sphere.md

Intersection of tangent planes

A sphere with center on the plane of the face ABC of a tetrahedron SABC passes through A, B and C, and meets the edges SA, SB, SC again at A₁, B₁, C₁, respectively. The planes through A₁, B₁, C₁ tangent to the sphere meet at a point O. Prove that O is the circumcenter of the tetrahedron SA₁B₁C₁. ^{1 (Problem 12)}

Of course we can define the sphere as , then pick arbitrary points A, B and C on the equator circle like , and . However, this will cause many radical calculations later.

So firstly, we define arbitrary points A, B and C on plane xy as A(a,0,0), B(0,b,0) and C(c,0,0), then calculate the sphere with equator on plane xy:

Because the sphere passes through A, B and C, parameters g, h and k can be solved by:

Next, given an arbitrary point S(d,e,f), we have straight line SA:

Apply on Eq. 1 then we get a quadratic equation about t. Because t₀ = 0 is the point A, we can reduce the order and get the other root t₁ for point A₁. Analogously, we can calculate points B₁ and C₁.

Now the sphere Eq. 1 and three points A₁, B₁ and C₁ are all rational expressions about a, b, c, d, e and f. We can convert them to homogeneous coordinates (then all expressions are polynomial) to speed up calculation.

When a quadric surface is denoted as , the tangent plane passing through a point on the surface is:

Then it's easy to calculate the intersection O of the three tangent planes, and calculate the distance to S, A₁, B₁ and C₁.

Here is the computational proof.

Eight vertices of a hexahedron lie on a sphere

If seven vertices of a (quadrilaterally-faced) hexahedron lie on a sphere, then so does the eighth vertex. ^{1 (Problem 11)}

Just like the previous problem, we define arbitrary points O(0,0,0), A(a,0,0), B(b,c,0) and C(d,e,f), then calculate the sphere passing through these four points:

Now we should set points D, E and F on this sphere to make OBCD, OCAE and OABF as three quadrilaterals (i.e. D, E and F are on plane OBC, OCA and OAB respectively).

Take F as example, it is on the sphere (Eq. 2) and plane OAB (), so we need to add another arbitrary parameter to fix this point.

Let's try which passes through O, then eliminate y and z in Eq. 2 and get a quadratic equation about x. Just like the previous problem, because x₀ = 0 is the point O, we can reduce the order and get the other root x₁ for point F.

Analogously, we can get E from plane OCA and ( or may not work for the special case C(0,0,f)), and get D from plane OBC and .

Now the sphere Eq. 2 and points A, B, C, D, E and F are all rational expressions about a, b, c, d, e, f, m, n and p. Convert them to homogeneous coordinates, calculate planes CDE, BFD and AEF, then calculate the intersection G of these three planes.

However, the homogeneous coordinates of G contain too many terms (x_G has more than 12 thousand terms so x_G² should have more than 70 million terms), which may not be feasible to check if G is on the sphere.

So we should pick D, E and F in a different way to make G simple. Take plane BFD as example, the coordinates of the plane will be simpler if coordinates of B and F have less differences. So we should choose a simple plane passing through B instead of O during picking F.

Let's try to get a quadratic equation about x, then calculate x_F from x_B by Vieta's formula. Now the plane BFD is much simpler than the previous trying.

Analogously, we can get E by and get F by .

Here is the computational proof.