Finding the centre of a sphere that touches three faces of a tetrahedron.

Question

The tetrahedron $OABC$ is defined by the vectors $\mathbf a=\vec{OA}, \mathbf b=\vec{OB}, \mathbf c=\vec{OC}$ with $\mathbf a\cdot(\mathbf b\times\mathbf c)>0$. A sphere $T_r$ with radius $r>0$ lies inside the tetrahedron and intersects each of the three faces $OAB, OAC,$ and $OBC$ in exactly one point. Show that the centre $P$ of $T_r$ satisfies $$\vec{OP}=r\frac{|\mathbf b\times \mathbf c|\mathbf a+|\mathbf c\times \mathbf a|\mathbf b+|\mathbf a\times \mathbf b|\mathbf c}{\mathbf a\cdot(\mathbf b\times \mathbf c)}$$.

Some useful things we've done so far:

$(1)$ The distance from a plane $\mathbf x\cdot \mathbf n=d$ is $d$ (assuming I've done that right).

$(2)$ The plane and the sphere $|\mathbf x-\mathbf p|^2=r^2$ intersect at exactly one point if $|\mathbf p\cdot\mathbf n-d|=r$.

So for the tetrahedron I considered each face as a plane through $O$ so using the previous result $d=0$ and so $|\mathbf p\cdot\mathbf n|=r$ where $\mathbf p=\vec{OP}$ where $\mathbf n$ could be any of the unit normals to the planes $(\frac{(\mathbf a\times\mathbf b)}{|\mathbf a\times\mathbf b|}, \frac{(\mathbf b\times\mathbf c)}{|\mathbf b\times\mathbf c|}, \frac{(\mathbf c\times\mathbf a)}{|\mathbf c\times\mathbf a|})$

(although I'm not sure I can just say this as what we have previously shown $((2))$ is not a necessary condition).

I can't really progress further other than just jumbling stuff around.

Help is much appreciated, thank you

(Forgive me if this has been asked before)

Answer 1

Set $\vec{OP}=\alpha \mathbf a+\beta \mathbf b +\gamma \mathbf c$ and take the dot product of that with the three unit vectors. You get for instance: $$ \vec{OP}\cdot{\mathbf a\times\mathbf b\over|\mathbf a\times\mathbf b|}= \gamma{\mathbf c\cdot(\mathbf a\times\mathbf b)\over|\mathbf a\times\mathbf b|}=r, \quad\hbox{whence:}\quad \gamma=r{|\mathbf a\times\mathbf b|\over\mathbf c\cdot(\mathbf a\times\mathbf b)}, $$ and so on.