What is the radius of S?

Question

Let $A$ and $B$ be opposite vertices of a cube with an edge length of $1$. Suppose S is a sphere which is tangent to the three faces meeting at A and the three edges meeting at $B$. What is the radius of $S$?

I honestly have no idea where to start. I tried imagining this as a cube with a sphere inside it. By doing so i got the radius of the sphere to be $\sqrt{6}$. I have no idea if that is correct. I simply used Pythagoras theorem.

Im not sure what it means to be a tangent to three faces meeting at$ A $ and tangent to three edges meeting at $B$. Please help.

Answer 1

Let us assume that the cube has its edges parallel to the coordinate axes with

$$A=(1,1,1) \ \ \text{and} \ \ B=(0,0,0).$$

The issue is symmetrical with respect to the 3 axes, therefore sphere (S) has its center $(x_0,y_0,z_0)$ along the first diagonal $x_0=y_0=z_0=a$;

A simple reasoning shows that $a=1-R$.

As a consequence the generic equation of (S) is:

$$(x-1+R)^2+(y-1+R)^2+(z-1+R)^2=R^2\tag{1}$$

The generic point of the $x$-axis can be described in the following way :

$$(x=X,y=0,z=0) \ \ \ \text{without constraint on} \ X\tag{2}$$

Plugging (2) into (1) gives an equation to be verified by abscissas of intersection points of (S) with $x$-axis :

$$(X-1+R)^2+(-1+R)^2+(-1+R)^2=R^2 \ \iff $$

$$X^2 + (2R - 2)X + (2R^2-6R+3)=0\tag{3}$$

The tangency condition is equivalent to the fact that quadratic equation (3) has a double root (this is classical : otherwise there would exist $2$ or $0$ solutions, i.e., $2$ or $0$ intersection points with the $x$ axis)

Having a double root for (2) is classically equivalent to the nullity of the discriminant :

$$\Delta=-R^2+4R-2=0$$

(yet another quadratic equation) with two solutions :

$$\color{red}{R=2-\sqrt{2}} \ \ \text{and} \ \ R=2+\sqrt{2}.$$

Only the first one is realistic for your question.

It took me a certain time before understanding the framework of validity of the second solution : (S) is tangent not to the edges issued from $B$ themselves, but on extensions of them (i.e., not between the vertices), not to the faces meeting in $A$ themselves, but to extensions of them.

A side remark : Due to the symmetry of this issue with respect to the different axes, we haven't had to consider the other tangency conditions, with $y$ or $z$ axes.

Here are two graphical illustrations for the 2 cases :

Fig. 1 : Case $R=2-\sqrt{2}$.

Fig. 2 : Case $R=2+\sqrt{2}$ : the sphere (S) is tangent to the extension of the edges and to the extension of the faces.

Answer 2

We can place the cube such that $A=(0,0,0)$ and $B=(1,1,1)$ just like Jean Marie did. Let $(c_1,c_2,c_3)$ be the coordinates of the center and $R$ the radius of the sphere.

The faces meeting $A$ are planes with equations: $\{x=1\}$, $\{y=1\}$, $\{z=1\}$. First, let us consider the plane $\pi_1=\{x=1\}$. Take the point $(1,c_2,c_3)\in\pi_1$, this is the orthogonal projection of the center in $\pi_1$. Since $\pi_1$ is tangent to the sphere, this point belongs to the sphere and thus, its distance from $(c_1,c_2,c_3)$ gives us the radius $R=1-c_1$. At this point we assumed the sphere is "under" the plane $\pi_1$, i.e. $c_1<1$.

Similar arguments about the other tangent planes give $R=1-c_2$ and $R=1-c_3$. Collectively they give \begin{align} c_1=c_2 &= c_3=c \text{, and}\\ c &= 1-R \end{align}

Now we consider the edges meeting $B$, they are lines with equations: $\{x=y=0\}$, $\{y=z=0\}$, $\{x=z=0\}$. First take the line $l_1=\{x=y=0\}$. Like before take the orthogonal projection of the center in $l_1$, this is the point $(0,0,c_3)$ and its distance from the center is

$$R=\sqrt{c_1^2+c_2^2+c_3^2}=\sqrt{c^2+c^2}= \left\{ \begin{array}{cc} c\sqrt{2}, & \text{ if } c>0\\ -c\sqrt{2}, & \text{ if } c<0 \end{array} \right.$$

