Ex-circum-center?

Assume that the excenter and the circumcenter are the same point of $\triangle ABC$ . Since the excenter is always outside the triangle, the circumcenter would have to be outside as well, meaning one of the angles would be obtuse. Let $A$ be the obtuse angle, $E$ be the excenter (and circumcenter), $D$ be the tangent point of the extension of $AB$ , and $F$ be the tangent point of the extension of $AC$ .

Let the circumradius be $R$ and the excenter opposite the obtuse angle (the one pictured) be $r_1$ . Then $EC = R$ and $EF = r_1$ , and since $EC$ is the hypotenuse and $EF$ is a leg of right triangle $\triangle CEF$ , $EF \leq EC$ , so $r_1 \leq R$ .

However, since the excenter $r_1$ is opposite the obtuse angle, it must be greater than the other excenters $r_2$ and $r_3$ , and $r_1$ is greater than the incenter $r$ . We also have the relationship $r_1 + r_2 + r_3 - r = 4R$ . But since $r_1 > r_2$ , $r_1 > r_3$ , and $r_1 > r$ , we have $4r_1 > r_1 + r_2 + r_3 - r = 4R$ , so $4r_1 > 4R$ or $r_1 > R$ .

Therefore, we have a contradiction that $r_1 \leq R$ and $r_1 > R$ , so reject our assumption that the excenter and incenter can be the same point.