Recently I played a little bit around with GeoGebra and I constructed the in- and circumcircle of a $\triangle ABC$ with $A=(0,0)$ and $B=(1,0)$ and I asked myself if it is possible to construct the area where all the points are for which the center of the circumcircle is still inside the incircle. With the trace option I figured that it almost looks like a circle but I don't think it is one so how could one approach a construction of it? I came to no real approach or start.
I baptize this curve with the name ,,Rocklage-curve". It is probably possible to create similar surfaces in other cases.
Best Answer
I will first find a condition the side lengths $a,b,c$ must satisfy for the circumcentre to lie within the incircle. In trilinear coordinates $x:y:z$ the incircle has equation $$ayz+xbz+xyc-(ax+by+cz)\left(\frac{x(b+c-a)^2}{4bc}+\frac{y(c+a-b)^2}{4ca}+\frac{z(a+b-c)^2}{4ab}\right)=0$$ Now substitute the circumcentre's coordinates $a(b^2+c^2-a^2):b(c^2+a^2-b^2):c(a^2+b^2-c^2)$ into this equation and factor over $\mathbb Q(\sqrt2)$. The factor that matters is $$a^3+b^3+c^3-(a^2(b+c)+b^2(c+a)+c^2(a+b))+2\sqrt2abc$$ Since the circumcenter lies in the incircle when $a=b=c=1$, our desired condition is when the above expression is nonpositive: $$a^3+b^3+c^3-(a^2(b+c)+b^2(c+a)+c^2(a+b))+2\sqrt2abc\le0$$ $$\iff\frac1{2abc}(a+b-c)(b+c-a)(c+a-b)\ge\sqrt2-1$$ You want the region for $C=(p,q)$ where the condition holds for $\triangle ABC$. In this triangle $c=1$, $a=\sqrt{p^2+q^2}$ and $b=\sqrt{(1-p)^2+q^2}$. Substituting these into the condition and plotting in the $pq$ plane reveals a very decidedly non-circular shape:
Numerical integration based on the cylindrical algebraic decomposition in Mathematica reveals the area and perimeter of this shape as $$A=0.9271379517\dots$$ $$L=3.4431978101\dots$$