Lastly, substitute $c$ with $1-R$ and solve for $R$ to get the two solutions. One for a sphere with center inside the cube ($0<c<1$) and another one with its center outside of the cube ($c<0$)

$$ R=\left\{\begin{array}{cc} 2-\sqrt{2}, & \text{ if } c>0 \\ 2+\sqrt{2}, & \text{ if } c<0 \end{array}\right. $$

I must thank Jean Marie and David K. I didn't see the alternative solution for $c<0$ before their comments and explanations

Note: In the first step we assumed the center of the cube to be "under" the planes. We show that assuming $c_i>1$ for some $i$ leads to a contradiction. Take $c_i>0$ for example, then as above, $R=c_i-1$ which follows $c_i=1+R$. Then choose $l$ one of the two lines parallel to $\{x_i=1\}$ again as above we have for the radius $R=\sqrt{c_i^2+something^2}>\sqrt{c_i^2}=\sqrt{(1+R)^2}= R+1$. This is a contradiction, and thus we can only find such a sphere if all $c_i<1$

Answer 3

It is easy to see how a sphere can be tangent to all six faces of a cube. Suppose we have a cube of side $a$; then the sphere has radius $\frac a2$ and the center of the sphere is at the center of the cube.

It is only slightly harder to see how a sphere can be tangent to all eight edges of a cube of side $b$. By symmetry, the center of the sphere is at the center of the cube. Take a plane $\pi_c$ through the center of the cube parallel to one pair of faces; the plane $\pi_c$ intersects the cube in a square of side $b$ and intersects the sphere in a great circle that passes through the four vertices of the square. (That is, the sphere is tangent to each edge of the cube at the midpoint of the edge, which is also the point on the edge that produces one of the vertices of the square where the cube intersects the plane $\pi_c$.) The radius of the sphere is therefore $\frac{\sqrt2}2 b.$

Given a sphere of radius $r$, then, we can construct two cubes around it: a cube $c_A$ of edge length $2 r$ whose faces are tangent to the sphere, and a cube $c_B$ of edge length $\frac2{\sqrt2} r$ whose edges are tangent to the sphere, oriented so all faces of one cube are parallel to faces of the other cube.

Let $A$ be one of the vertices of cube $c_A$. Let $B$ be the vertex of cube $c_B$ farthest from $A.$ If we now construct a cube $c_0$ with edges and faces parallel to the faces and edges of $c_A$ and $c_B$, with one vertex at $A$ and the opposite vertex at $B,$ the three edges of cube $c_0$ meeting at $B$ are collinear with the three edges of $c_B$ meeting at $B$, and the three faces of cube $c_0$ meeting at $A$ are coplanar with the three faces of $c_A$ meeting at $A$.

So the cube $c_0$ has all the properties you want, other than the edge length: $A$ and $B$ are opposite vertices, the three faces meeting at $A$ are tangent to the sphere, and the three edges meeting at $B$ are tangent to the sphere.

Now measure the distance from the center of the sphere to one of the faces that meets at $A.$ This is $r$. Measure the distance to the opposite face of $c_0,$ which is coplanar with one of the faces of cube $c_B,$ so this distance is half the edge of the cube, $\frac1{\sqrt2} r$.

The distance between opposite faces of cube $c_0$ is therefore $$ r + \frac1{\sqrt2} r = \frac{2 + \sqrt2}{2} r.$$

But you want a unit cube, so the edge length of $c_0$ needs to be $1.$ This implies that $\frac{1 + \sqrt2}{2} r = 1,$ so

$$ r = \frac{2}{1 + \sqrt2} = 2 - \sqrt2. $$

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As an aside, if you take vertex $A$ of cube $c_A$ as before, and let $B'$ be the vertex of cube $c_B$ nearest to $A,$ and let cube $c_1$ have opposite vertices $A$ and $B'$ with sides and edges parallel to those of the other cubes. Then the sphere is tangent to the straight-line extensions of the three edges of $c_1$ meeting at $B'$, and tangent to the planar extensions of the three planes of $c_1$ meeting at $A.$ The edge length of $c_1$ is $r - \frac1{\sqrt2} r,$ and therefore $r$ is $2 + \sqrt2$ times the edge of the cube. That is where Jean-Marie's other solution comes from.