**Solution (outline). **We will use oriented lines, planes and angles. For arbitrary, non-collinear points *X*,*Y*,*Z*, the orinetation of the line *XY* the same as the direction of the vector , and the vector points into the positive side of the plane *XYZ*,

*Lemma.* Let be an oriented line which is contained by the oriented planes , and . Let be the angle by which the plane can be rotated to around (*i*=1,2; these angles are considered modulo 360^{o}). Suppose that some point *X* has signed distances *d*,*r*,-*r* from the planes , , , respectively. Then we have

*Proof.* Denote by the unit normal vector of the plane which points to the positive side of the plane and let be the direction vector of the line Let be a vector from a point of to *X*. Then

Let .Then

Taking the dot-product with ,

sin _{2}^{.}*r*-sin _{1}^{.}(-*r*)=sin (_{2}-_{1})^{.}*d*

From this we get

The lemma is proved.

Let *X* and *r*_{X} be the center and radius of the sphere, respectively, which is circumcribed by the planes *OA*_{i+1}*A*_{i+2} and *B*_{i}*A*_{i+1}*A*_{i+2}. Let *d*_{X} be the signed distance between *X* and the plane *A*_{1}*A*_{2}*A*_{3} síktól *d*_{X}. The signed distances between *X* and the planes *B*_{i}*A*_{i+1}*A*_{i+2} are all *r*_{X}, and the signed distances between *X* and the planes *OA*_{i+1}*A*_{i+2} are all -*r*_{X}.

Analogously, let *Y* and *r*_{Y} be the center and radius of the sphere which is circumscribed by the planes *B*_{i}*A*_{i+1}*A*_{i+2} and *C*_{i}*A*_{i+1}*A*_{i+2}. let *d*_{Y} by the signed distance between *Y* and the plane *A*_{1}*A*_{2}*A*_{3}. The signed distances between *Y* and the planes *C*_{i}*A*_{i+1}*A*_{i+2} are all *r*_{Y}, and the signed distances between *X* and the planes *B*_{i}*A*_{i+1}*A*_{i+2} are all -*r*_{Y}.

For every *i*=1,2,3 take the external angle bisector of the planes *A*_{i+1}*A*_{i+2}*O* and *A*_{i+1}*A*_{i+2}*C*_{i}, and let *Z* be the intersection point of the three angle bisector planes --- *Z* will be center of the desired sphere. (Now assume that none of the angle bisectors coincides with the plane *A*_{1}*A*_{2}*A*_{3}; then the point *Z* is unique and does not lie on *A*_{1}*A*_{2}*A*_{3}.)

Let *d*_{Z} and be the signed distance between *Z* and *A*_{1}*A*_{2}*A*_{3} and *A*_{1}*A*_{2}*C*_{3}, respectively. Since *Z* lies on the external angle bisector between *A*_{1}*A*_{2}*C*_{3} and *A*_{1}*A*_{2}*O*, the signed distance between *Z* and *A*_{1}*A*_{2}*O* is -.

Let =(*A*_{1}*A*_{2}*A*_{3},*A*_{1}*A*_{2}*O*), =(*A*_{1}*A*_{2}*A*_{3},*A*_{1}*A*_{2}*B*_{3}) and =(*A*_{1}*A*_{2}*A*_{3},*A*_{1}*A*_{2}*C*_{3}), where the angles are measured around the oriented line *A*_{1}*A*_{2}.

Applying the Lemma for the line *A*_{1}*A*_{2}, the planes *A*_{1}*A*_{2}*A*_{3}, *A*_{1}*A*_{2}*B*_{3}, *A*_{1}*A*_{2}*O* and the point *X* we get