Nine congruent spheres are packed inside a unit cube in such

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Nine congruent spheres are packed inside a unit cube in such a way that one of them has its center at the center of the cube and each of the others is tangent to the center sphere and to three faces of the cube. What is the radius of each sphere?

Nice little puzzle but I am not really sure why the silent treatment of the correct answer by Bernard Leak - here is the upper middle school treatment.

Let ABGFDEHC$ABGFDEHC$ be the given cube (Fig. 1):

If we cut the given cube with a plane p$p$ that passes through the points A,B,C$A,B,C$ and D$D$ then the five congruent spheres of the same radius r$r$ inscribed into the cube as specified, will be seen as shown below (Fig. 2):

Observe that since all the linear measurements of the rectangle ABCD$ABCD$ are known:

|AB|=|DC|=1,|AD|=|BC|=2–√,|BD|=|AC|=3–√$\begin{array}{}& |AB|=|DC|=1,|AD|=|BC|=\sqrt{2},|BD|=|AC|=\sqrt{3}\end{array}$

the planar angles ABD$ABD$ and DBC$DBC$ are also known:

tan∠DBC=12–√(1)$\begin{array}{}\text{(1)}& \tan \mathrm{\angle }DBC={\displaystyle \frac{1}{\sqrt{2}}}\end{array}$

and

tan∠ABD=2–√(2)$\begin{array}{}\text{(2)}& \tan \mathrm{\angle }ABD=\sqrt{2}\end{array}$

We, thus, find ourselves in a rather peculiar situation. Namely:

we already know how to construct the radius of the congruent spheres r$r$ even though we technically haven’t solved the problem yet!

How is this possible?

Let the sphere so$s_o$ centered at the point O$O$ and the lower left sphere sp$s_p$ centered at the point P$P$ touch at the point T$T$. Then, the common interior tangent straight line drawn at right angles with respect to the diagonal AD$AD$ through the point T$T$ will intersect the side AB$AB$ at the point S$S$ (Fig. 3):

such that:

∠ABD=α=∠BST(3)$\begin{array}{}\text{(3)}& \mathrm{\angle }ABD=\alpha =\mathrm{\angle }BST\end{array}$

But, located on a perpendicular bisector ST$ST$ of the line segment OP$OP$, the point S$S$ is equidistant from the centers O$O$ and P$P$ (fig. 4):

Let the sphere sp$s_p$ and the side BC$BC$ touch at the point R$R$. Being two tangent straight lines, SR$SR$ and ST$ST$, drawn to the same circle (sphere) sp$s_p$ through a single point, S$S$, the lengths of the line segments SR$SR$ and ST$ST$ are equal: |SR|=|ST|$|SR|=|ST|$ (Fig. 5):

Since the triangles PRS$PRS$ and PTS$PTS$ share the same hypotenuse SP$SP$ and have the remaining sides PT$PT$ and PR$PR$ , being the radii of the same circle, of equal length, it follows that these triangles are congruent by SSS. Hence, the straight line SP$SP$ is actually the bisector of the angle BST$BST$ whose magnitude is known to us.

But the triangle PSO$PSO$, as we have shown, is isosceles, with |SP|=|SO|$|SP|=|SO|$, and the straight line ST$ST$ is both its height and the bisector of its interior angle at the vertex S$S$. Hence:

∠PST=∠OST=α2$\begin{array}{}& \mathrm{\angle }PST=\mathrm{\angle }OST={\displaystyle \frac{\alpha }{2}}\end{array}$

which means that the magnitude of the angle BOS$BOS$ is known to us and is constructible with Euclidean tools alone to boot (Fig. 6):

and the game is, essentially, over since in this particular case we can construct the answer before we obtain its analytical form:

• bisect the angle ABD$ABD$ to obtain the angular measure of α/2$\alpha /2$
• from a right angle subtract the magnitude of α/2$\alpha /2$ to obtain the angular measure of (π−α)/2$(\pi -\alpha )/2$
• let the diagonals AC$AC$ and BD$BD$ of our rectangle intersect at the point O$O$
• using the diagonal BD$BD$ as one of its sides and the point O$O$ as its vertex, draw an angle of magnitude (π−α)/2$(\pi -\alpha )/2$ until its remaining side intersects the line segment BC$BC$ at the point S$S$ (Fig. 7):

• and, lastly, draw a straight line τ$\tau$ perpendicular to the diagonal BD$BD$ through the point S$S$ until τ$\tau$ intersects BD$BD$ at the point T$T$ (Fig. 8):

• the length of the line segment TO$TO$ is the length of the radius sought-after: |OT|=r$|OT|=r$

But, OK.

Assume that you do not really see that the center P$P$ of the sphere sp$s_p$ is located at a vertex of a cube whose side length is equal to r$r$. Then, let us brute-force it: let the straight line BO$BO$ intersect the sphere sp$s_p$ at the second point Q$Q$ and let ∠OBR=β$\mathrm{\angle }OBR=\beta$ (Fig. 9):

From the right triangle BRP$BRP$ via (1) we find the length |RB|$|RB|$:

|RB|=rtanβ=r2–√(4)$\begin{array}{}\text{(4)}& |RB|={\displaystyle \frac{r}{\tan \beta }}=r\sqrt{2}\end{array}$

Then, on the one hand, from the same triangle via the Pythagorean theorem we have:

(r+|QB|)2=r2+2r2=3r2$\begin{array}{}& (r+|QB|{)}^2=r^2+2r^2=3r^2\end{array}$

from where:

|QB|=r⋅(−1+3–√)(5)$\begin{array}{}\text{(5)}& |QB|=r\cdot (-1+\sqrt{3})\end{array}$

While on the other hand, our half-diagonal OB$OB$ is composite and clearly:

|QB|+r+r+r=3–√2$\begin{array}{}& |QB|+r+r+r={\displaystyle \frac{\sqrt{3}}{2}}\end{array}$

from where with (5) we have:

r⋅(−1+3–√)+3r=3–√2$\begin{array}{}& r\cdot (-1+\sqrt{3})+3r={\displaystyle \frac{\sqrt{3}}{2}}\end{array}$

which is to say that:

r⋅(2+3–√)=3–√2$\begin{array}{}& r\cdot (2+\sqrt{3})={\displaystyle \frac{\sqrt{3}}{2}}\end{array}$

Multiplying both sides of the above relation by the (conjugate) number (2−3–√)$(2-\sqrt{3})$, we find:

r=3–√−32(6)$\begin{array}{}\text{(6)}& r=\sqrt{3}-{\displaystyle \frac{3}{2}}\end{array}$

the answer.

Which gives us yet another, simpler, way to recover the length of the radius r$r$ of all nine congruent spheres with Euclidean tools alone: given a cube with the side length of a$a$, from the length of its main diagonal a3–√$a\sqrt{3}$ subtract the length of its side a$a$ scaled by 3/2:$3/2:$

ra=a3–√−a32