I'm trying to find the largest sphere that lies inside the tetrahedron whose vertices are the origin and the points A(1, 0, 0), B(0, 1, 0) and C(0, 0,−2).
Any ideas? Thanks!
If it had been a regular tetrahedron with equilateral triangles, the distance between each side would represent the diameter of the
sphere. In this case, the tetreahdron is not even isoceles and I
have no idea where to start, since I am having a hard time even
visualizng the sphere in the shape.
I first drew the tetrahedron and tried to see if I could visulalize
where the centre of the sphere could be and tried to see if I could
see some formula to find its radius. Also, I calculated the
distances between the points to discover it was not a regular
tetrahedron but instead had distances of 1, sqrt(2), sqrt(5) and 2.
I am thinking it may have to do with the sphere being
tangent to the sides after doing some research, but really have no
idea where to begin testing this. Thanks!
June 26th 2009, 03:09 AM
The sphere will be the one for which each of the planes of the tetrahedron are tangential (to it), i.e. the one for which the perpendicular distances from the sphere centre to each of the planes of the tetrahedron are equal.
Flipping the tetrahedron upside down, so as to avoid the irritating negative z co-ordinates, the equation of the ABC (new C) plane will be x+y+z/2=1, and the perpendicular distance from an arbitrary point (X,Y,Z) on the origin side onto this plane will be (1-X-Y-Z/2)/sqrt(1+1+1/4) = (2/3)(1-X-Y-Z/2). The distances from (X,Y,Z) onto the other three planes will be X, Y and Z. Equating the four leads to a radius of 1/4.