Suppose the vertices of the Miquel's triangle are , , . We want to minimize the area of , which we split into , , .

Let's first consider .
Let , , . The triangle's area is equal to . Note that, since are concyclic, and are supplementary, so . Therefore, the area of is minimized when is minimized, which is when and are the feet of the perpendiculars dropped from to and .

This condition is symmetric and satisfies the Miquel's triangle condition, so the area of is minimized at the same time the areas of the triangles and are minimized, which is the same time the area of the big triangle is minimized.

Apparently this triangle is called the pedal triangle of . Okay, that's the answer.

Assume the same variable and point names from Part 1. Let , , . The area we want to maximize is



(abuse of cyclic sum notation here). By the Law of Sines, that is maximized when this expression is maximized:



Let's switch variables: , , (barycentric coordinates normalized to sum to the area of the triangle). The constraint is , where is the area of . The expression we want to maximize can be rewritten as



We drop the outside fraction and bash to get



and symmetric expressions for and . This turns out to be proportional to (Law of Cosines, and note that the denominator is symmetric), which becomes trilinear coordinates , so is the circumcenter.