When can we inscribed sphere in tetrahedron?

Question

I'm wondering is there any conditions to do that. I believe propably there are but what ?

Answer 1

Given any tetrahedron with vertices $v_1,\ldots,v_4$. For $k = 1,\ldots 4$, let

• $h_k$ be the distance of $v_k$ to its opposite face.
• $\Delta_k$ be the area of the face opposite to $v_k$.
• $V$ and $\Delta = \sum_{k=1}^4 \Delta_k$ be the volume and surface area of the tetrahedron.

Consider following area weighted average of the vertices. $$v = \frac{1}{\Delta}\sum_{k=1}^4 \Delta_k v_k$$ It is easy to see its distance to the face opposite to vertex $v_k$ equals to $$\frac1\Delta \Delta_k h_k = \frac{3V}{\Delta}$$ Since this is independent of $k$, this point $v$ is equal distance to all faces of the tetrahedron. You can inscribe a sphere of radius $\frac{3V}{\Delta}$, centered at $v$ to the tetrahedron.

Answer 2

Here a "physical" proof that any tetrahedron has an inscribed sphere.

Let $\Delta$ be a tetrahedron.

Consider a vertex $v$ and the three planes containing the faces meeting at $v$. Consider the triangular infinite cone they form at $v$. Think of having it in a vertical position. Now put a unit sphere inside the cone. The sphere will fall until it touches the three faces.

Now consider the fourth face of $\Delta$, the one opposite to $v$, and consider the family of planes parallel to it. The higher planes won't touch the sphere, wile planes closer to the vertex will slice a circle on the sphere. By continuity there is one of such plane $P$, which touches the sphere in one point (and having the sphere below $P$). It determines a tetrahedron where the unit sphere is inscribed. Since $P$ is parallel to the original face of $\Delta$, then the obtained tetrahedron is similar to $\Delta$. So by scaling you obtain a sphere inscribed in $\Delta$.