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]

Consider a tetrahedron ABCD. Determine the center and radius of the circumscribing sphere.
Find the midpoint of each edge of the tetrahedron.
There is a plane perpendicular to each edge of the tetrahedron that passes through the midpoint of that edge. Call the plane that is perpendicular to the edge connecting vertices X and Y, P(X,Y). Every point on that plane is equidistant to the two vertices of the tetrahedron that determine that edge – e.g. every point on P(A,B) is equidistant from A and B.
The intersection of two such planes determines a line – e.g. every point on the line of intersection of P(A,B) and P(B,C) is equidistant to A and B and C. This line is perpendicular to the circle that circumscribes triangle ABC and passes through the center of that circle.
Similarly, every point on the line of intersection of P(B,C) and P(B,D) is equidistant to B and C and D. This line is perpendicular to the circle that circumscribes triangle BCD and passes through the center of that circle.
The six lines of intersection – coming from the intersections of
P(A,B) and P(A,C),
P(A,D) and P(B,A),
P(B,C) and P(B,D),
P(C,A) and P(C,B),
P(C,D) and P(D,A),
P(D,B) and P(D,C)
all pass through a point O, which is the center of the circumscribing sphere. The radius of the sphere is AO = BO = CO = DO
[one point of intersection comes from the intersection of two lines – each line of intersection comes from the intersection of two perpendicular bisecting planes – thus only three perpendicular planes to edges of the tetrahedron are needed to locate the center of the sphere]
CHALLENGE - Can any tetrahedron have an inscribed sphere - i.e., a sphere that is tangent to all four faces? If your answer is yes - how would you construct such a sphere? If your answer is no - can you give an example of a tetrahedron in which there cannot be an inscribed sphere?