ a parameter. By definition, $b+c=2s-a$

Now, from Heron's formula, $\begin{aligned} s(s-a)(s-b)(s-c) &=T^2 \\ s(s-a)\left( s^2-(b+c)s+bc\right) &= T^2 \\ s(s-a)\left( s^2-(2s-a)s+bc\right) &= T^2 \\ s(s-a)( bc - s(s-a)) &= T^2 \\ bc &= s(s-a)+\frac{T^2}{s(s-a)} \end{aligned}$

Knowing $b+c$ and $bc$ means we can find $b$ and $c$ as the roots of a quadratic (up to a choice of sign).

We also have $T=rs$ and $4RT=abc$ where $r$ and $R$ are the inradius and circumradius respectively. So $r=\frac{T}{s},\;\;\;\;R=\frac{abc}{4T}$

Now, it's possible to proceed in a few ways. We can embed everything in a Cartesian coordinate system (which works but is messy); we can use trilinear coordinates to find the distance between the incentre and the Euler line (this involves a lot of trig); or we can try to stick with geometry as long as possible.

Recall the Euler line goes through the circumcentre $O$ and the centroid $G$ . We want the distance to the incentre $I$ . With a bit of internet based research, we can find the following: $OI=\sqrt{R(R-2r)},\;\;\;GI=\frac13 \sqrt{5r^2-16Rr+s^2},\;\;\; GO=\sqrt{R^2-\frac19 \left(a^2+b^2+c^2\right)}$

ie we have all the sidelengths of the triangle $\Delta IGO$ in terms of quantities we've already found; and the distance we want is the altitude of this triangle from $I$ .

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I didn't fancy the algebra here so just proceeded numerically (from the above we can try values of $a$ and calculate the distance); I used a golden section search to find the maximum distance, which came out to be $\boxed{0.2794\ldots}$ when the sides of the triangle were approximately $3.9734,5.5766,6.4500)$ .

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Bonus question (although, really, it's the original question again): in what way is this the most scalene triangle satisfying the conditions? Is there some simple geometric intuition here?