circumsphere of a tetrahedron

Topic:
Geometry
It is well known that any triangle can be circumscribed by a circle. Here is a description of the procedure to do this. In this applet you are asked to consider a related problem – e.g., can any tetrahedron be inscribed in a sphere. Read the description of the triangle procedure; explore the problem using the interactive applet. THEN – read the description of the tetrahedron procedure that appears below the applet window. Consider a triangle ABC. Determine the center and radius of the circumscribing circle. Find the midpoint of each edge of the triangle. There is a line perpendicular to each edge of the triangle that passes through the midpoint of that edge. Call the line that is perpendicular to the edge connecting vertices X and Y, L(X,Y). Every point on that line is equidistant to the two vertices of the triangle that determine that edge – e.g. every point on L(A,B) is equidistant from A and B. The intersection of two such lines determines a point – e.g. the point of intersection of L(A,B) and L(B,C) is equidistant to A and B and C. This point is the center of the circle that circumscribes triangle ABC The three points of intersection – coming from the intersections of L(A,B) and L(A,C), L(B,A) and L(B,C), L(C,A) and L(C,B) are all concurrent at a point O, which is the center of the circumscribing circle The radius of the circle is AO = BO = CO [one point of intersection comes from two perpendicular bisecting lines – thus only two perpendicular lines to the edges of the triangle are needed to locate the center of the circle]