Optimal triangle (minimal area) surrounding a disk

The reason might be that Mathematica fails to find an initial point that satisfies these complex conditions.

a = RegionWithin[dreieck, Disk[]];
FindInstance[a, {x1, y1, x2, y2, x3, y3}, Reals]

Also doesn't terminate. Mathematica doesn't seem to have an option for giving initial points. The more interesting question might be, what can you do solve this problem instead.

You simplify the search for feasible points a lot by choosing different variables. Note that any minimal triangle will touch the circle at 3 points and the lines will be tangents of the circle.

Each point on the unit circle is defined via one real number $b, c, d$ for the sake of rotational symmetry we can assume b == 0.

$$x cos(b) - y sin(b) + 1 = 0$$ defines a set of $(x,y)$ which is the tangent at the point $(sin(b),cos(b))$.

We can calculate the intersection of two tangents by finding a point $(x,y)$ that satisfies both equations.

Solve[{x Cos[b] - y Sin[b] + 1 == 0,  x Cos[c] - y Sin[c] + 1 == 0}, {x, y}]

Has the solution

{x -> -((Sin[b] - Sin[d])/(Cos[c] Sin[b] - Cos[b] Sin[d])), 
 y -> -((-Cot[b] + Cos[c] Csc[b])/(-Cos[c] + Cot[b] Sin[d]))}

Using those points we can now construct a triangle and calculate the area as you did above. However we also need a constraint that the center of the triangle (average of edge points) must lay in the circle. Futhermore only want to consider triangles bigger than the circle itself.

This results in this query:

NMinimize[{Area[
   Triangle[{{-((Sin[b] - Sin[c])/(
       Cos[c] Sin[b] - Cos[b] Sin[c])), -((-Cot[c] + 
        Cos[b] Csc[c])/(-Cos[b] + Cot[c] Sin[b]))}, {-((
       Sin[b] - Sin[d])/(
       Cos[d] Sin[b] - Cos[b] Sin[d])), -((-Cot[d] + 
        Cos[b] Csc[d])/(-Cos[b] + Cot[d] Sin[b]))}, {-((
       Sin[c] - Sin[d])/(
       Cos[d] Sin[c] - Cos[c] Sin[d])), -((-Cot[d] + 
        Cos[c] Csc[d])/(-Cos[c] + Cot[d] Sin[c]))}}]], ({-((
        Sin[b] - Sin[c])/(
        Cos[c] Sin[b] - Cos[b] Sin[c])), -((-Cot[c] + 
         Cos[b] Csc[c])/(-Cos[b] + Cot[c] Sin[b]))} + {-((
        Sin[b] - Sin[d])/(
        Cos[d] Sin[b] - Cos[b] Sin[d])), -((-Cot[d] + 
         Cos[b] Csc[d])/(-Cos[b] + Cot[d] Sin[b]))} + {-((
        Sin[c] - Sin[d])/(
        Cos[d] Sin[c] - Cos[c] Sin[d])), -((-Cot[d] + 
         Cos[c] Csc[d])/(-Cos[c] + Cot[d] Sin[c]))})/3 [Element] 
   Disk[], Area[
    Triangle[{{-((Sin[b] - Sin[c])/(
        Cos[c] Sin[b] - Cos[b] Sin[c])), -((-Cot[c] + 
         Cos[b] Csc[c])/(-Cos[b] + Cot[c] Sin[b]))}, {-((
        Sin[b] - Sin[d])/(
        Cos[d] Sin[b] - Cos[b] Sin[d])), -((-Cot[d] + 
         Cos[b] Csc[d])/(-Cos[b] + Cot[d] Sin[b]))}, {-((
        Sin[c] - Sin[d])/(
        Cos[d] Sin[c] - Cos[c] Sin[d])), -((-Cot[d] + 
         Cos[c] Csc[d])/(-Cos[c] + Cot[d] Sin[c]))}}]] >= 
   Area[Disk[]], b == 0}, {b, c, d}]

which gives us

{5.19615, {b -> 0, c -> 2.0944, d -> -2.0944}}

which when plotted using RegionPlot looks like:

An exact solution using Herons Formula

Wikipedia gives an exact formula for the length of the incircle based on the side length a,b,c. Herons formula gives the area of the triangle using the same information.

s := (a + b + c)/2; Heron = Sqrt[s*(s - a) (s - b) (s - c)];

Minimize[{Heron, 1 == Sqrt[(s - a) (s - b) (s - c)/s], a > 0, b > 0, c > 0}, {a, b, c}] Minimizes the area of a triangle with a circle of radius 1 inside. If the resulting expression is Fully Simplified the objective turns out to be 3 Sqrt[3] (matching our numerical calculation before) while all lengths are 2 Sqrt[2]